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✓ 35 results · all verified · 22 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 13 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Divergence Theorem and Classical Stokes

1 · Prerequisites

2 · Summary

regular-surfaces-and-surface-integrals supplies regular patches, orientations, flux, finitely patched surfaces, and the graph-based surface-integral formulas that this page repeatedly reuses. Through the prerequisite closure already established on disk, the proofs also use Jordan-set change of variables, Green's theorem on elementary plane regions, line integrals, and the star-shaped equivalence between closed and conservative fields. Those inputs are exactly what let the page work with explicit Euclidean patches and explicit boundary chains, without appealing to any unbuilt manifold or homology machinery.

The page first defines divergence, curl, the Laplacian, and vector potentials, then proves the standard first-order identities and the two degree-two identities $curl \nabla = 0$ and $div curl = 0$ . It next builds the coordinate-direction solid machinery needed for an honest elementary class of three-dimensional regions, proves the divergence theorem first for one elementary solid and then for finite gluings, and extracts the vector forms, Green identities, and the flux interpretation of divergence. The last block defines induced boundary chains for $C^{2}$ patches, proves classical Stokes in that setting, and recovers the planar Green formulas and the circulation interpretation of curl.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Divergence and curl of a $C^{1}$ vector field

Definition

Let $n \geq 1$ , let $U \subseteq R^{n}$ be open and let $F = (F_{0}, \dots, F_{n - 1}) : U \to R^{n}$ be $C^{1}$ in the componentwise Euclidean sense of $C^{k}$ Euclidean maps and diffeomorphisms. Then the divergence of $F$ is $div F : = \sum_{i < n} \partial_{i} F_{i}$ , the function $U \to R$ whose value at $p$ is $\sum_{i < n} \partial_{i} F_{i} (p)$ . The partial derivatives are those of Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , and the sum is the finite sum used throughout The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ . Since each $\partial_{i} F_{i}$ is continuous on $U$ , so is $div F$ .

Now let $n = 3$ and let $F : U \to R^{3}$ be $C^{1}$ on an open $U \subseteq R^{3}$ . Following The cross product in $R^{3}$ , write the three coordinates of a point and of a vector as $x, y, z$ rather than $0, 1, 2$ , so that $F = (F_{x}, F_{y}, F_{z})$ means $F = (F_{0}, F_{1}, F_{2})$ and $\partial_{x}, \partial_{y}, \partial_{z}$ are $\partial_{0}, \partial_{1}, \partial_{2}$ . With that naming, the curl of $F$ is $curl F : = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ , a map $U \to R^{3}$ each of whose coordinates is continuous on $U$ . In this naming the divergence reads $div F = \partial_{x} F_{x} + \partial_{y} F_{y} + \partial_{z} F_{z}$ .

Both operators are defined pointwise from the first partial derivatives of the components, so no differentiability of $F$ beyond $C^{1}$ is used and no orientation or metric structure enters beyond the standard coordinates of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. For a $C^{1}$ scalar function $f$ on $U$ , the gradient $\nabla f = (\partial_{0} f, \dots, \partial_{n - 1} f)$ is that of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case; in the three-coordinate naming, $\nabla f = (\partial_{x} f, \partial_{y} f, \partial_{z} f)$ .

Remarks

Why the curl is only defined in three coordinates. If $A = J F - (J F)^{T}$ with the row-component Jacobian convention, then the curl coordinates are $A_{z y}$ , $A_{x z}$ and $A_{y x}$ . Thus the curl is encoded, with fixed signs, by the three independent off-diagonal entries of $A$ (or by twice those entries if “antisymmetric part” means $A / 2$ ). In $n$ coordinates there are $n (n - 1) / 2$ independent entries. Only at $n = 3$ is that number again $n$ , which is what allows the collection to be read as a vector in the same space. The divergence has no such restriction and is defined for every $n \geq 1$ .
The word "divergence" here is about vector fields. It has nothing to do with the divergence of a sequence or of a series; the two senses share only the word.
Placement of the minus sign in the second coordinate. Some presentations write the middle coordinate as $- (\partial_{x} F_{z} - \partial_{z} F_{x})$ . That is the same real number as $\partial_{z} F_{x} - \partial_{x} F_{z}$ , and the form displayed above is the one whose three coordinates read off the coordinate formula of The cross product in $R^{3}$ in the same cyclic pattern.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Divergence and curl are linear and satisfy the scalar product rules

Statement

Let $n \geq 1$ , let $U \subseteq R^{n}$ be open, let $F, G : U \to R^{n}$ be $C^{1}$ , let $f : U \to R$ be $C^{1}$ and let $a, b \in R$ . Then $a F + b G$ and $f F$ are $C^{1}$ on $U$ and

$div (a F + b G) = a div F + b div G,$

$div (f F) = ⟨ \nabla f, F ⟩ + f div F .$

If moreover $n = 3$ , then

$curl (a F + b G) = a curl F + b curl G, curl (f F) = \nabla f \times F + f curl F .$

Here $div$ , $curl$ and the coordinate naming are those of Divergence and curl of a $C^{1}$ vector field, $\nabla f$ is the gradient of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, $⟨ \cdot, \cdot ⟩$ is the inner product of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\times$ is the cross product of The cross product in $R^{3}$ .

Facts & Assumptions

Given: The open set $U$ , the $C^{1}$ maps $F, G : U \to R^{n}$ , the $C^{1}$ scalar $f : U \to R$ and the reals $a, b$ of the Statement.

[F1]

The divergence of a $C^{1}$ field $F$ on an open $U \subseteq R^{n}$ is $div F = \sum_{i < n} \partial_{i} F_{i}$ , and for $n = 3$ its curl is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

For scalar-valued $f$ on an open subset of $R^{m}$ , the gradient is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F3]

For $u = (u_{x}, u_{y}, u_{z})$ and $v = (v_{x}, v_{y}, v_{z})$ in $R^{3}$ , $u \times v = (u_{y} v_{z} - u_{z} v_{y}, u_{z} v_{x} - u_{x} v_{z}, u_{x} v_{y} - u_{y} v_{x})$ (The cross product in $R^{3}$ ).

[F4]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{k < m} x_{k} y_{k}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F5]

A map $f : U \to R^{q}$ is of class $C^{k}$ when each component is of class $C^{k}$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

[L1]

For real functions of one real variable differentiable at a point, $f + g$ is differentiable there with $(f + g)^{'} = f^{'} + g^{'}$ , $α f$ is differentiable there with $(α f)^{'} = α f^{'}$ , and $f g$ is differentiable there with $(f g)^{'} = f^{'} g + f g^{'}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · direct

1.1givenL1F5

A partial derivative $\partial_{j}$ at a point $p$ is the ordinary one-variable derivative at $0$ of $t \mapsto (\cdot) (p + t e_{j})$ , so [L1] applies to it verbatim: for scalar $C^{1}$ functions $u, v$ on $U$ and reals $a, b$ one has $\partial_{j} (a u + b v) = a \partial_{j} u + b \partial_{j} v$ and $\partial_{j} (u v) = (\partial_{j} u) v + u \partial_{j} v$ pointwise on $U$ , and the right-hand sides are continuous, so $a u + b v$ and $u v$ are again $C^{1}$ .

1.2givenL1F5

Applying 1.1 componentwise, $a F + b G$ and $f F$ have $C^{1}$ components, hence are $C^{1}$ by [F5]; so all four expressions in the Statement are defined.

2.1step 1.1F1algebra

By [F1], $div (a F + b G) = \sum_{i < n} \partial_{i} (a F_{i} + b G_{i})$ , and step 1.1 rewrites each summand as $a \partial_{i} F_{i} + b \partial_{i} G_{i}$ ; summing gives $a \sum_{i < n} \partial_{i} F_{i} + b \sum_{i < n} \partial_{i} G_{i} = a div F + b div G$ .

2.2step 1.1F1algebra

By [F1], the first coordinate of $curl (a F + b G)$ is $\partial_{y} (a F_{z} + b G_{z}) - \partial_{z} (a F_{y} + b G_{y})$ , which step 1.1 rewrites as $a (\partial_{y} F_{z} - \partial_{z} F_{y}) + b (\partial_{y} G_{z} - \partial_{z} G_{y})$ ; the second coordinate is $\partial_{z} (a F_{x} + b G_{x}) - \partial_{x} (a F_{z} + b G_{z}) = a (\partial_{z} F_{x} - \partial_{x} F_{z}) + b (\partial_{z} G_{x} - \partial_{x} G_{z})$ and the third is $\partial_{x} (a F_{y} + b G_{y}) - \partial_{y} (a F_{x} + b G_{x}) = a (\partial_{x} F_{y} - \partial_{y} F_{x}) + b (\partial_{x} G_{y} - \partial_{y} G_{x})$ . The three coordinates are those of $a curl F + b curl G$ .

2.3step 1.1F1F2F4algebra

By [F1], $div (f F) = \sum_{i < n} \partial_{i} (f F_{i})$ , and step 1.1 rewrites each summand as $(\partial_{i} f) F_{i} + f \partial_{i} F_{i}$ . Splitting the sum gives $\sum_{i < n} (\partial_{i} f) F_{i} + f \sum_{i < n} \partial_{i} F_{i}$ , whose first term is $⟨ \nabla f, F ⟩$ by [F2] and [F4] and whose second is $f div F$ by [F1].

2.4step 1.1F1F2F3algebra

By [F1], the first coordinate of $curl (f F)$ is $\partial_{y} (f F_{z}) - \partial_{z} (f F_{y})$ , which step 1.1 rewrites as $((\partial_{y} f) F_{z} - (\partial_{z} f) F_{y}) + f (\partial_{y} F_{z} - \partial_{z} F_{y})$ . By [F2] and [F3] with $u = \nabla f$ and $v = F$ , the first bracket is the first coordinate $u_{y} v_{z} - u_{z} v_{y}$ of $\nabla f \times F$ , and the second summand is $f$ times the first coordinate of $curl F$ .

2.5step 1.1F1F2F3algebra

By [F1], the second coordinate of $curl (f F)$ is $\partial_{z} (f F_{x}) - \partial_{x} (f F_{z})$ , which step 1.1 rewrites as $((\partial_{z} f) F_{x} - (\partial_{x} f) F_{z}) + f (\partial_{z} F_{x} - \partial_{x} F_{z})$ . By [F2] and [F3] the first bracket is the second coordinate $u_{z} v_{x} - u_{x} v_{z}$ of $\nabla f \times F$ , and the second summand is $f$ times the second coordinate of $curl F$ .

2.6step 1.1F1F2F3algebra

By [F1], the third coordinate of $curl (f F)$ is $\partial_{x} (f F_{y}) - \partial_{y} (f F_{x})$ , which step 1.1 rewrites as $((\partial_{x} f) F_{y} - (\partial_{y} f) F_{x}) + f (\partial_{x} F_{y} - \partial_{y} F_{x})$ . By [F2] and [F3] the first bracket is the third coordinate $u_{x} v_{y} - u_{y} v_{x}$ of $\nabla f \times F$ , and the second summand is $f$ times the third coordinate of $curl F$ .

3.1step 2.1step 2.2step 2.3step 2.4step 2.5step 2.6∎

Steps 2.4, 2.5 and 2.6 give the three coordinates of $\nabla f \times F + f curl F$ , so $curl (f F) = \nabla f \times F + f curl F$ ; with steps 2.1, 2.2 and 2.3 this is every assertion of the Statement.

Remarks

The scalar $f$ need only be $C^{1}$ . No mixed second derivative of $f$ appears in either product rule, so nothing here needs $C^{2}$ ; the identities of The curl of the gradient of a $C^{2}$ function vanishes and The divergence of the curl of a $C^{2}$ field vanishes are where the second-order hypothesis becomes necessary.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The divergence and curl of a cross product

Statement

Let $U \subseteq R^{3}$ be open and let $F, G : U \to R^{3}$ be $C^{1}$ . Then $F \times G$ is $C^{1}$ on $U$ and

$div (F \times G) = ⟨ curl F, G ⟩ - ⟨ F, curl G ⟩,$

$curl (F \times G) = (div G) F - (div F) G + D F G - D G F .$

Here $D F G$ denotes the map $U \to R^{3}$ whose $i$ th coordinate at $p$ is $\sum_{j < 3} \partial_{j} F_{i} (p) G_{j} (p)$ , that is, the Jacobian matrix of $F$ at $p$ applied to the vector $G (p)$ , and $D G F$ is defined the same way with the roles of $F$ and $G$ exchanged. The operators are those of Divergence and curl of a $C^{1}$ vector field, the cross product is that of The cross product in $R^{3}$ , the inner product that of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and the Jacobian matrix that of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case.

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ and the $C^{1}$ maps $F, G : U \to R^{3}$ of the Statement, with coordinates named $x, y, z$ .

[F1]

[F2]

The divergence of a $C^{1}$ field $F$ on an open $U \subseteq R^{n}$ is $div F = \sum_{i < n} \partial_{i} F_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

The curl of a $C^{1}$ field $F$ on an open $U \subseteq R^{3}$ is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F4]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{k < m} x_{k} y_{k}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F5]

If every partial derivative $\partial_{j} f_{i} (a)$ of $f : U \to R^{n}$ exists, the Jacobian matrix is $J f (a) = (\partial_{j} f_{i} (a))_{i < n, j < m}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L1]

For real functions of one real variable differentiable at a point, $f g$ is differentiable there and $(f g)^{'} = f^{'} g + f g^{'}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L2]

If $f$ is totally differentiable at $a$ then $\partial_{j} f (a) = D f (a) e_{j}$ , and the matrix of $D f (a)$ is $J f (a)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L3]

The cross product is bilinear and alternating (The cross product is bilinear, alternating, and orthogonal to both factors).

Proof

technique · direct

1.1givenF1L1

By [F1] the three coordinates of $F \times G$ are $F_{y} G_{z} - F_{z} G_{y}$ , $F_{z} G_{x} - F_{x} G_{z}$ and $F_{x} G_{y} - F_{y} G_{x}$ . Each is a difference of products of $C^{1}$ scalars, so by [L1] applied in each coordinate direction each has continuous first partial derivatives, given by $\partial_{j} (F_{a} G_{b}) = (\partial_{j} F_{a}) G_{b} + F_{a} \partial_{j} G_{b}$ ; hence $F \times G$ is $C^{1}$ and both sides of both identities are defined.

2.1step 1.1L1F2F3F4algebra

Expanding $div (F \times G) = \partial_{x} (F_{y} G_{z} - F_{z} G_{y}) + \partial_{y} (F_{z} G_{x} - F_{x} G_{z}) + \partial_{z} (F_{x} G_{y} - F_{y} G_{x})$ by step 1.1 gives twelve terms. Those carrying a derivative of $F$ are $(\partial_{y} F_{z} - \partial_{z} F_{y}) G_{x} + (\partial_{z} F_{x} - \partial_{x} F_{z}) G_{y} + (\partial_{x} F_{y} - \partial_{y} F_{x}) G_{z}$ , which is $⟨ curl F, G ⟩$ by [F3] and [F4]; those carrying a derivative of $G$ are $- F_{x} (\partial_{y} G_{z} - \partial_{z} G_{y}) - F_{y} (\partial_{z} G_{x} - \partial_{x} G_{z}) - F_{z} (\partial_{x} G_{y} - \partial_{y} G_{x})$ , which is $- ⟨ F, curl G ⟩$ . This is the first identity.

2.2step 1.1L1F3F2F5L2algebra

By [F3] and step 1.1 the first coordinate of $curl (F \times G)$ is $\partial_{y} (F_{x} G_{y} - F_{y} G_{x}) - \partial_{z} (F_{z} G_{x} - F_{x} G_{z})$ , that is $(\partial_{y} F_{x}) G_{y} + F_{x} \partial_{y} G_{y} - (\partial_{y} F_{y}) G_{x} - F_{y} \partial_{y} G_{x} - (\partial_{z} F_{z}) G_{x} - F_{z} \partial_{z} G_{x} + (\partial_{z} F_{x}) G_{z} + F_{x} \partial_{z} G_{z}$ . Adding and subtracting $F_{x} \partial_{x} G_{x}$ and $G_{x} \partial_{x} F_{x}$ regroups this as $F_{x} div G - G_{x} div F + ((\partial_{x} F_{x}) G_{x} + (\partial_{y} F_{x}) G_{y} + (\partial_{z} F_{x}) G_{z}) - (F_{x} \partial_{x} G_{x} + F_{y} \partial_{y} G_{x} + F_{z} \partial_{z} G_{x})$ , using [F2] for the two divergences and [F5] for the two bracketed sums.

2.3step 1.1L1F3F2F5L2algebra

The same computation in the second coordinate gives $\partial_{z} (F_{y} G_{z} - F_{z} G_{y}) - \partial_{x} (F_{x} G_{y} - F_{y} G_{x})$ , which after adding and subtracting $F_{y} \partial_{y} G_{y}$ and $G_{y} \partial_{y} F_{y}$ is $F_{y} div G - G_{y} div F + \sum_{j < 3} (\partial_{j} F_{y}) G_{j} - \sum_{j < 3} F_{j} \partial_{j} G_{y}$ ; in the third coordinate it gives $\partial_{x} (F_{z} G_{x} - F_{x} G_{z}) - \partial_{y} (F_{y} G_{z} - F_{z} G_{y})$ , which after adding and subtracting $F_{z} \partial_{z} G_{z}$ and $G_{z} \partial_{z} F_{z}$ is $F_{z} div G - G_{z} div F + \sum_{j < 3} (\partial_{j} F_{z}) G_{j} - \sum_{j < 3} F_{j} \partial_{j} G_{z}$ .

3.1step 2.2step 2.3F5L2algebra

In steps 2.2 and 2.3 the sums $\sum_{j < 3} (\partial_{j} F_{i}) G_{j}$ and $\sum_{j < 3} F_{j} \partial_{j} G_{i}$ are the $i$ th coordinates of $D F G$ and of $D G F$ : by [F5] the $i$ th row of the Jacobian matrix of $F$ is $(\partial_{j} F_{i})_{j < 3}$ , and by [L2] that matrix is the matrix of the total derivative, so applying it to the vector $G$ produces exactly that sum coordinate by coordinate.

4.1step 2.1step 3.1L3∎

Substituting step 3.1 into steps 2.2 and 2.3 gives the three coordinates of $(div G) F - (div F) G + D F G - D G F$ , which is the second identity; with step 2.1 both assertions hold at every point of $U$ , and by [L3] both sides of each are unchanged in form when $F$ and $G$ are replaced by linear combinations, since the cross product is bilinear.

Remarks

Where the alternating law is visible. Taking $G = F$ makes $F \times F = 0$ by [L3], and both identities then read $0 = 0$ : in the first because $⟨ curl F, F ⟩ - ⟨ F, curl F ⟩ = 0$ , and in the second because the four terms cancel in pairs.
Only first derivatives are used. Both identities hold for $C^{1}$ fields; nothing here interchanges two partial derivatives, which is why no $C^{2}$ hypothesis appears.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A $C^{1}$ field on an open subset of $R^{3}$ is closed exactly when its curl vanishes

Statement

Let $U \subseteq R^{3}$ be open and let $F : U \to R^{3}$ be $C^{1}$ . Then a $C^{1}$ field on an open subset of $R^{3}$ is closed if and only if its curl vanishes identically: $F$ is closed in the sense of Exact and closed C1 vector fields exactly when $curl F (p) = 0$ for every $p \in U$ .

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ and the $C^{1}$ field $F : U \to R^{3}$ of the Statement, with the three coordinates named $x, y, z$ as on this page.

[F1]

[F2]

A $C^{1}$ field $F = (F_{0}, \dots, F_{n - 1})$ on an open $U \subseteq R^{n}$ , with coordinates and partial derivatives indexed from $0$ , is closed when $\partial_{j} F_{i} = \partial_{i} F_{j}$ for all $i, j < n$ (Exact and closed C1 vector fields).

[F3]

For $x \in R^{n}$ one writes $x_{k} : = x (k)$ for $k < n$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

technique · direct

1.1givenF2F3

By [F3] the coordinates of a point of $R^{3}$ are indexed $0, 1, 2$ , and the names $x, y, z$ used on this page are those three indices in that order; so the closedness condition of [F2] at $n = 3$ is the system of equations $\partial_{j} F_{i} = \partial_{i} F_{j}$ ranging over all pairs $i, j$ drawn from ${x, y, z}$ .

1.2F2algebra

In that system the equations with $i = j$ read $\partial_{i} F_{i} = \partial_{i} F_{i}$ and hold for every field, and the equation indexed $(i, j)$ is the same equation as the one indexed $(j, i)$ . Hence the system is equivalent to its three equations indexed by the unordered pairs ${y, z}$ , ${z, x}$ and ${x, y}$ .

2.1step 1.1step 1.2F1algebra

Written out, those three equations are $\partial_{y} F_{z} = \partial_{z} F_{y}$ , $\partial_{z} F_{x} = \partial_{x} F_{z}$ and $\partial_{x} F_{y} = \partial_{y} F_{x}$ . Their left-minus-right differences $\partial_{y} F_{z} - \partial_{z} F_{y}$ , $\partial_{z} F_{x} - \partial_{x} F_{z}$ and $\partial_{x} F_{y} - \partial_{y} F_{x}$ are, by [F1], exactly the first, second and third coordinates of $curl F$ .

3.1step 2.1F2F1

For the forward direction, suppose $F$ is closed. By steps 1.1 and 1.2 the three equations of step 2.1 hold at every $p \in U$ , so by [F1] each of the three coordinates of $curl F (p)$ is zero; hence $curl F$ vanishes identically on $U$ .

3.2step 2.1F1F2

For the converse direction, suppose $curl F (p) = 0$ for every $p \in U$ . By [F1] each of the three differences of step 2.1 is zero at every $p$ , so the three equations of step 2.1 hold on $U$ ; by steps 1.1 and 1.2 these are equivalent to the full system of [F2], so $F$ is closed.

4.1step 3.1step 3.2∎

Steps 3.1 and 3.2 are the two implications, so $F$ is closed if and only if its curl vanishes identically.

Remarks

Why the count of equations matters. Closedness in $R^{n}$ is a condition on all ordered pairs of indices, and the curl in $R^{3}$ has three coordinates. Step 1.2 is what shows that these are the same amount of information: the diagonal equations are automatic and each off-diagonal equation is listed twice. In $R^{n}$ with $n \neq 3$ the number of independent equations is $n (n - 1) / 2$ , so there is no vector of that many coordinates in the same space to collect them into, and closedness is then stated only as the system itself.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The curl of the gradient of a $C^{2}$ function vanishes

Statement

Let $U \subseteq R^{3}$ be open and let $ϕ : U \to R$ be $C^{2}$ . Then $\nabla ϕ$ is a $C^{1}$ field on $U$ and

$curl \nabla ϕ = 0 on U .$

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ and the $C^{2}$ function $ϕ : U \to R$ of the Statement, with the three coordinates named $x, y, z$ .

[F1]

[F2]

For scalar-valued $f$ , its gradient is $\nabla f (a) = (\partial_{0} f (a), \dots, \partial_{m - 1} f (a))$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F3]

A scalar $f$ is of class $C^{k}$ on $U$ when, for every word $(i_{1}, \dots, i_{r})$ of coordinate indices with $0 \leq r \leq k$ , the iterated derivative $\partial_{i_{r}} \dots \partial_{i_{1}} f$ exists and is continuous on $U$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F4]

A map $f : U \to R^{q}$ is of class $C^{k}$ when each component is of class $C^{k}$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

[L1]

If $f$ is $C^{2}$ on an open subset of $R^{m}$ , then $\partial_{i} \partial_{j} f = \partial_{j} \partial_{i} f$ for every pair of coordinate indices (Clairaut--Schwarz theorem for continuous second partial derivatives).

Proof

technique · direct

1.1givenF2F3F4

By [F2] the components of $\nabla ϕ$ are the three first partial derivatives $\partial_{x} ϕ, \partial_{y} ϕ, \partial_{z} ϕ$ . Since $ϕ$ is $C^{2}$ , [F3] with $k = 2$ says that every iterated derivative $\partial_{i} \partial_{j} ϕ$ exists and is continuous on $U$ ; so each component of $\nabla ϕ$ has continuous first partial derivatives, and by [F4] the field $\nabla ϕ$ is $C^{1}$ on $U$ and its curl is defined.

2.1step 1.1F1F2L1

By [F1] and [F2] the first coordinate of $curl \nabla ϕ$ is $\partial_{y} (\partial_{z} ϕ) - \partial_{z} (\partial_{y} ϕ)$ , and by [L1] applied to $ϕ$ with the index pair $y, z$ these two iterated derivatives are equal, so this coordinate is zero at every point of $U$ .

2.2step 1.1F1F2L1

By [F1] and [F2] the second coordinate of $curl \nabla ϕ$ is $\partial_{z} (\partial_{x} ϕ) - \partial_{x} (\partial_{z} ϕ)$ , and by [L1] applied with the index pair $z, x$ these are equal, so this coordinate is zero at every point of $U$ .

2.3step 1.1F1F2L1

By [F1] and [F2] the third coordinate of $curl \nabla ϕ$ is $\partial_{x} (\partial_{y} ϕ) - \partial_{y} (\partial_{x} ϕ)$ , and by [L1] applied with the index pair $x, y$ these are equal, so this coordinate is zero at every point of $U$ .

3.1step 2.1step 2.2step 2.3∎

All three coordinates vanish at every point of $U$ , so $curl \nabla ϕ = 0$ on $U$ . The hypothesis that $ϕ$ is $C^{2}$ was used twice: in step 1.1, so that $\nabla ϕ$ is $C^{1}$ and its curl is defined at all, and in steps 2.1 to 2.3 as the hypothesis of [L1].

Remarks

Why $C^{1}$ would not do. With $ϕ$ merely $C^{1}$ the field $\nabla ϕ$ need not be differentiable, so $curl \nabla ϕ$ need not be defined; the statement would have no content rather than a weaker one.
What the converse would say. This theorem says every gradient of a $C^{2}$ function is curl-free. Which curl-free fields are gradients is a separate question, answered on a star-shaped open set by A $C^{1}$ field with vanishing curl on a star-shaped open subset of $R^{3}$ is conservative; the hypothesis on the domain there is not decorative, and the companion examples page exhibits a curl-free field with no potential.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The divergence of the curl of a $C^{2}$ field vanishes

Statement

Let $U \subseteq R^{3}$ be open and let $F : U \to R^{3}$ be $C^{2}$ . Then $curl F$ is a $C^{1}$ field on $U$ and

$div (curl F) = 0 on U .$

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ and the $C^{2}$ field $F : U \to R^{3}$ of the Statement, with the three coordinates named $x, y, z$ .

[F1]

The divergence of a $C^{1}$ field $G$ on an open $U \subseteq R^{n}$ is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

[F3]

[F4]

A map $f : U \to R^{q}$ is of class $C^{k}$ when each component is of class $C^{k}$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

[L1]

Proof

technique · direct

1.1givenF2F3F4

By [F2] each coordinate of $curl F$ is a difference of two first partial derivatives of components of $F$ . Since $F$ is $C^{2}$ , [F4] and [F3] with $k = 2$ give that every iterated derivative $\partial_{i} \partial_{j} F_{a}$ exists and is continuous on $U$ , so each coordinate of $curl F$ has continuous first partial derivatives; by [F4] again, $curl F$ is $C^{1}$ on $U$ and its divergence is defined.

2.1step 1.1F1F2algebra

By [F1] and [F2], $div (curl F) = \partial_{x} (\partial_{y} F_{z} - \partial_{z} F_{y}) + \partial_{y} (\partial_{z} F_{x} - \partial_{x} F_{z}) + \partial_{z} (\partial_{x} F_{y} - \partial_{y} F_{x})$ , which written out is the sum of the six terms $\partial_{x} \partial_{y} F_{z}$ , $- \partial_{x} \partial_{z} F_{y}$ , $\partial_{y} \partial_{z} F_{x}$ , $- \partial_{y} \partial_{x} F_{z}$ , $\partial_{z} \partial_{x} F_{y}$ and $- \partial_{z} \partial_{y} F_{x}$ .

3.1step 2.1L1

Each component of $F$ is $C^{2}$ , so [L1] gives $\partial_{x} \partial_{y} F_{z} = \partial_{y} \partial_{x} F_{z}$ , $\partial_{y} \partial_{z} F_{x} = \partial_{z} \partial_{y} F_{x}$ and $\partial_{z} \partial_{x} F_{y} = \partial_{x} \partial_{z} F_{y}$ . Pairing the six terms of step 2.1 accordingly, $\partial_{x} \partial_{y} F_{z}$ cancels $- \partial_{y} \partial_{x} F_{z}$ , $\partial_{y} \partial_{z} F_{x}$ cancels $- \partial_{z} \partial_{y} F_{x}$ , and $\partial_{z} \partial_{x} F_{y}$ cancels $- \partial_{x} \partial_{z} F_{y}$ .

4.1step 3.1∎

The six terms therefore sum to zero at every point of $U$ , so $div (curl F) = 0$ on $U$ . The hypothesis that $F$ is $C^{2}$ is used in step 1.1, so that $curl F$ is $C^{1}$ and has a divergence, and in step 3.1 as the hypothesis of [L1].

Remarks

Where the hypothesis bites. If $F$ is only $C^{1}$ , then $curl F$ is merely continuous and its partial derivatives need not exist, so $div (curl F)$ is not defined; there is nothing to assert, rather than a weaker assertion.
The converse. A divergence-free $C^{1}$ field on a star-shaped open subset of $R^{3}$ is the curl of something: that is A divergence-free $C^{1}$ field on a star-shaped open subset of $R^{3}$ has a vector potential.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field

Definition

Let $n \geq 1$ , let $U \subseteq R^{n}$ be open and let $f : U \to R$ be $C^{2}$ in the sense of $C^{k}$ maps and multi-index derivative notation in Euclidean space. Then $\nabla f$ is a $C^{1}$ field on $U$ by $C^{k}$ Euclidean maps and diffeomorphisms, since each of its components $\partial_{i} f$ has continuous first partial derivatives, so its divergence is defined; the Laplacian of $f$ is

$Δ f : = div \nabla f = \sum_{i < n} \partial_{i} \partial_{i} f,$

with the gradient of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case and the divergence of Divergence and curl of a $C^{1}$ vector field. A $C^{2}$ function with $Δ f = 0$ on $U$ is called harmonic on $U$ .

For a $C^{2}$ map $F = (F_{0}, \dots, F_{q - 1}) : U \to R^{q}$ , whose components are $C^{2}$ by $C^{k}$ Euclidean maps and diffeomorphisms, $Δ F$ is the field whose $i$ th coordinate is $Δ F_{i}$ . In the three-coordinate naming of this page, $Δ f = \partial_{x} \partial_{x} f + \partial_{y} \partial_{y} f + \partial_{z} \partial_{z} f$ and $Δ F = (Δ F_{x}, Δ F_{y}, Δ F_{z})$ .

Remarks

The vector case is componentwise by convention, and the convention is stated because sources leave it implicit. Nothing forces a single reading of $Δ$ on a field; the componentwise one is the one that makes the curl-of-a-curl identity of The curl of a curl is the gradient of the divergence minus the Laplacian true as written, and it is the reading in force everywhere on this page.
Why $C^{2}$ and not $C^{1}$ . Forming $\nabla f$ consumes one degree of differentiability, so $div \nabla f$ needs $\nabla f$ to be $C^{1}$ ; that is exactly $f$ being $C^{2}$ . Nothing here interchanges two partial derivatives, so no appeal to a mixed-partials theorem is made in the definition itself, and $Δ f$ is defined by the displayed sum in the fixed order $\partial_{i} \partial_{i}$ .
The planar equation. For $n = 2$ the condition $Δ f = 0$ reads $\partial_{x} \partial_{x} f + \partial_{y} \partial_{y} f = 0$ . That is the equation written out in The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair ↗ for the real and imaginary parts of a holomorphic function; a reader meeting the word "harmonic" in either place is meeting one notion.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The curl of a curl is the gradient of the divergence minus the Laplacian

Statement

Let $U \subseteq R^{3}$ be open and let $F : U \to R^{3}$ be $C^{2}$ . Then $curl curl F$ , $\nabla div F$ and $Δ F$ are all defined on $U$ and

$curl curl F = \nabla div F - Δ F .$

Here $Δ F$ is the componentwise Laplacian of The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field.

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ and the $C^{2}$ field $F : U \to R^{3}$ of the Statement, with the three coordinates named $x, y, z$ .

[F1]

[F2]

The divergence of a $C^{1}$ field $G$ on an open $U \subseteq R^{n}$ is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

For a $C^{2}$ map $F$ , $Δ F$ is the field whose $i$ th coordinate is $Δ F_{i}$ , and $Δ f = \sum_{i < n} \partial_{i} \partial_{i} f$ for a $C^{2}$ scalar $f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F4]

For scalar-valued $f$ , its gradient is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F5]

[L1]

Proof

technique · direct

1.1givenF1F2F3F4F5

Every component of $F$ is $C^{2}$ , so by [F5] every iterated derivative $\partial_{i} \partial_{j} F_{a}$ exists and is continuous on $U$ . Hence each coordinate of $curl F$ , being a difference of first partial derivatives of components of $F$ by [F1], has continuous first partial derivatives, so $curl F$ is $C^{1}$ and $curl curl F$ is defined; likewise $div F$ is $C^{1}$ by [F2], so $\nabla div F$ is defined by [F4]; and $Δ F$ is defined by [F3].

2.1step 1.1F1algebra

By [F1] applied twice, the first coordinate of $curl curl F$ is $\partial_{y} (curl F)_{z} - \partial_{z} (curl F)_{y} = \partial_{y} (\partial_{x} F_{y} - \partial_{y} F_{x}) - \partial_{z} (\partial_{z} F_{x} - \partial_{x} F_{z})$ , that is $\partial_{y} \partial_{x} F_{y} - \partial_{y} \partial_{y} F_{x} - \partial_{z} \partial_{z} F_{x} + \partial_{z} \partial_{x} F_{z}$ .

3.1step 2.1algebra

Adding and subtracting the single term $\partial_{x} \partial_{x} F_{x}$ rewrites step 2.1 as $(\partial_{x} \partial_{x} F_{x} + \partial_{y} \partial_{x} F_{y} + \partial_{z} \partial_{x} F_{z}) - (\partial_{x} \partial_{x} F_{x} + \partial_{y} \partial_{y} F_{x} + \partial_{z} \partial_{z} F_{x})$ .

4.1step 3.1L1F2F3F4

By [L1], $\partial_{y} \partial_{x} F_{y} = \partial_{x} \partial_{y} F_{y}$ and $\partial_{z} \partial_{x} F_{z} = \partial_{x} \partial_{z} F_{z}$ , so the first bracket of step 3.1 is $\partial_{x} (\partial_{x} F_{x} + \partial_{y} F_{y} + \partial_{z} F_{z}) = \partial_{x} div F$ , the first coordinate of $\nabla div F$ by [F2] and [F4]; the second bracket is $Δ F_{x}$ , the first coordinate of $Δ F$ by [F3]. Hence the first coordinate of $curl curl F$ is that of $\nabla div F - Δ F$ .

4.2step 2.1step 3.1L1F1F2F3F4

In the second coordinate, [F1] gives $\partial_{z} (curl F)_{x} - \partial_{x} (curl F)_{z} = \partial_{z} \partial_{y} F_{z} - \partial_{z} \partial_{z} F_{y} - \partial_{x} \partial_{x} F_{y} + \partial_{x} \partial_{y} F_{x}$ ; adding and subtracting $\partial_{y} \partial_{y} F_{y}$ and applying [L1] to $\partial_{z} \partial_{y} F_{z} = \partial_{y} \partial_{z} F_{z}$ and $\partial_{x} \partial_{y} F_{x} = \partial_{y} \partial_{x} F_{x}$ turns it into $\partial_{y} div F - Δ F_{y}$ . In the third coordinate, [F1] gives $\partial_{x} (curl F)_{y} - \partial_{y} (curl F)_{x} = \partial_{x} \partial_{z} F_{x} - \partial_{x} \partial_{x} F_{z} - \partial_{y} \partial_{y} F_{z} + \partial_{y} \partial_{z} F_{y}$ ; adding and subtracting $\partial_{z} \partial_{z} F_{z}$ and applying [L1] to $\partial_{x} \partial_{z} F_{x} = \partial_{z} \partial_{x} F_{x}$ and $\partial_{y} \partial_{z} F_{y} = \partial_{z} \partial_{y} F_{y}$ turns it into $\partial_{z} div F - Δ F_{z}$ .

5.1step 4.1step 4.2∎

All three coordinates of $curl curl F$ agree with those of $\nabla div F - Δ F$ at every point of $U$ , which is the asserted identity. The hypothesis that $F$ is $C^{2}$ is used in step 1.1, so that all three expressions are defined, and in steps 4.1 and 4.2 as the hypothesis of [L1].

Remarks

The added and subtracted term is what makes the identity close. The expansion of $(curl curl F)_{x}$ contains no pure second derivative $\partial_{x} \partial_{x} F_{x}$ , while both $\nabla div F$ and $Δ F$ do; that one term belongs to both groups and cancels between them, which is why it can be inserted at will and why neither side alone matches the expansion.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A $C^{1}$ field with vanishing curl on a star-shaped open subset of $R^{3}$ is conservative

Statement

Let $U \subseteq R^{3}$ be open and star-shaped and let $F : U \to R^{3}$ be $C^{1}$ with $curl F = 0$ on $U$ . Then a $C^{1}$ field with vanishing curl on a star-shaped open subset of $R^{3}$ is exact, conservative and path-independent: there is a $C^{2}$ function $ϕ : U \to R$ with $F = \nabla ϕ$ , any two piecewise- $C^{1}$ paths in $U$ with the same endpoints give $F$ the same vector line integral, and

$\int_{γ} F \cdot d r = 0$

for every closed piecewise- $C^{1}$ path $γ$ in $U$ . Conversely, a field exact on such a set has vanishing curl, so on a star-shaped open subset of $R^{3}$ vanishing curl and exactness are equivalent.

Facts & Assumptions

Given: The star-shaped open set $U \subseteq R^{3}$ with a star centre $a \in U$ , and the $C^{1}$ field $F : U \to R^{3}$ with $curl F = 0$ on $U$ .

[F1]

A nonempty open set $U \subseteq R^{n}$ is star-shaped with respect to $a \in U$ when $a + t (x - a) \in U$ for every $x \in U$ and $0 \leq t \leq 1$ (Star-shaped open subsets of Euclidean space).

[F2]

For a continuous field $F$ on an open $U \subseteq R^{n}$ , a $C^{1}$ function $ϕ : U \to R$ is a potential when $F = \nabla ϕ$ ; $F$ is conservative when it has a potential, and path-independent when any two piecewise- $C^{1}$ paths in $U$ with the same initial and terminal points have equal vector line integrals (Piecewise-C1 path-connected domains, potential functions, conservative fields, and path independence).

[F3]

[L1]

A $C^{1}$ field on an open subset of $R^{3}$ is closed if and only if its curl vanishes identically (A $C^{1}$ field on an open subset of $R^{3}$ is closed exactly when its curl vanishes).

[L2]

Let $U \subseteq R^{n}$ be open and star-shaped and let $F : U \to R^{n}$ be $C^{1}$ . Then the five conditions that $F$ be closed, exact, conservative, path-independent, and give every closed piecewise- $C^{1}$ path in $U$ zero integral are equivalent (On a star-shaped open domain, closed, exact, conservative, path-independent, and zero-loop are equivalent).

Proof

technique · direct

1.1givenL1F3

The field $F$ is $C^{1}$ on the open set $U \subseteq R^{3}$ and its curl vanishes identically, so by the reverse direction of [L1] it is closed.

1.2givenF1

By [F1] the set $U$ is nonempty, open and star-shaped with respect to its centre $a$ . With $n = 3$ these are exactly the hypotheses [L2] places on the domain, and $F$ is $C^{1}$ as [L2] requires of the field.

2.1step 1.1step 1.2L2F2

By steps 1.1 and 1.2, [L2] applies and its first condition holds, so all five hold: $F$ is exact, hence there is a $C^{2}$ function $ϕ$ on $U$ with $F = \nabla ϕ$ ; $F$ is conservative, so it has a potential in the sense of [F2]; $F$ is path-independent; and every closed piecewise- $C^{1}$ path in $U$ gives $F$ integral zero.

3.1step 2.1L1L2F2∎

For the converse reading, suppose instead that $F$ is exact on $U$ . Then the first condition of [L2] holds by the same equivalence, so $F$ is closed, and the forward direction of [L1] makes $curl F$ vanish identically. Together with step 2.1 this gives the stated equivalence between vanishing curl and exactness on a star-shaped open subset of $R^{3}$ .

Remarks

The hypothesis on the domain is doing work. Star-shapedness is not a convenience: the companion examples page gives a $C^{1}$ field with vanishing curl on a connected open subset of $R^{3}$ that has no potential. What fails there is exactly [F1], since no point of the complement of a line is a star centre for it.
Why the potential is $C^{2}$ and not merely $C^{1}$ . Exactness in Exact and closed C1 vector fields asks for a $C^{2}$ potential, which is what makes all mixed second partial derivatives of $ϕ$ available and continuous; a conservative field in the sense of [F2] is only required to have a $C^{1}$ one. Step 2.1 supplies the stronger form because [L2] does.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Vector potentials of a continuous field on an open subset of $R^{3}$

Definition

Let $U \subseteq R^{3}$ be open and let $B : U \to R^{3}$ be continuous. Given a map $A : U \to R^{3}$ , we say $A$ is a vector potential for $B$ when $A$ is $C^{1}$ on $U$ and $curl A = B$ at every point of $U$ , with the curl of Divergence and curl of a $C^{1}$ vector field and the class $C^{1}$ of $C^{k}$ Euclidean maps and diffeomorphisms. A field admitting a vector potential is said to have a vector potential on $U$ .

Remarks

This is the curl analogue of exactness, not the same notion. A field is exact when it is the gradient of a $C^{2}$ scalar (Exact and closed C1 vector fields); it has a vector potential when it is the curl of a $C^{1}$ field. The two conditions constrain a field in different ways: on an open subset of $R^{3}$ a gradient of a $C^{2}$ function has vanishing curl and a curl of a $C^{2}$ field has vanishing divergence.
Nonuniqueness on a nonempty domain. If $U$ is nonempty and $A$ is a vector potential for $B$ , take the $C^{2}$ coordinate function $ϕ (x) = x_{0}$ . Then $A + \nabla ϕ = A + e_{0}$ is distinct from $A$ , is $C^{1}$ , and has the same curl by the linearity of curl (Divergence and curl are linear and satisfy the scalar product rules) and $curl \nabla ϕ = 0$ (The curl of the gradient of a $C^{2}$ function vanishes). On the empty open set there is only one map to $R^{3}$ , so the nonempty hypothesis is essential to this remark.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A divergence-free $C^{1}$ field on a star-shaped open subset of $R^{3}$ has a vector potential

Statement

Let $U \subseteq R^{3}$ be open and star-shaped with star centre $a$ , and let $B : U \to R^{3}$ be $C^{1}$ with $div B = 0$ on $U$ . Then $B$ has a vector potential on $U$ in the sense of Vector potentials of a continuous field on an open subset of $R^{3}$ : the map

$A (x) : = \int_{0}^{1} t B (a + t (x - a)) \times (x - a) d t (x \in U),$

understood coordinatewise, is $C^{1}$ on $U$ and satisfies $curl A = B$ .

Facts & Assumptions

Given: The star-shaped open set $U \subseteq R^{3}$ with centre $a$ , and the $C^{1}$ field $B : U \to R^{3}$ with $div B = 0$ on $U$ . Throughout, $w : = x - a$ and $z_{t} : = a + t w$ .

[F1]

Given a continuous $B$ on an open $U \subseteq R^{3}$ , a map $A$ is a vector potential for $B$ when $A$ is $C^{1}$ on $U$ and $curl A = B$ (Vector potentials of a continuous field on an open subset of $R^{3}$ ).

[F2]

A nonempty open $U \subseteq R^{n}$ is star-shaped with respect to $a \in U$ when $a + t (x - a) \in U$ for every $x \in U$ and $0 \leq t \leq 1$ (Star-shaped open subsets of Euclidean space).

[F3]

[F4]

[F5]

If every partial derivative $\partial_{j} f_{i} (a)$ of $f$ exists, the Jacobian matrix is $J f (a) = (\partial_{j} f_{i} (a))_{i < n, j < m}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L1]

For $C^{1}$ fields $F, G$ on an open subset of $R^{3}$ , $curl (F \times G) = (div G) F - (div F) G + D F G - D G F$ (The divergence and curl of a cross product).

[L2]

Let $α < β$ and $c < d$ , let $g, h : [α, β] \times [c, d] \to R$ be continuous, and suppose for every fixed $t$ that $s \mapsto g (s, t)$ is differentiable on $(α, β)$ with derivative $h (s, t)$ . Then $G (s) = \int_{c}^{d} g (s, t) d t$ is differentiable on $[α, β]$ with $G^{'} (s) = \int_{c}^{d} h (s, t) d t$ (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral).

[L3]

If $G$ is continuous on $[α, β]$ and differentiable on $(α, β)$ , and $f$ is Riemann integrable on $[α, β]$ with $f = G^{'}$ on $(α, β)$ , then $\int_{α}^{β} f = G (β) - G (α)$ (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative).

[L4]

If $f$ is totally differentiable at $p$ and $g$ at $f (p)$ , then $D (g \circ f) (p) = D g (f (p)) \circ D f (p)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L5]

If $f$ is totally differentiable at $p$ then $D_{v} f (p) = D f (p) v$ for every $v$ ; in particular $\partial_{j} f (p) = D f (p) e_{j}$ , and the matrix of $D f (p)$ is $J f (p)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L6]

If every partial derivative of $f$ exists on a neighbourhood of $p$ and is continuous at $p$ , then $f$ is totally differentiable at $p$ and $D f (p)$ is the linear map with matrix $J f (p)$ (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L7]

[L8]

A continuous map from a compact metric space to a metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous); a closed bounded subset of $R^{m}$ is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L9]

If $f$ and $g$ are integrable between $u$ and $v$ and $∣ f - g ∣ \leq η$ throughout the closed interval with those endpoints, then $∣ \int_{u}^{v} f - \int_{u}^{v} g ∣ \leq η ∣ v - u ∣$ (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error); a continuous function on a closed bounded interval is Riemann integrable (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

technique · constructive

1.1givenF2L9construct

Take $A$ to be the map displayed in the Statement. By [F2] every $z_{t} = a + t w$ with $0 \leq t \leq 1$ lies in $U$ when $x \in U$ , so $B (z_{t})$ is defined there; by [L9] each coordinate of the integrand, being continuous in $t$ , is integrable on $[0, 1]$ , so $A (x)$ is defined for every $x \in U$ .

1.2givenF3L4L5L6

By [F3] each coordinate of $t B (z_{t}) \times w$ is a sum of terms $\pm t B_{k} (z_{t}) w_{l}$ . The map $(x, t) \mapsto z_{t}$ is continuous, so each such term is continuous in $(x, t)$ ; and since $B$ is $C^{1}$ , [L6], [L4] and [L5] give $\partial_{j} (B_{k} (z_{t})) = t (\partial_{j} B_{k}) (z_{t})$ , which is again continuous in $(x, t)$ , while $\partial_{j} w_{l}$ is $1$ if $l = j$ and $0$ otherwise. Hence each coordinate of the integrand has, in each coordinate of $x$ , a partial derivative that is continuous in $(x, t)$ .

2.1step 1.2L2L6L8L9F4F5

Fix $p \in U$ , choose a closed box $Q \subseteq U$ with $p$ in its interior, and fix indices $i, j$ . Applying [L2] with $[α, β]$ the $j$ th edge of $Q$ , the other coordinates of $x$ held at those of $p$ , and $[c, d] = [0, 1]$ , using step 1.2 for the continuity of $g$ and $h$ and for the derivative hypothesis, gives that $\partial_{j} A_{i}$ exists at $p$ with $\partial_{j} A_{i} (p) = \int_{0}^{1} \partial_{j} (t (B (z_{t}) \times w)_{i}) d t$ . The set $Q \times [0, 1]$ is closed and bounded in $R^{4}$ , hence compact by [L8], so the integrand of that formula is uniformly continuous on it by [L8]; given $ε > 0$ this supplies $δ > 0$ such that points of $Q$ within $δ$ make the two integrands differ by at most $ε$ at every $t$ , and [L9] then bounds the difference of the two integrals by $ε$ . So $\partial_{j} A_{i}$ is continuous on the interior of $Q$ , and as $p$ was arbitrary, $A$ is $C^{1}$ on $U$ and $curl A$ is defined by [F4] and [F5].

2.2step 1.1L4L5F4F5given

Fix $t$ with $0 \leq t \leq 1$ and consider the two fields $x \mapsto B (z_{t})$ and $x \mapsto w = x - a$ on $U$ . For the first, step 1.2 gives $\partial_{j} (B_{k} (z_{t})) = t (\partial_{j} B_{k}) (z_{t})$ , so by [F5] its Jacobian matrix is $t J B (z_{t})$ and by [F4] its divergence is $t \sum_{k < 3} (\partial_{k} B_{k}) (z_{t}) = t (div B) (z_{t}) = 0$ . For the second, $\partial_{j} w_{l}$ is $1$ if $l = j$ and $0$ otherwise, so its Jacobian matrix is the identity and its divergence is $3$ ; both fields are $C^{1}$ since these derivatives are continuous.

3.1step 2.2L1F3algebra

Applying [L1] to those two fields at a fixed $t$ , and multiplying by $t$ , gives ${curl}_{x} (t B (z_{t}) \times w) = t (3 B (z_{t}) - 0 \cdot w + t J B (z_{t}) w - B (z_{t})) = 2 t B (z_{t}) + t^{2} J B (z_{t}) w,$ where by step 2.2 the term $(div G) F$ contributes $3 B (z_{t})$ , the term $- (div F) G$ contributes $0$ , the term $D F G$ contributes $t J B (z_{t}) w$ and the term $- D G F$ contributes $- B (z_{t})$ . At $x = a$ this reads $0 = 0$ in the second and fourth terms, since $w = 0$ there.

4.1step 2.1step 3.1L2

By [F4] each coordinate of $curl A$ is a difference of two of the partial derivatives produced in step 2.1, and each of those is an integral over $[0, 1]$ ; subtracting the two integrals and using step 3.1 for the resulting integrand gives $curl A (x) = \int_{0}^{1} (2 t B (z_{t}) + t^{2} J B (z_{t}) w) d t,$ again coordinatewise.

5.1step 3.1L7L4L5L6L9

For fixed $x$ , put $Γ (t) : = t^{2} B (z_{t})$ on $[0, 1]$ . The map $t \mapsto z_{t}$ is differentiable with derivative $w$ , and $B$ is totally differentiable by [L6], so [L4] and [L5] give $\frac{d}{d t} B (z_{t}) = D B (z_{t}) w = J B (z_{t}) w$ ; with [L7] applied to the product of $t^{2}$ and each coordinate of $B (z_{t})$ this yields $Γ^{'} (t) = 2 t B (z_{t}) + t^{2} J B (z_{t}) w$ , the integrand of step 4.1, which is continuous on $[0, 1]$ and hence integrable by [L9].

6.1step 4.1step 5.1L3

By step 5.1 the function $Γ$ is continuous on $[0, 1]$ and differentiable there, and its derivative is the integrand of step 4.1, so [L3] applied coordinate by coordinate on $[0, 1]$ evaluates that integral as $Γ (1) - Γ (0) = 1^{2} B (z_{1}) - 0^{2} B (z_{0}) = B (x)$ , using $z_{1} = x$ and the factor $t^{2}$ at $t = 0$ .

7.1step 2.1step 6.1F1discharge-construct: the displayed formula∎

Steps 4.1 and 6.1 give $curl A = B$ on $U$ , and step 2.1 gives that $A$ is $C^{1}$ on $U$ ; by [F1] the constructed $A$ is a vector potential for $B$ .

Remarks

Where each hypothesis enters. Star-shapedness is used exactly once, in step 1.1, to know that the segment from the centre to $x$ stays in $U$ so that the integral is defined. The vanishing of $div B$ is used exactly once, in step 2.2, to kill the term $- (div F) G$ ; without it the curl of $A$ would carry an extra term $- t^{2} (div B) (z_{t}) w$ and the integrand would not be an exact derivative in $t$ .
The potential is not unique and the formula is not canonical. Adding the gradient of any $C^{2}$ function leaves the curl unchanged by The curl of the gradient of a $C^{2}$ function vanishes, so the displayed $A$ is one witness among many; it is the one that vanishes at the star centre.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The curl measures the antisymmetric part of the total derivative

Statement

Let $U \subseteq R^{3}$ be open, let $F : U \to R^{3}$ be $C^{1}$ and let $p \in U$ . Then for all $u, v \in R^{3}$ ,

$⟨ D F (p) u, v ⟩ - ⟨ D F (p) v, u ⟩ = ⟨ curl F (p), u \times v ⟩,$

where $D F (p)$ is the total derivative of $F$ at $p$ , whose matrix is the Jacobian matrix $J F (p) = (\partial_{j} F_{i} (p))_{i, j < 3}$ .

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ , the $C^{1}$ field $F : U \to R^{3}$ , the point $p \in U$ and vectors $u, v \in R^{3}$ , with the three coordinates named $x, y, z$ .

[F1]

[F2]

[F3]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{k < m} x_{k} y_{k}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

[L1]

If $f$ is totally differentiable at $a$ then $D_{v} f (a)$ exists for every $v$ and equals $D f (a) v$ ; in particular $\partial_{j} f (a) = D f (a) e_{j}$ , and the matrix of $D f (a)$ is $J f (a)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L2]

If every partial derivative of $f$ exists on a neighbourhood of $a$ and is continuous at $a$ , then $f$ is totally differentiable at $a$ and $D f (a)$ is the linear map with matrix $J f (a)$ (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L3]

The cross product is bilinear and alternating (The cross product is bilinear, alternating, and orthogonal to both factors).

Proof

technique · direct

1.1givenL1L2F4

Since $F$ is $C^{1}$ on $U$ , its partial derivatives exist on $U$ and are continuous, so [L2] makes $F$ totally differentiable at $p$ with $D F (p)$ the linear map of matrix $J F (p)$ ; by [L1] and [F4] the entries of that matrix are $\partial_{j} F_{i} (p)$ , so $(D F (p) u)_{i} = \sum_{j < 3} \partial_{j} F_{i} (p) u_{j}$ .

1.2F3F4algebra

Hence, by [F3] and [F4], $⟨ D F (p) u, v ⟩ - ⟨ D F (p) v, u ⟩ = \sum_{i < 3} \sum_{j < 3} \partial_{j} F_{i} (p) u_{j} v_{i} - \sum_{i < 3} \sum_{j < 3} \partial_{j} F_{i} (p) v_{j} u_{i},$ and exchanging the names of the two summation indices in the second double sum turns it into $\sum_{i < 3} \sum_{j < 3} \partial_{i} F_{j} (p) v_{i} u_{j}$ , so the difference equals $\sum_{i < 3} \sum_{j < 3} (\partial_{j} F_{i} (p) - \partial_{i} F_{j} (p)) u_{j} v_{i}$ .

2.1step 1.2algebra

In the double sum of step 1.2 the terms with $i = j$ have coefficient $\partial_{i} F_{i} (p) - \partial_{i} F_{i} (p) = 0$ , so only the six terms with $i \neq j$ contribute, that is the three unordered index pairs ${y, z}$ , ${z, x}$ and ${x, y}$ , each occurring twice.

3.1step 1.2step 2.1F1algebra

Grouping the two terms of the pair ${y, z}$ gives $(\partial_{y} F_{z} (p) - \partial_{z} F_{y} (p)) u_{y} v_{z} + (\partial_{z} F_{y} (p) - \partial_{y} F_{z} (p)) u_{z} v_{y}$ , that is $(\partial_{y} F_{z} (p) - \partial_{z} F_{y} (p)) (u_{y} v_{z} - u_{z} v_{y})$ . The pair ${z, x}$ gives $(\partial_{z} F_{x} (p) - \partial_{x} F_{z} (p)) (u_{z} v_{x} - u_{x} v_{z})$ and the pair ${x, y}$ gives $(\partial_{x} F_{y} (p) - \partial_{y} F_{x} (p)) (u_{x} v_{y} - u_{y} v_{x})$ .

4.1step 3.1F1F2F3

By [F1] the three coefficients in step 3.1 are the first, second and third coordinates of $curl F (p)$ , and by [F2] the three bracketed factors are the first, second and third coordinates of $u \times v$ . By [F3] their sum is therefore $⟨ curl F (p), u \times v ⟩$ , which with step 1.2 is the asserted identity.

5.1step 4.1L3F2∎

As a check on the signs, take $u = e_{x}$ and $v = e_{y}$ : the left side is $⟨ D F (p) e_{x}, e_{y} ⟩ - ⟨ D F (p) e_{y}, e_{x} ⟩ = \partial_{x} F_{y} (p) - \partial_{y} F_{x} (p)$ and the right side is the third coordinate of $curl F (p)$ , since $e_{x} \times e_{y} = e_{z}$ by [F2]; the pairs $(e_{y}, e_{z})$ and $(e_{z}, e_{x})$ give the first and second coordinates in the same way. When $u = v$ both sides vanish, the left by inspection and the right because the cross product is alternating by [L3].

Remarks

The identity is what makes the curl coordinate-free enough for Stokes. Its left side is built from the total derivative and two vectors, with no reference to a coordinate system beyond the one the inner product carries; the right side reads off the coordinates. That is exactly the form in which the curl enters The curl flux integrand of a $C^{2}$ patch is a two-dimensional curl of the pulled-back field, where $u$ and $v$ are the two parameter derivatives of a patch.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A $C^{1}$ map sends a compact set of content zero to a set of content zero

Statement

Let $m \geq 1$ . Then if $ψ$ is $C^{1}$ on an open $W \subseteq R^{m}$ with values in $R^{m}$ and $A \subseteq W$ is compact with content zero, then $ψ [A]$ is compact and has content zero.

Content zero and nullity are those of Measure zero and content zero in $R^{m}$ by countable and finite cube covers.

Facts & Assumptions

Given: The integer $m \geq 1$ , the open set $W \subseteq R^{m}$ , the $C^{1}$ map $ψ : W \to R^{m}$ , and the compact set $A \subseteq W$ of content zero.

[F1]

A set $E \subseteq R^{m}$ is null when, for every $ε > 0$ , it is covered by a sequence of closed cubes whose nonnegative volume series converges with sum at most $ε$ ; it has content zero when such a cover can be finite (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[F2]

Padding a finite cover with degenerate zero-volume cubes proves that content zero implies null (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[F3]

A map $f : X \to Y$ between metric spaces is Lipschitz with constant $L$ , where $L \in R$ and $L \geq 0$ , when $d_{Y} (f (x), f (x^{'})) \leq L d_{X} (x, x^{'})$ for all $x, x^{'} \in X$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[F4]

A metric space is compact when every open cover of it has a finite subcover (Open cover, subcover, compact metric space, and compact subset of a metric space).

[F5]

A map $f : U \to R^{q}$ is of class $C^{k}$ when each component is of class $C^{k}$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

[F6]

For $x \in R^{m}$ , $∥ x ∥_{2} = \sqrt{\sum_{k < m} x_{k}^{2}}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L1]

Every subset of a null subset of $R^{m}$ is null (Subsets and countable unions of null subsets of $R^{m}$ are null).

[L2]

If $T : R^{m} \to R^{m}$ is Lipschitz and $E$ is null, then $T [E]$ is null (A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets).

[L3]

If $f : [α, β] \to R^{m}$ is continuous and differentiable on $(α, β)$ with $∥ f^{'} (t) ∥_{2} \leq M$ there, then $∥ f (β) - f (α) ∥_{2} \leq M (β - α)$ (The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ).

[L4]

If $f$ is totally differentiable at $a$ and $g$ at $f (a)$ , then $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L5]

If $f$ is totally differentiable at $a$ then $D_{v} f (a)$ exists for every $v \in R^{m}$ and equals $D f (a) v$ , and the matrix of $D f (a)$ is $J f (a)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L6]

If every partial derivative of $f$ exists on a neighbourhood of $a$ and is continuous at $a$ , then $f$ is totally differentiable at $a$ with $D f (a)$ the linear map of matrix $J f (a)$ (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative).

[L7]

For a continuous real-valued $f$ on a nonempty compact metric space, the image $f [X]$ is bounded above and below (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L8]

For continuous $f : X \to Y$ between metric spaces, if $K \subseteq X$ is a compact subset of $X$ , then $f [K]$ is a compact subset of $Y$ (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L9]

A closed box in $R^{m}$ is compact, and a subset $K \subseteq R^{m}$ is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L10]

A compact subset of $R^{m}$ is null if and only if it has content zero (For compact subsets of $R^{m}$ , measure zero and content zero coincide).

Proof

technique · direct

1.1givenF1

If $A = \emptyset$ then $ψ [A] = \emptyset$ , which is covered by the single degenerate cube $\prod_{j < m} [0, 0]$ of volume $0$ , so it has content zero by [F1] and the assertion holds. For the rest of the proof assume $A \neq \emptyset$ .

1.2givenF2

Since $A$ has content zero, [F2] makes $A$ null.

1.3givenF5L6L8

A $C^{1}$ map is continuous, since by [F5] and [L6] each component is totally differentiable and hence continuous at every point of $W$ . So $ψ [A]$ is a compact subset of $R^{m}$ by [L8].

2.1step 1.1givenF4L9

Every point $c \in A$ lies in the open set $W$ , so some closed cube $Q$ centred at $c$ with positive edge is contained in $W$ , and the interior of $Q$ contains $c$ . Those interiors form an open cover of the compact $A$ , so by [F4] and [L9] finitely many of them cover $A$ : there are closed cubes $Q_{1}, \dots, Q_{N} \subseteq W$ with $N \geq 1$ whose union contains $A$ .

3.1step 2.1L5L6L7L9F5F6

Fix $i$ with $1 \leq i \leq N$ . The $m^{2}$ functions $\partial_{j} ψ_{k}$ are continuous on $W$ by [F5], and $Q_{i}$ is a nonempty compact subset of $W$ by [L9], so [L7] bounds each of them on $Q_{i}$ : there is $C \geq 0$ with $∣ \partial_{j} ψ_{k} (c) ∣ \leq C$ for all $c \in Q_{i}$ and all $j, k < m$ . Put $M : = m^{3 / 2} C$ . For $c \in Q_{i}$ and $v \in R^{m}$ , [L5] and [L6] give $D ψ (c) v = J ψ (c) v$ , whose $k$ th coordinate is $\sum_{j < m} \partial_{j} ψ_{k} (c) v_{j}$ , of absolute value at most $m C ∥ v ∥_{2}$ because $∣ v_{j} ∣ \leq ∥ v ∥_{2}$ by [F6]; hence $∥ D ψ (c) v ∥_{2} \leq \sqrt{m} m C ∥ v ∥_{2} = M ∥ v ∥_{2}$ , again by [F6].

4.1step 3.1L3L4L5L6F3F5

Let $x, y \in Q_{i}$ . A cube is convex, so $γ (t) : = x + t (y - x)$ lies in $Q_{i} \subseteq W$ for $0 \leq t \leq 1$ . The map $γ$ is differentiable with $γ^{'} (t) = y - x$ , and $ψ$ is totally differentiable on $W$ by [F5] and [L6], so [L4] and [L5] make $t \mapsto ψ (γ (t))$ differentiable on $[0, 1]$ with derivative $D ψ (γ (t)) (y - x)$ , of norm at most $M ∥ y - x ∥_{2}$ by step 3.1. Hence [L3] on $[0, 1]$ gives $∥ ψ (y) - ψ (x) ∥_{2} \leq M ∥ y - x ∥_{2}$ , so the restriction $ψ ∣_{Q_{i}}$ is Lipschitz with constant $M$ in the sense of [F3].

5.1step 4.1F3F6

Write $Q_{i} = \prod_{j < m} [α_{j}, β_{j}]$ and let $ρ_{i} : R^{m} \to Q_{i}$ be the coordinatewise clamp, $ρ_{i} (v)_{j} = \min {\max {v_{j}, α_{j}}, β_{j}}$ . Each scalar clamp satisfies $∣ \min {\max {s, α}, β} - \min {\max {s^{'}, α}, β} ∣ \leq ∣ s - s^{'} ∣$ , so $∥ ρ_{i} (v) - ρ_{i} (v^{'}) ∥_{2} \leq ∥ v - v^{'} ∥_{2}$ by [F6] and $ρ_{i}$ is Lipschitz with constant $1$ ; therefore $T_{i} : = ψ \circ ρ_{i}$ is defined on all of $R^{m}$ , agrees with $ψ$ on $Q_{i}$ since $ρ_{i}$ fixes $Q_{i}$ pointwise, and is Lipschitz with constant $M$ by step 4.1 and [F3].

6.1step 1.2step 5.1L1L2

For each $i$ , the set $A \cap Q_{i}$ is a subset of the null set $A$ of step 1.2, hence null by [L1]; so [L2] applied to the Lipschitz map $T_{i}$ of step 5.1 makes $T_{i} [A \cap Q_{i}]$ null, and that set is $ψ [A \cap Q_{i}]$ because $T_{i}$ agrees with $ψ$ on $Q_{i}$ .

7.1step 2.1step 6.1F1

By step 2.1 the union of the $Q_{i}$ contains $A$ , so $ψ [A] = ⋃_{i = 1}^{N} ψ [A \cap Q_{i}]$ . Let $ε > 0$ . By step 6.1 and [F1] each of the $N$ sets admits a sequence of closed cubes covering it with volume sum at most $ε / N$ ; concatenating those $N$ sequences gives one sequence of closed cubes covering $ψ [A]$ with volume sum at most $ε$ , so $ψ [A]$ is null by [F1]. The index set is finite, so only finitely many covers are named and no choice principle is used.

8.1step 1.3step 7.1L10∎

The set $ψ [A]$ is compact by step 1.3 and null by step 7.1, so [L10] gives that it has content zero.

Remarks

Why the published Lipschitz theorem is not enough on its own. [L2] is stated for a Lipschitz map defined on all of $R^{m}$ , and $ψ$ is defined only on $W$ and need not be Lipschitz there — its derivative may be unbounded near $\partial W$ . Steps 2.1 to 5.1 exist to manufacture, on each of finitely many cubes, a genuinely global Lipschitz map that agrees with $ψ$ where it matters.
Compactness is used twice, for different things. It supplies the finite subcover in step 2.1, and in step 8.1 it converts nullity back into content zero; a null set need not have content zero without it.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Additivity of the integral over finitely many Jordan pieces that fill a Jordan set up to content zero

Statement

Let $m \geq 1$ , let $A \subseteq R^{m}$ be bounded and Jordan measurable, let $N \geq 1$ , and let $A_{1}, \dots, A_{N} \subseteq A$ be bounded Jordan measurable sets such that $A_{i} \cap A_{j}$ has content zero whenever $i \neq j$ and such that $A ∖ ⋃_{i = 1}^{N} A_{i}$ has content zero. Let $f : A \to R$ be bounded, Riemann integrable over $A$ and Riemann integrable over each $A_{i}$ . Then

$\int_{A} f = \sum_{i = 1}^{N} \int_{A_{i}} f .$

Facts & Assumptions

Given: The sets $A$ and $A_{1}, \dots, A_{N}$ with $N \geq 1$ , the content-zero hypotheses on the pairwise intersections and on the residual set $A ∖ ⋃_{i} A_{i}$ , and the bounded function $f$ integrable over $A$ and over each $A_{i}$ , all as in the Statement.

[F1]

For bounded Jordan measurable $E$ and bounded $f : E \to R$ , choosing a nondegenerate rectangle $Q \supseteq E$ and writing ${\tilde{f}}_{Q}$ for the extension of $f$ by $0$ on $Q ∖ E$ , the function $f$ is Riemann integrable over $E$ when ${\tilde{f}}_{Q}$ is integrable over $Q$ , and then $\int_{E} f = \int_{Q} {\tilde{f}}_{Q}$ (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[F2]

A set has content zero when it can be covered by finitely many closed cubes of arbitrarily small total volume, and both nullity and content zero pass to subsets (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[L1]

The definition of $\int_{E} f$ is independent of the chosen bounding rectangle (The Riemann integral over a Jordan set is independent of the bounding rectangle).

[L2]

For integrable $f, g$ on a nondegenerate rectangle $Q$ and scalars $α, β$ , the function $α f + β g$ is integrable and its integral is $α \int_{Q} f + β \int_{Q} g$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L3]

Let $E$ be bounded and Jordan measurable and let $f, g : E \to R$ be bounded with ${x \in E : f (x) \neq g (x)}$ of content zero. Then $f$ is Riemann integrable over $E$ if and only if $g$ is, and when they are integrable their integrals are equal (Changing a bounded integrand on a content-zero set does not change its Riemann integral).

[L4]

A metric-bounded set $E \subseteq R^{m}$ is Jordan measurable if and only if its boundary $\partial E$ is null, equivalently has content zero (A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero).

[L5]

A continuous graph over a compact nondegenerate rectangle has content zero (The graph of a continuous function on a closed nondegenerate rectangle in $R^{m}$ has content zero in $R^{m + 1}$ ).

Proof

technique · direct

1.1givenF1F2L1L4L5

Fix one nondegenerate rectangle $Q \supseteq A$ ; since each $A_{i} \subseteq A$ , the same $Q$ bounds every one of the $N + 1$ sets. Write $\tilde{f}$ for the zero extension of $f$ from $A$ to $Q$ and ${\tilde{f}}_{i}$ for the zero extension of $f ∣_{A_{i}}$ from $A_{i}$ to $Q$ . By hypothesis and [F1], with [L1] licensing the common choice of $Q$ , all $N + 1$ of these functions are integrable over $Q$ , with $\int_{Q} \tilde{f} = \int_{A} f$ and $\int_{Q} {\tilde{f}}_{i} = \int_{A_{i}} f$ . If $m = 1$ , the boundary of $Q = [u, v]$ is the two-point set ${u, v}$ , and each point has content zero because for every $ε > 0$ it lies in a closed interval of length below $ε$ ; if $m > 1$ , the boundary of $Q$ is the finite union of its coordinate faces, each a continuous graph over a compact nondegenerate rectangle, so [L5] makes every face content zero. Thus $\partial Q$ has content zero by [F2] in every dimension, and therefore $Q$ is Jordan measurable by [L4].

2.1step 1.1L2given

Put $g : = \sum_{i = 1}^{N} {\tilde{f}}_{i} - \tilde{f}$ on $Q$ . By [L2] it is integrable over $Q$ , being a finite linear combination of the integrable functions of step 1.1. It is bounded as well: if $A = \emptyset$ then every zero extension and hence $g$ is identically zero, while if $A \neq \emptyset$ the boundedness of $f$ supplies a real $M \geq 0$ with $∣ f (x) ∣ \leq M$ on $A$ , and then $∣ g ∣ \leq (N + 1) M$ on $Q$ .

3.1step 2.1givenalgebra

Let $S : = (⋃_{i \neq j} (A_{i} \cap A_{j})) \cup (A ∖ ⋃_{i = 1}^{N} A_{i})$ and let $x \in Q ∖ S$ . If $x \notin A$ then $\tilde{f} (x) = 0$ and every ${\tilde{f}}_{i} (x) = 0$ , because $A_{i} \subseteq A$ , so $g (x) = 0$ . If $x \in A$ then $x$ lies in some $A_{i}$ , since otherwise it would lie in the residual set, and in exactly one, since otherwise it would lie in one of the pairwise intersections; hence $\sum_{i} {\tilde{f}}_{i} (x) = f (x) = \tilde{f} (x)$ and again $g (x) = 0$ . So ${x \in Q : g (x) \neq 0} \subseteq S$ .

4.1step 3.1F2

The set $S$ is the union of the $N (N - 1)$ pairwise intersections and the residual set, each of content zero by hypothesis. Given $ε > 0$ , cover each of those finitely many sets by finitely many closed cubes of total volume at most $ε / (N (N - 1) + 1)$ and take all of those cubes together: this is a finite cover of $S$ by closed cubes of total volume at most $ε$ , so $S$ has content zero by [F2], and so does its subset ${x \in Q : g (x) \neq 0}$ .

5.1step 2.1step 4.1L3

By step 1.1 the set $Q$ is bounded and Jordan measurable and $g$ is bounded on it, and by step 4.1 the set where $g$ differs from the zero function has content zero; so [L3] applies with the zero function and gives $\int_{Q} g = 0$ .

6.1step 5.1L2F1∎

Expanding $\int_{Q} g$ by [L2] and using step 1.1, $0 = \sum_{i = 1}^{N} \int_{Q} {\tilde{f}}_{i} - \int_{Q} \tilde{f} = \sum_{i = 1}^{N} \int_{A_{i}} f - \int_{A} f$ , which is the asserted identity. For $N = 1$ there is no pairwise intersection and $S$ is the residual set alone; the hypothesis $N \geq 1$ excludes the empty index set, for which the right-hand side would be $0$ while the left need not be.

Remarks

An individual piece may be empty. Nothing above requires $A_{i} \neq \emptyset$ : an empty piece contributes the integral $0$ and creates no exceptional point, so the hypothesis constrains only the overlaps and the residue.
Why integrability over each piece is stated explicitly. For Jordan measurable $A_{i} \subseteq A$ , this integrability follows from the other hypotheses by restricting the zero extension of $f$ to the integrable indicator of $A_{i}$ . The proof records it as a hypothesis because step 1.1 starts from the piece integrals, rather than inserting that standard product argument into the additivity calculation.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Change of variables for a $C^{1}$ map injective and regular only on the interior of a compact Jordan set

Statement

Let $n \geq 1$ , let $W \subseteq R^{n}$ be open, let $ψ : W \to R^{n}$ be $C^{1}$ , and let $D \subseteq W$ be compact and Jordan measurable. Suppose $ψ$ is injective on the interior of $D$ and has nonvanishing Jacobian determinant there, and put $V : = ψ [D^{\circ}]$ . Then

$ψ [D]$ is compact and Jordan measurable, $V$ is bounded, open and Jordan measurable, and $ψ [D] ∖ V$ has content zero;
for every continuous $h : ψ [D] \to R$ the three integrals below exist and

$\int_{D} h (ψ (x)) ∣ \det D ψ (x) ∣ d x = \int_{V} h (y) d y = \int_{ψ [D]} h (y) d y .$

No injectivity and no invertibility of the derivative is assumed at any point of $\partial D$ .

Facts & Assumptions

Given: The data of the Statement: $W$ , $ψ$ , the compact Jordan set $D \subseteq W$ , the injectivity and nonvanishing Jacobian determinant of $ψ$ on $D^{\circ}$ , the set $V = ψ [D^{\circ}]$ , and a continuous $h : ψ [D] \to R$ .

[F1]

The boundary of $A$ is $\partial A : = \overline{A} ∖ int (A)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[F2]

A set has content zero when it can be covered by finitely many closed cubes of arbitrarily small total volume, and content zero passes to subsets (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[F3]

For a $C^{1}$ map $g$ of an open subset of $R^{n}$ into $R^{n}$ , its Jacobian determinant is $\det D g (x)$ (The Jacobian determinant of a square-dimensional $C^{1}$ map is the determinant of its Jacobian matrix).

[F4]

For bounded Jordan measurable $E$ , bounded $f : E \to R$ and a nondegenerate rectangle $Q \supseteq E$ , the function $f$ is Riemann integrable over $E$ when its zero extension ${\tilde{f}}_{Q}$ is integrable over $Q$ , and then $\int_{E} f = \int_{Q} {\tilde{f}}_{Q}$ (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

If $f : U \to R^{n}$ is $C^{1}$ on an open $U$ and $D f (a)$ is invertible, then there are open sets $V^{'}, W^{'}$ with $a \in V^{'} \subseteq U$ and $f (a) \in W^{'}$ such that $f ∣_{V^{'}} : V^{'} \to W^{'}$ is bijective, and its inverse is $C^{1}$ (The Euclidean inverse function theorem).

[L2]

[L3]

If $ψ$ is $C^{1}$ on an open $W \subseteq R^{m}$ with values in $R^{m}$ and $A \subseteq W$ is compact with content zero, then $ψ [A]$ is compact and has content zero (A $C^{1}$ map sends a compact set of content zero to a set of content zero).

[L4]

For a bounded, open, Jordan measurable $V^{'} \subseteq R^{n}$ there are compact Jordan sets $K_{1} \subseteq K_{2} \subseteq \dots \subseteq V^{'}$ , each a finite union of closed grid rectangles, such that every compact $C \subseteq V^{'}$ lies in some $K_{j}$ and $cont (V^{'} ∖ K_{j}) \to 0$ (A bounded open Jordan set has an increasing exhaustion by compact finite unions of grid rectangles with vanishing content remainder).

[L5]

Let $U \subseteq R^{n}$ be open, let $g : U \to R^{n}$ be injective and $C^{1}$ with $D g (x)$ invertible for every $x \in U$ , and let $K \subseteq U$ be compact and Jordan measurable. For bounded $f : g (K) \to R$ , integrability of $f$ on $g (K)$ is equivalent to integrability of $x \mapsto f (g (x)) ∣ \det D g (x) ∣$ on $K$ , and when either holds $\int_{g (K)} f (y) d y = \int_{K} f (g (x)) ∣ \det D g (x) ∣ d x$ (Change of variables for an injective $C^{1}$ map on a compact Jordan set).

[L6]

Under the hypotheses of [L5], if $K \subseteq U$ is compact and Jordan measurable then $g (K)$ is compact and Jordan measurable (An injective $C^{1}$ map with invertible derivative sends compact Jordan sets to compact Jordan sets).

[L7]

For integrable $f, g$ on a nondegenerate rectangle $Q$ and scalars $α, β$ : $α f + β g$ is integrable with integral $α \int_{Q} f + β \int_{Q} g$ ; if $f \leq g$ then $\int_{Q} f \leq \int_{Q} g$ ; and $∣ f ∣$ is integrable with $∣ \int_{Q} f ∣ \leq \int_{Q} ∣ f ∣$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L8]

Every continuous real function on a compact Jordan measurable set $E \subseteq R^{m}$ is Riemann integrable over $E$ (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

[L9]

If bounded Jordan measurable $E, F$ have $E \cap F$ of content zero, then $cont (E \cup F) = cont (E) + cont (F)$ (Jordan content is finitely additive when the overlap has content zero).

[L10]

For continuous $f : X \to Y$ between metric spaces, the image of a compact subset of $X$ is a compact subset of $Y$ (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L11]

A metric-bounded $E \subseteq R^{m}$ is Jordan measurable if and only if its indicator $1_{E}$ is Riemann integrable on a fixed nondegenerate bounding rectangle $Q$ , and then $\int_{Q} 1_{E} = cont (E)$ (A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content).

[L12]

Let $E$ be bounded and Jordan measurable and let $f, g : E \to R$ be bounded with ${x \in E : f (x) \neq g (x)}$ of content zero. Then $f$ is integrable over $E$ if and only if $g$ is, and their integrals then agree (Changing a bounded integrand on a content-zero set does not change its Riemann integral).

[L13]

A closed box in $R^{m}$ is compact, and a subset of $R^{m}$ is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L14]

For every real square matrix $A$ , $\det (A) \neq 0$ if and only if $A$ is invertible (A finite square real matrix is invertible if and only if its determinant is nonzero).

Proof

technique · direct

1.1givenF1L2L7L8L11

Suppose first $D^{\circ} = \emptyset$ . Then $V = \emptyset$ and, by [F1], $D = \partial D$ , which has content zero by [L2]; so $cont (D) = 0$ by [L11], the parameter integrand is continuous on the compact Jordan $D$ and hence integrable by [L8], and [L7] with [L11] bounds its integral in absolute value by $\sup_{D} ∣ h (ψ) \det D ψ ∣ \cdot cont (D) = 0$ . All three integrals are then $0$ and both assertions hold. Assume $D^{\circ} \neq \emptyset$ for the rest of the proof.

1.2givenF1L2L3L13

The set $\partial D$ is a closed subset of the compact $D$ by [F1], hence compact by [L13], and it has content zero by [L2] since $D$ is Jordan measurable. So [L3] gives that $ψ [\partial D]$ is compact and has content zero.

2.1step 1.2F1F2L2

By [F1] the interior $D^{\circ}$ is open and bounded, and $\partial (D^{\circ}) = \overline{D^{\circ}} ∖ D^{\circ} \subseteq D ∖ D^{\circ} = \partial D$ because $\overline{D^{\circ}} \subseteq \overline{D} = D$ . So $\partial (D^{\circ})$ has content zero by step 1.2 and [F2], and $D^{\circ}$ is Jordan measurable by [L2].

2.2step 1.1givenF3L1L14

On $D^{\circ}$ the map $ψ$ is injective and $\det D ψ \neq 0$ , so [F3] and [L14] make each $D ψ (c)$ invertible, and then [L1] makes $ψ$ carry an open neighbourhood of each $c \in D^{\circ}$ onto an open set. Hence $ψ [D^{\circ}] = V$ is open, and $ψ ∣_{D^{\circ}} : D^{\circ} \to V$ is a bijection whose inverse is $C^{1}$ , in particular continuous, on $V$ .

3.1step 1.2step 2.2F1F2L2L10L13

By [L10] the set $ψ [D]$ is compact, hence closed and bounded by [L13]. Since $D = D^{\circ} \cup \partial D$ by [F1], $ψ [D] = V \cup ψ [\partial D]$ , so $ψ [D] ∖ V \subseteq ψ [\partial D]$ has content zero by step 1.2 and [F2]. As $V$ is open with $V \subseteq ψ [D]$ , we get $\partial V = \overline{V} ∖ V \subseteq ψ [D] ∖ V$ and $\partial (ψ [D]) = ψ [D] ∖ int (ψ [D]) \subseteq ψ [D] ∖ V$ ; both therefore have content zero, and [L2] makes $V$ and $ψ [D]$ Jordan measurable.

3.2step 2.1step 2.2L4L5L6L8

Apply [L4] to the bounded open Jordan set $D^{\circ}$ of step 2.1, obtaining compact Jordan sets $K_{1} \subseteq K_{2} \subseteq \dots \subseteq D^{\circ}$ with every compact subset of $D^{\circ}$ contained in some $K_{j}$ and $cont (D^{\circ} ∖ K_{j}) \to 0$ . By step 2.2 the hypotheses of [L5] hold with $U = D^{\circ}$ and $g = ψ ∣_{D^{\circ}}$ , so for each $j$ the set $ψ [K_{j}]$ is compact and Jordan measurable by [L6] and $\int_{ψ [K_{j}]} h = \int_{K_{j}} h (ψ (x)) ∣ \det D ψ (x) ∣ d x,$ both integrals existing because $h$ is continuous on the compact Jordan $ψ [K_{j}]$ , hence integrable there by [L8].

4.1step 3.1F4L8L12

The set $ψ [D]$ is compact and Jordan measurable by step 3.1 and $h$ is continuous on it, so [L8] makes $h$ integrable over $ψ [D]$ and, $ψ [D]$ being compact, $∣ h ∣ \leq M$ there for some $M \geq 0$ . Fix a nondegenerate rectangle $Q \supseteq ψ [D]$ . The zero extensions of $h ∣_{V}$ and of $h$ from $ψ [D]$ differ only on $ψ [D] ∖ V$ , which has content zero by step 3.1, so [L12] applied on $Q$ makes the first integrable too, with $\int_{V} h = \int_{ψ [D]} h$ by [F4].

4.2step 1.2step 3.2F1L2L7L8L9L11

The map $x \mapsto h (ψ (x)) ∣ \det D ψ (x) ∣$ is continuous on the compact Jordan $D$ , hence integrable over $D$ and over each compact Jordan $K_{j}$ by [L8], and bounded there by some $M^{'} \geq 0$ . Fix a nondegenerate rectangle $Q^{'} \supseteq D$ . Because $\partial (D ∖ K_{j}) \subseteq \partial D \cup \partial K_{j}$ — a point outside both boundaries lies either in $int K_{j}$ , whose neighbourhood misses $D ∖ K_{j}$ , or outside $\overline{K_{j}}$ and inside $int D$ , whose neighbourhood lies in $D ∖ K_{j}$ — the set $D ∖ K_{j}$ is Jordan measurable by [L2] and [F1]. The two zero extensions differ only on $D ∖ K_{j}$ and by at most $M^{'}$ , so [L7] and [L11] give $∣ \int_{D} h (ψ) ∣ \det D ψ ∣ - \int_{K_{j}} h (ψ) ∣ \det D ψ ∣ ∣ \leq M^{'} cont (D ∖ K_{j}) .$ Now $D ∖ K_{j} = \partial D \cup (D^{\circ} ∖ K_{j})$ is a union of two disjoint Jordan sets, $\partial D$ having content zero by step 1.2, so [L9] gives $cont (D ∖ K_{j}) = cont (D^{\circ} ∖ K_{j})$ , which tends to $0$ by step 3.2. Hence those integrals converge to $\int_{D} h (ψ) ∣ \det D ψ ∣$ .

5.1step 2.2step 3.1step 3.2L4L7L10L11

Apply [L4] to the bounded open Jordan set $V$ of step 3.1, obtaining compact Jordan $C_{1} \subseteq C_{2} \subseteq \dots \subseteq V$ with $cont (V ∖ C_{l}) \to 0$ and every compact subset of $V$ inside some $C_{l}$ . Fix $l$ . By step 2.2 the inverse of $ψ ∣_{D^{\circ}}$ is continuous, so $(ψ ∣_{D^{\circ}})^{- 1} [C_{l}]$ is a compact subset of $D^{\circ}$ by [L10], and step 3.2 puts it inside some $K_{j (l)}$ ; applying $ψ$ gives $C_{l} \subseteq ψ [K_{j (l)}]$ and hence $V ∖ ψ [K_{j}] \subseteq V ∖ C_{l}$ for every $j \geq j (l)$ , the sets $K_{j}$ being increasing. Both sets are Jordan measurable by step 3.1, step 3.2 and the boundary inclusion of step 4.2, so [L7] and [L11] give $cont (V ∖ ψ [K_{j}]) \leq cont (V ∖ C_{l})$ ; letting $l$ grow, $cont (V ∖ ψ [K_{j}]) \to 0$ .

6.1step 4.1step 5.1L7L11

With $M$ and $Q$ as in step 4.1, the zero extensions of $h ∣_{V}$ and of $h ∣_{ψ [K_{j}]}$ differ only on $V ∖ ψ [K_{j}]$ and by at most $M$ , so [L7] and [L11] give $∣ \int_{V} h - \int_{ψ [K_{j}]} h ∣ \leq M cont (V ∖ ψ [K_{j}])$ , which tends to $0$ by step 5.1. Hence $\int_{ψ [K_{j}]} h \to \int_{V} h$ .

7.1step 4.2step 6.1∎

By step 3.2 the two sequences of integrals agree term by term; by step 4.2 the parameter side converges to $\int_{D} h (ψ) ∣ \det D ψ ∣$ and by step 6.1 the image side converges to $\int_{V} h$ , so those two numbers are equal, and step 4.1 identifies $\int_{V} h$ with $\int_{ψ [D]} h$ . With step 3.1 this is both assertions of the Statement.

Remarks

What the published compact theorem cannot do here. [L5] requires the derivative to be invertible at every point of an open set containing the compact domain. A spherical octant, parametrized by polar angle and azimuth, has vanishing projected Jacobian determinant along the parameter boundary, so no such open set exists and [L5] does not apply to it. Everything above is the work of pushing the degeneracy into $\partial D$ , where [L3] makes its image negligible.
The conclusion is about the open image, and that is not a defect. The set $ψ [D]$ may fold its boundary onto itself, and no injectivity is assumed there; what the identity says is that the fold contributes nothing, because $ψ [D] ∖ V$ has content zero.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

A cyclic permutation of the coordinates of $R^{3}$ preserves Jordan measurability and integrals

Statement

For $k \in {x, y, z}$ let $σ_{k} : R^{3} \to R^{3}$ be given by

$σ_{x} (p) = (p_{y}, p_{z}, p_{x}), σ_{y} (p) = (p_{z}, p_{x}, p_{y}), σ_{z} (p) = (p_{x}, p_{y}, p_{z}) .$

Each $σ_{k}$ is a linear bijection with $\det D σ_{k} = 1$ everywhere. Let $E \subseteq R^{3}$ be compact and Jordan measurable. Then $σ_{k} [E]$ is compact and Jordan measurable, and for every bounded $H : σ_{k} [E] \to R$ the function $H$ is Riemann integrable over $σ_{k} [E]$ if and only if $H \circ σ_{k}$ is Riemann integrable over $E$ ; when either holds, the integral over the permuted set equals the integral of the composite with the permutation over the original set,

$\int_{σ_{k} [E]} H = \int_{E} H \circ σ_{k} .$

Facts & Assumptions

Given: The index $k \in {x, y, z}$ , the map $σ_{k}$ displayed in the Statement, the compact Jordan measurable set $E \subseteq R^{3}$ and the bounded function $H$ on $σ_{k} [E]$ .

[F1]

For a commutative ring $R$ , $n \geq 1$ and $A = (a_{i j}) \in M_{n} (R)$ , $\det A = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ , with columns indexed by $i < n$ and rows by $σ (i)$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F2]

An inversion of $σ \in S_{n}$ is a pair $(i, j)$ with $i < j < n$ and $σ (i) > σ (j)$ , and $sgn (σ) = (- 1)^{inv (σ)}$ , where $inv (σ)$ is the number of inversions (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

[F3]

For a $C^{1}$ map $g$ of an open subset of $R^{n}$ into $R^{n}$ , the Jacobian determinant is $\det D g (x)$ , and the change-of-variables scale factor is $∣ \det D g (x) ∣$ (The Jacobian determinant of a square-dimensional $C^{1}$ map is the determinant of its Jacobian matrix).

[F4]

If every partial derivative $\partial_{j} f_{i} (a)$ exists, the Jacobian matrix is $J f (a) = (\partial_{j} f_{i} (a))_{i < n, j < m}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F5]

Integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

[L2]

Under those hypotheses, if $K \subseteq U$ is compact and Jordan measurable then $g (K)$ is compact and Jordan measurable (An injective $C^{1}$ map with invertible derivative sends compact Jordan sets to compact Jordan sets).

Proof

technique · direct

1.1givenF4

Each $σ_{k}$ is linear: writing coordinates as indices $0, 1, 2$ for $x, y, z$ , the map $σ_{x}$ sends $p$ to the point with coordinates $(p_{1}, p_{2}, p_{0})$ , so by [F4] its partial derivatives are the constants $\partial_{j} (σ_{x})_{i}$ , and its Jacobian matrix $A_{x}$ at every point has $(A_{x})_{i j} = 1$ exactly for $(i, j) \in {(0, 1), (1, 2), (2, 0)}$ and $0$ elsewhere. Likewise $σ_{y}$ sends $p$ to $(p_{2}, p_{0}, p_{1})$ , with matrix $A_{y}$ having entry $1$ exactly at $(0, 2), (1, 0), (2, 1)$ , and $σ_{z}$ is the identity with matrix the identity matrix. All three matrices have exactly one entry $1$ in each row and in each column, so each $σ_{k}$ is a bijection of $R^{3}$ with $D σ_{k}$ constant and invertible.

1.2F1F2algebra

In the Leibniz sum [F1] for $\det A_{x}$ , a term is nonzero only when $(A_{x})_{σ (i), i} = 1$ for every $i < 3$ , that is when $σ (0) = 2$ , $σ (1) = 0$ and $σ (2) = 1$ ; exactly one permutation does this. Its inversions are $(0, 1)$ , since $2 > 0$ , and $(0, 2)$ , since $2 > 1$ , while $(1, 2)$ is not one, since $0 < 1$ ; so $inv (σ) = 2$ and $sgn (σ) = + 1$ by [F2], giving $\det A_{x} = 1$ .

1.3F1F2algebra

In the Leibniz sum for $\det A_{y}$ , the only nonzero term has $σ (0) = 1$ , $σ (1) = 2$ and $σ (2) = 0$ ; its inversions are $(0, 2)$ , since $1 > 0$ , and $(1, 2)$ , since $2 > 0$ , while $(0, 1)$ is not one, since $1 < 2$ ; so again $inv (σ) = 2$ and $\det A_{y} = 1$ by [F1] and [F2]. For $A_{z}$ the identity matrix, the only nonzero term is the identity permutation, with no inversion, so $\det A_{z} = 1$ .

2.1step 1.1step 1.2step 1.3F3L2

By steps 1.1, 1.2 and 1.3 each $σ_{k}$ is a $C^{1}$ injection of the open set $R^{3}$ into $R^{3}$ whose derivative is invertible at every point, with $\det D σ_{k} = 1$ and hence $∣ \det D σ_{k} ∣ = 1$ by [F3]. So [L2] applies with $U = R^{3}$ , $g = σ_{k}$ and $K = E$ , and $σ_{k} [E]$ is compact and Jordan measurable.

3.1step 2.1L1F3F5∎

With the same data, [L1] gives that $H$ is integrable over $σ_{k} [E]$ if and only if $x \mapsto H (σ_{k} (x)) ∣ \det D σ_{k} (x) ∣ = H (σ_{k} (x))$ is integrable over $E$ , that is if and only if $H \circ σ_{k}$ is, and that in that case $\int_{σ_{k} [E]} H = \int_{E} H (σ_{k} (x)) \cdot 1 d x = \int_{E} H \circ σ_{k}$ , the integrals being those of [F5].

Remarks

Why the cyclic order and not the increasing one. The three maps above send the coordinate $k$ to the last slot and keep the other two in the cyclic order $x \to y \to z \to x$ . Taking instead the two surviving coordinates in increasing order would transpose them in the case $k = y$ , and a transposition has one inversion and hence determinant $- 1$ ; every identity on this page that treats the three directions alike depends on the cyclic choice.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Simple solid regions in a coordinate direction and their cyclic coordinate projection

Definition

Coordinates on $R^{3}$ are named $x, y, z$ for the indices $0, 1, 2$ of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ . For each $k \in {x, y, z}$ the cyclic coordinate projection $π_{k} : R^{3} \to R^{2}$ drops the $k$ th coordinate and keeps the other two in cyclic order:

$π_{x} (p) = (p_{y}, p_{z}), π_{y} (p) = (p_{z}, p_{x}), π_{z} (p) = (p_{x}, p_{y}) .$

A simple description of a solid in the direction $k$ is a quadruple $(k, D, γ_{1}, γ_{2})$ in which $D \subseteq R^{2}$ is compact, Jordan measurable and has nonempty interior, and $γ_{1}, γ_{2} : D \to R$ are continuous with $γ_{1} \leq γ_{2}$ on $D$ and $γ_{1} < γ_{2}$ on the interior of $D$ . The simple solid region it describes is

$E = {p \in R^{3} : π_{k} (p) \in D, γ_{1} (π_{k} (p)) \leq p_{k} \leq γ_{2} (π_{k} (p))} .$

The set $D$ is the base, $γ_{2}$ the upper graph function and $γ_{1}$ the lower graph function of the description. A solid is simple in the direction $k$ when some such description of it is supplied; the description is part of the data and is not inferred from the set $E$ .

Writing $σ_{k}$ for the cyclic permutation of A cyclic permutation of the coordinates of $R^{3}$ preserves Jordan measurability and integrals, so that $σ_{k} (p) = (π_{k} (p), p_{k})$ , the image $σ_{k} [E]$ is exactly the solid between the graphs of $γ_{1}$ and $γ_{2}$ over the base $D$ in the sense of A solid between continuous graphs over a compact Jordan base. That set is compact and Jordan measurable by A solid between continuous graphs over a compact Jordan base is Jordan measurable and integrates by vertical sections, and $σ_{k}^{- 1}$ is again a cyclic coordinate permutation, so $E$ is compact and Jordan measurable as well; integration over $E$ is that of The Riemann integral of a bounded function over a bounded Jordan measurable set, and interiors, closures and boundaries are those of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space.

Remarks

Weak inequality on the base, strict inside. The graphs are allowed to meet on $\partial D$ , so a vertical section of $E$ over a boundary point of the base may be a single point; that is what lets a ball be described in every direction, since its two hemispherical graph functions agree exactly on the equatorial circle. The strictness on the interior of $D$ is what makes the interior of $E$ nonempty and is used where the outward normal is identified.
The cyclic order is not cosmetic. With $π_{y} (p) = (p_{z}, p_{x})$ rather than $(p_{x}, p_{z})$ , each $σ_{k}$ has determinant $1$ and each coordinate of an oriented area vector is the Jacobian determinant of the matching projection; taking the surviving coordinates in increasing order would reverse both signs in the case $k = y$ and no statement on this page would hold uniformly in $k$ .
Nonempty interior of the base. A base with empty interior need not make $E$ a graph: a line-segment base with $γ_{1} < γ_{2}$ produces a vertical rectangle. It does, however, make $E$ three-dimensionally content zero and makes the strictness condition on the interior vacuous. Requiring nonempty interior keeps every simple solid region a genuine solid. The boundary of $D$ has content zero by A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero, which is what makes the base the closure of its interior up to a negligible set in the arguments that follow.

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Each coordinate of the oriented area vector is the Jacobian determinant of the matching cyclic projection

Statement

Let $O \subseteq R^{2}$ be open and let $φ = (φ_{x}, φ_{y}, φ_{z}) : O \to R^{3}$ be $C^{1}$ , with parameters named $u, v$ and $φ_{u} : = \partial_{u} φ$ , $φ_{v} : = \partial_{v} φ$ . Then at every point of $O$ ,

$(φ_{u} \times φ_{v})_{k} = \det D (π_{k} \circ φ)$

for each of the three coordinate directions $k \in {x, y, z}$ , where $π_{k}$ is the cyclic coordinate projection of Simple solid regions in a coordinate direction and their cyclic coordinate projection.

Facts & Assumptions

Given: The open set $O \subseteq R^{2}$ and the $C^{1}$ map $φ : O \to R^{3}$ of the Statement.

[F1]

[F2]

For a $C^{1}$ map $g$ of an open subset of $R^{n}$ into $R^{n}$ , its Jacobian determinant is $\det D g (x)$ , the determinant of its Jacobian matrix (The Jacobian determinant of a square-dimensional $C^{1}$ map is the determinant of its Jacobian matrix).

[F3]

[F4]

For a commutative ring $R$ , $n \geq 1$ and $A = (a_{i j}) \in M_{n} (R)$ , $\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ , with columns indexed by $i < n$ and rows by $σ (i)$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F5]

An inversion of $σ \in S_{n}$ is a pair $(i, j)$ with $i < j < n$ and $σ (i) > σ (j)$ , and $sgn (σ) = (- 1)^{inv (σ)}$ (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

[F6]

The cyclic coordinate projections are $π_{x} (p) = (p_{y}, p_{z})$ , $π_{y} (p) = (p_{z}, p_{x})$ and $π_{z} (p) = (p_{x}, p_{y})$ (Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F7]

A map $f : U \to R^{q}$ is of class $C^{k}$ when each component is of class $C^{k}$ ( $C^{k}$ Euclidean maps and diffeomorphisms).

Proof

technique · direct

1.1givenF2F3F4F5F7

Each $π_{k} \circ φ$ is a map of the two parameters into $R^{2}$ whose two components are components of $φ$ , hence $C^{1}$ by [F7], so by [F2] and [F3] it has a Jacobian matrix $(\begin{array}{cc} a_{00} & a_{01} \\ a_{10} & a_{11} \end{array})$ with $a_{i 0} = \partial_{u}$ and $a_{i 1} = \partial_{v}$ of its $i$ th component. There are exactly two elements of $S_{2}$ : the identity, with no inversion and sign $+ 1$ , contributing $a_{00} a_{11}$ , and the transposition sending $0$ to $1$ and $1$ to $0$ , with the single inversion $(0, 1)$ and sign $- 1$ , contributing $- a_{10} a_{01}$ . So [F4] and [F5] give $\det D (π_{k} \circ φ) = a_{00} a_{11} - a_{10} a_{01}$ .

1.2givenF1F3

By [F1] with $u = φ_{u}$ and $v = φ_{v}$ , whose coordinates are the partial derivatives named in [F3], the oriented area vector has coordinates $(φ_{u} \times φ_{v})_{x} = \partial_{u} φ_{y} \partial_{v} φ_{z} - \partial_{u} φ_{z} \partial_{v} φ_{y},$ $(φ_{u} \times φ_{v})_{y} = \partial_{u} φ_{z} \partial_{v} φ_{x} - \partial_{u} φ_{x} \partial_{v} φ_{z},$ $(φ_{u} \times φ_{v})_{z} = \partial_{u} φ_{x} \partial_{v} φ_{y} - \partial_{u} φ_{y} \partial_{v} φ_{x} .$

2.1step 1.1step 1.2F6

By [F6] the projection $π_{x}$ retains the coordinates $y$ then $z$ , so $π_{x} \circ φ = (φ_{y}, φ_{z})$ and step 1.1 gives $\det D (π_{x} \circ φ) = \partial_{u} φ_{y} \partial_{v} φ_{z} - \partial_{u} φ_{z} \partial_{v} φ_{y}$ , which is the first coordinate computed in step 1.2.

2.2step 1.1step 1.2F6

By [F6] the projection $π_{y}$ retains the coordinates $z$ then $x$ , in that cyclic order, so $π_{y} \circ φ = (φ_{z}, φ_{x})$ and step 1.1 gives $\det D (π_{y} \circ φ) = \partial_{u} φ_{z} \partial_{v} φ_{x} - \partial_{u} φ_{x} \partial_{v} φ_{z}$ , the second coordinate computed in step 1.2. Retaining $x$ then $z$ in increasing order instead would exchange the two rows and give the opposite sign, which is why the cyclic order is part of the projection.

2.3step 1.1step 1.2F6

By [F6] the projection $π_{z}$ retains the coordinates $x$ then $y$ , so $π_{z} \circ φ = (φ_{x}, φ_{y})$ and step 1.1 gives $\det D (π_{z} \circ φ) = \partial_{u} φ_{x} \partial_{v} φ_{y} - \partial_{u} φ_{y} \partial_{v} φ_{x}$ , the third coordinate computed in step 1.2.

3.1step 2.1step 2.2step 2.3∎

Steps 2.1, 2.2 and 2.3 are the three asserted identities, valid at every point of $O$ ; in particular all three determinants vanish exactly where the oriented area vector does.

Remarks

The identity holds where the patch is not regular. Nothing above uses $φ_{u} \times φ_{v} \neq 0$ . That matters because the lateral faces of a boundary presentation are exactly the patches whose $k$ th projected Jacobian determinant vanishes, and the identity is what turns that analytic condition into a geometric one.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The outward unit normal at a boundary point of a compact solid

Definition

Let $E \subseteq R^{3}$ be compact and let $p \in \partial E$ , the boundary of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space. For a unit vector $ν \in R^{3}$ , that is one with $∥ ν ∥_{2} = 1$ in the norm of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , a unit vector $ν$ is outward at $p$ when there is a real $ε > 0$ with $p + t ν \notin E$ and $p - t ν \in E$ for every $t$ with $0 < t < ε$ .

A plane of unit normals at $p$ is a two-dimensional linear subspace $T \subseteq R^{3}$ ; the two unit vectors orthogonal to $T$ are $\pm ν$ for a single $ν$ , and when one of them is outward at $p$ the other is not, since replacing $ν$ by $- ν$ exchanges the two displayed conditions. In that situation the outward one is called the outward unit normal to $T$ at $p$ .

Remarks

Outwardness alone does not single out one vector. Take $E$ the closed unit ball and $p$ a point of the unit sphere. Every unit vector $ν$ with $⟨ p, ν ⟩ > 0$ satisfies the definition, because $∥ p \pm t ν ∥_{2}^{2} = 1 \pm 2 t ⟨ p, ν ⟩ + t^{2}$ is above $1$ for small $t > 0$ with the plus sign and below $1$ with the minus sign. So the definition is a condition on a unit vector and not a construction of one; what makes "the outward unit normal" a definite object is the second paragraph, where a plane is supplied and only two candidates remain.
Existence is not asserted. A boundary point of an arbitrary compact set need admit no outward unit vector: if $E = {p}$ is a singleton, then $p - t ν \notin E$ for every unit vector $ν$ and every $t > 0$ . Nothing below claims outwardness at seams and edges; the claim is made at the interior parameter points of a graph face whose projection lands in the interior of the base.
Why the condition is one-sided on each side. Requiring only $p + t ν \notin E$ would admit a vector tangent to a spike of $E$ ; requiring only $p - t ν \in E$ would admit a vector pointing along the surface. Both halves are used where outwardness is proved.

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Boundary presentations adapted to a simple solid region in a coordinate direction

Definition

Let $(k, D, γ_{1}, γ_{2})$ be a simple description of a solid $E \subseteq R^{3}$ in the direction $k$ (Simple solid regions in a coordinate direction and their cyclic coordinate projection), and write

$Γ_{2} : = {p : π_{k} (p) \in D, p_{k} = γ_{2} (π_{k} (p))}, Γ_{1} : = {p : π_{k} (p) \in D, p_{k} = γ_{1} (π_{k} (p))}$

for the upper and lower graph of the description. Let $Σ = ((D_{1}, φ_{1}), \dots, (D_{P}, φ_{P}))$ be a compatible finite patch presentation in the sense of Finitely patched regular surfaces, their area, scalar integrals, and flux, each $(D_{j}, φ_{j})$ a regular parametrized surface patch of Regular parametrized surface patches on compact Jordan parameter regions, whose patch images cover $\partial E$ and are contained in $\partial E$ . Write $φ_{j, u}, φ_{j, v}$ for the two parameter derivatives, and $φ_{j, u} \times φ_{j, v}$ for the oriented area vector of Unit normal fields, orientations, and flux through a regular surface patch, whose coordinates are those of The cross product in $R^{3}$ .

The presentation $Σ$ is adapted to the description $(k, D, γ_{1}, γ_{2})$ when the index set ${1, \dots, P}$ is partitioned into three sublists $Σ^{+}$ , $Σ^{-}$ and $Σ^{0}$ , supplied with the presentation, such that all of the following hold.

Upper faces. For $j \in Σ^{+}$ , the image of $φ_{j}$ lies in the graph of $γ_{2}$ and the $k$ th coordinate of $φ_{j, u} \times φ_{j, v}$ is positive on the interior of $D_{j}$ .
Lower faces. For $j \in Σ^{-}$ , the image of $φ_{j}$ lies in $Γ_{1}$ and the $k$ th coordinate of $φ_{j, u} \times φ_{j, v}$ is negative on the interior of $D_{j}$ .
Lateral faces. For $j \in Σ^{0}$ , the $k$ th coordinate of $φ_{j, u} \times φ_{j, v}$ vanishes on the interior of $D_{j}$ .
The graph faces cover the base. Writing $V_{j} : = π_{k} [φ_{j} [D_{j}^{\circ}]]$ for the projected image of the $j$ th patch, the projected images of the upper sublist are pairwise disjoint and fill $D$ up to content zero, and the same holds for the lower sublist: for each of $Σ^{+}$ and $Σ^{-}$ the sets $V_{j}$ with $j$ in that sublist are pairwise disjoint and $D ∖ ⋃ V_{j}$ has content zero in the sense of Measure zero and content zero in $R^{m}$ by countable and finite cube covers.
Both graph sublists are nonempty. $Σ^{+} \neq \emptyset$ and $Σ^{-} \neq \emptyset$ ; the lateral sublist $Σ^{0}$ may be empty.

The partition into the three sublists is part of the supplied data, exactly as the description $(k, D, γ_{1}, γ_{2})$ is; nothing here is inferred from the set $E$ or from the unordered collection of patch images. Interiors are those of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space and integrals over the projected images are those of The Riemann integral of a bounded function over a bounded Jordan measurable set.

Remarks

The conditions are on the sign of one coordinate, not on outwardness. Clauses 1 to 3 are analytic: by Each coordinate of the oriented area vector is the Jacobian determinant of the matching cyclic projection the $k$ th coordinate of the oriented area vector is the Jacobian determinant of $π_{k} \circ φ_{j}$ , so clause 1 says that the projection of an upper patch is orientation-preserving on the parameter interior and clause 3 says that a lateral patch projects with vanishing Jacobian determinant. That the induced normals of the graph faces then point out of $E$ is a theorem, At interior base points, the graph faces of an adapted presentation induce the outward unit normal, rather than part of this definition.
A graph face is not required to be a graph patch. Clause 1 asks only that the patch image lie in $Γ_{2}$ ; it does not ask that $φ_{j}$ be the map $w \mapsto (w, γ_{2} (w))$ read in the projected coordinates. That is what admits the eight spherical octants and the four quarter-cylinders: their graph functions have unbounded gradient at the equator or at the silhouette, so they are not $C^{1}$ on a neighbourhood of the closed base and could not parametrize a patch, while the octants and quarters themselves are patches in every direction at once.
Why the lateral condition is imposed on the interior. The parameter region of a patch is the closure of its interior, and the $k$ th coordinate of the oriented area vector is continuous on the whole region, so a vanishing condition on the interior already forces vanishing everywhere on the region. Stating it on the interior keeps the three clauses in the same form and matches where clauses 1 and 2 can be stated at all, since the oriented area vector may vanish on a parameter boundary.
What clause 4 is for. It is the only clause that ties the presentation to the base quantitatively: without it, one tiny upper patch in $Γ_{2}$ together with one tiny lower patch in $Γ_{1}$ could satisfy clauses 1, 2, 3 and 5 while covering almost none of either graph. Pairwise disjointness and the content-zero residue are what make the sum of the graph-face fluxes an integral over the whole of $D$ .

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The flux of a single-component field through a graph face is a base integral of its trace

Statement

Let $(k, D, γ_{1}, γ_{2})$ be a simple description of a solid $E$ in the direction $k$ and let $Σ$ be a boundary presentation adapted to it, with sublists $Σ^{+}, Σ^{-}, Σ^{0}$ (Boundary presentations adapted to a simple solid region in a coordinate direction). Let $R : E \to R$ be continuous and let $R e_{k}$ be the field on $E$ whose $k$ th coordinate is $R$ and whose other two coordinates are zero. For $j \in Σ^{+} \cup Σ^{-}$ put $ψ_{j} : = π_{k} \circ φ_{j}$ and $V_{j} : = ψ_{j} [D_{j}^{\circ}]$ , and write $σ_{k}^{- 1} (w, t)$ for the point of $R^{3}$ with $π_{k}$ -projection $w$ and $k$ th coordinate $t$ .

Then each $V_{j}$ is a bounded open Jordan measurable subset of $D$ , the displayed base integrand is integrable over $V_{j}$ , and the flux of $R e_{k}$ through an upper face is the integral of the trace of $R$ on the upper graph over the projected image, and through a lower face it is the negative of the corresponding integral:

$\int_{D_{j}} ⟨ R e_{k} (φ_{j}), φ_{j, u} \times φ_{j, v} ⟩ = \int_{V_{j}} R (σ_{k}^{- 1} (w, γ_{2} (w))) d w (j \in Σ^{+}),$

$\int_{D_{j}} ⟨ R e_{k} (φ_{j}), φ_{j, u} \times φ_{j, v} ⟩ = - \int_{V_{j}} R (σ_{k}^{- 1} (w, γ_{1} (w))) d w (j \in Σ^{-}) .$

Facts & Assumptions

Given: The simple description $(k, D, γ_{1}, γ_{2})$ of $E$ , the adapted presentation $Σ$ with its supplied sublists, the continuous $R : E \to R$ , and an index $j$ in $Σ^{+}$ or in $Σ^{-}$ .

[F1]

For a regular patch $(D^{'}, φ)$ and a continuous vector field $F$ , the flux in the orientation induced by $φ$ is $\int_{D^{'}} (F \circ φ) \cdot (φ_{u} \times φ_{v})$ (Unit normal fields, orientations, and flux through a regular surface patch).

[F2]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ , and the standard unit vector $e_{k}$ has $k$ th coordinate $1$ and the others $0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[F3]

A simple solid region in the direction $k$ is $E = {p \in R^{3} : π_{k} (p) \in D, γ_{1} (π_{k} (p)) \leq p_{k} \leq γ_{2} (π_{k} (p))}$ , with $π_{k}$ the cyclic coordinate projection and $γ_{1}, γ_{2}$ continuous on the compact Jordan base $D$ (Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F4]

For $j \in Σ^{+}$ the image of $φ_{j}$ lies in the graph of $γ_{2}$ and the $k$ th coordinate of $φ_{j, u} \times φ_{j, v}$ is positive on the interior of $D_{j}$ ; for $j \in Σ^{-}$ the image lies in the graph of $γ_{1}$ and that coordinate is negative on the interior of $D_{j}$ (Boundary presentations adapted to a simple solid region in a coordinate direction).

[F5]

A regular patch has a compact Jordan parameter region that is the closure of its nonempty interior, its parametrization is $C^{1}$ on an open neighbourhood of that region, and no point of $D^{\circ}$ has the same image as a distinct point of $D$ (Regular parametrized surface patches on compact Jordan parameter regions).

[F6]

Integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

For a $C^{1}$ map $φ$ of two variables into $R^{3}$ , $(φ_{u} \times φ_{v})_{k} = \det D (π_{k} \circ φ)$ for each of the three coordinate directions (Each coordinate of the oriented area vector is the Jacobian determinant of the matching cyclic projection).

[L2]

Let $ψ$ be $C^{1}$ on an open $W \supseteq D^{'}$ with $D^{'}$ compact Jordan, and suppose $ψ$ is injective on the interior of $D^{'}$ and has nonvanishing Jacobian determinant there. Then $ψ [D^{' \circ}]$ is bounded, open and Jordan measurable, and for continuous $h$ on $ψ [D^{'}]$ , $\int_{D^{'}} h (ψ (w)) ∣ \det D ψ (w) ∣ d w = \int_{ψ [D^{' \circ}]} h$ (Change of variables for a $C^{1}$ map injective and regular only on the interior of a compact Jordan set).

[L3]

Let $E^{'}$ be bounded Jordan measurable and let $f, g : E^{'} \to R$ be bounded with ${x \in E^{'} : f (x) \neq g (x)}$ of content zero. Then $f$ is integrable over $E^{'}$ if and only if $g$ is, and their integrals then agree (Changing a bounded integrand on a content-zero set does not change its Riemann integral).

[L4]

A metric-bounded set is Jordan measurable if and only if its boundary has content zero (A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero).

[L5]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

Proof

technique · direct

1.1givenF1F2

By [F2] the inner product of $R e_{k} (p)$ with any vector $ν$ is $R (p) ν_{k}$ , so by [F1] the flux integrand of $R e_{k}$ through the patch $(D_{j}, φ_{j})$ is $w \mapsto R (φ_{j} (w)) (φ_{j, u} \times φ_{j, v})_{k} (w)$ on $D_{j}$ .

1.2givenF3F4

Suppose $j \in Σ^{+}$ and put $h (w) : = R (σ_{k}^{- 1} (w, γ_{2} (w)))$ for $w \in D$ ; by [F3] the point $σ_{k}^{- 1} (w, γ_{2} (w))$ lies in $E$ and $h$ is continuous on $D$ , being $R$ composed with a continuous map. By [F4] the image of $φ_{j}$ lies in the graph of $γ_{2}$ , so for $w \in D_{j}$ the point $φ_{j} (w)$ has $π_{k}$ -projection $ψ_{j} (w) \in D$ and $k$ th coordinate $γ_{2} (ψ_{j} (w))$ ; hence $φ_{j} (w) = σ_{k}^{- 1} (ψ_{j} (w), γ_{2} (ψ_{j} (w)))$ and $R (φ_{j} (w)) = h (ψ_{j} (w))$ . For $j \in Σ^{-}$ the same computation with $γ_{1}$ in place of $γ_{2}$ defines a continuous $h$ on $D$ with $R (φ_{j} (w)) = h (ψ_{j} (w))$ .

2.1step 1.1L1F5

By [L1] the factor $(φ_{j, u} \times φ_{j, v})_{k}$ in step 1.1 is $\det D ψ_{j}$ , so the flux integrand is $w \mapsto R (φ_{j} (w)) \det D ψ_{j} (w)$ . The map $ψ_{j}$ is $C^{1}$ on an open neighbourhood of $D_{j}$ by [F5], since $φ_{j}$ is and $π_{k}$ is linear.

2.2step 1.2F5

The map $ψ_{j}$ is injective on $D_{j}^{\circ}$ . Indeed let $a, b \in D_{j}^{\circ}$ with $ψ_{j} (a) = ψ_{j} (b)$ . By step 1.2 both $φ_{j} (a)$ and $φ_{j} (b)$ are determined by their common $π_{k}$ -projection through the same graph function, so $φ_{j} (a) = φ_{j} (b)$ ; by [F5] no point of $D_{j}^{\circ}$ shares its image with a distinct point of $D_{j}$ , so $a = b$ .

3.1step 2.1F4

By [F4] and step 2.1, $\det D ψ_{j}$ is positive on $D_{j}^{\circ}$ when $j \in Σ^{+}$ and negative there when $j \in Σ^{-}$ ; in either case it is nonvanishing on $D_{j}^{\circ}$ .

4.1step 2.2step 3.1L2L5F5F6

By [F5] the parameter region $D_{j}$ is compact and Jordan measurable, so steps 2.2 and 3.1 put the data $(ψ_{j}, D_{j})$ under the hypotheses of [L2]. Hence $V_{j} = ψ_{j} [D_{j}^{\circ}]$ is bounded, open and Jordan measurable, it is contained in $D$ by step 1.2, and with the continuous $h$ of step 1.2 $\int_{D_{j}} h (ψ_{j} (w)) ∣ \det D ψ_{j} (w) ∣ d w = \int_{V_{j}} h .$ Both sides exist, the left by [L5] on the compact Jordan $D_{j}$ and the right as part of [L2], with integrals read as in [F6].

5.1step 4.1L3L4L5F5

On $D_{j}^{\circ}$ the two functions $∣ \det D ψ_{j} ∣$ and $ε_{j} \det D ψ_{j}$ coincide, where $ε_{j} = + 1$ for $j \in Σ^{+}$ and $ε_{j} = - 1$ for $j \in Σ^{-}$ , by step 3.1. They can differ only on $\partial D_{j}$ , which has content zero by [F5] and [L4]; both are continuous on the compact Jordan $D_{j}$ , hence bounded and integrable by [L5], so multiplying each by the bounded continuous $h \circ ψ_{j}$ and applying [L3] on $D_{j}$ gives $\int_{D_{j}} h (ψ_{j}) ∣ \det D ψ_{j} ∣ = ε_{j} \int_{D_{j}} h (ψ_{j}) \det D ψ_{j} .$

6.1step 1.1step 1.2step 2.1step 4.1step 5.1

Let $j \in Σ^{+}$ , so $ε_{j} = + 1$ . Combining steps 1.1, 1.2 and 2.1 the flux integral is $\int_{D_{j}} h (ψ_{j}) \det D ψ_{j}$ , which by step 5.1 equals $\int_{D_{j}} h (ψ_{j}) ∣ \det D ψ_{j} ∣$ and by step 4.1 equals $\int_{V_{j}} h = \int_{V_{j}} R (σ_{k}^{- 1} (w, γ_{2} (w))) d w$ . That is the first asserted identity.

7.1step 1.1step 1.2step 2.1step 4.1step 5.1∎

Let $j \in Σ^{-}$ , so $ε_{j} = - 1$ and $h (w) = R (σ_{k}^{- 1} (w, γ_{1} (w)))$ . Steps 1.1, 1.2 and 2.1 again make the flux integral $\int_{D_{j}} h (ψ_{j}) \det D ψ_{j}$ , and step 5.1 now reads $\int_{D_{j}} h (ψ_{j}) ∣ \det D ψ_{j} ∣ = - \int_{D_{j}} h (ψ_{j}) \det D ψ_{j}$ , so the flux integral is $- \int_{D_{j}} h (ψ_{j}) ∣ \det D ψ_{j} ∣$ , which by step 4.1 is $- \int_{V_{j}} R (σ_{k}^{- 1} (w, γ_{1} (w))) d w$ . This is the second asserted identity, and the sign comes from that replacement of the absolute determinant and from nothing else.

Remarks

Injectivity of the projection is forced, not assumed. Step 2.2 uses only that the patch image lies in a graph over the base: two interior parameter points with the same projection are then carried to the same point of $R^{3}$ , which the patch definition forbids. Nothing in the adapted-presentation conditions had to say it.
Where the absolute value is paid for. Change of variables produces $∣ \det D ψ_{j} ∣$ , while the flux integrand carries $\det D ψ_{j}$ with its sign. Step 5.1 is the whole difference between the two faces of a solid: the upper one contributes with a plus sign and the lower one with a minus, and that is what makes the two contributions add to an increment of $R$ across the solid rather than cancel.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The single-direction flux identity on a simple solid region

Statement

Let $(k, D, γ_{1}, γ_{2})$ be a simple description of a solid $E$ in the direction $k$ and let $Σ = ((D_{1}, φ_{1}), \dots, (D_{P}, φ_{P}))$ be a boundary presentation adapted to it (Boundary presentations adapted to a simple solid region in a coordinate direction). Let $R$ be a real function of class $C^{1}$ on an open set containing $E$ and let $R e_{k}$ be the field whose $k$ th coordinate is $R$ and whose other two coordinates are zero. Then the flux of $R e_{k}$ over the presentation equals the integral of the $k$ th partial derivative of $R$ over $E$ :

$\sum_{j = 1}^{P} \int_{D_{j}} ⟨ R e_{k} (φ_{j}), φ_{j, u} \times φ_{j, v} ⟩ = \int_{E} \partial_{k} R .$

Facts & Assumptions

Given: The simple description $(k, D, γ_{1}, γ_{2})$ of $E$ , the adapted presentation $Σ$ with its supplied sublists $Σ^{+}, Σ^{-}, Σ^{0}$ , and the function $R$ of class $C^{1}$ on an open $O \supseteq E$ . Write $ψ_{j} = π_{k} \circ φ_{j}$ , $V_{j} = ψ_{j} [D_{j}^{\circ}]$ , $σ_{k}^{- 1} (w, t)$ for the point with $π_{k}$ -projection $w$ and $k$ th coordinate $t$ , and $R^{γ_{i}} (w) : = R (σ_{k}^{- 1} (w, γ_{i} (w)))$ for $i = 1, 2$ .

[F1]

For a compatible finite patch presentation, the oriented flux is the sum of the patch values, each patch value being $\int_{D_{j}} (F \circ φ_{j}) \cdot (φ_{j, u} \times φ_{j, v})$ (Finitely patched regular surfaces, their area, scalar integrals, and flux, Unit normal fields, orientations, and flux through a regular surface patch).

[F2]

For $j \in Σ^{0}$ the $k$ th coordinate of $φ_{j, u} \times φ_{j, v}$ vanishes on the interior of $D_{j}$ ; the projected images of the upper sublist are pairwise disjoint and fill $D$ up to content zero, and the same holds for the lower sublist (Boundary presentations adapted to a simple solid region in a coordinate direction).

[F3]

A regular patch has a compact Jordan parameter region that is the closure of its nonempty interior, and its parametrization is $C^{1}$ on an open neighbourhood of that region (Regular parametrized surface patches on compact Jordan parameter regions).

[F4]

The simple solid region described by $(k, D, γ_{1}, γ_{2})$ is $E = {p \in R^{3} : π_{k} (p) \in D, γ_{1} (π_{k} (p)) \leq p_{k} \leq γ_{2} (π_{k} (p))}$ with $D$ compact Jordan and $γ_{1} \leq γ_{2}$ continuous on $D$ , and $σ_{k} (p) = (π_{k} (p), p_{k})$ carries $E$ onto the solid between the graphs of $γ_{1}$ and $γ_{2}$ over $D$ (Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F5]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ); a $C^{1}$ function has continuous first partial derivatives ( $C^{k}$ Euclidean maps and diffeomorphisms); and $\partial_{k}$ is the $k$ th partial derivative appearing in the divergence of Divergence and curl of a $C^{1}$ vector field.

[F6]

Integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

For $j \in Σ^{+}$ the flux of $R e_{k}$ through $(D_{j}, φ_{j})$ is $\int_{V_{j}} R^{γ_{2}}$ , and for $j \in Σ^{-}$ it is $- \int_{V_{j}} R^{γ_{1}}$ ; each $V_{j}$ is a bounded open Jordan measurable subset of $D$ and the base integrand is integrable over it (The flux of a single-component field through a graph face is a base integral of its trace).

[L2]

Let $A$ be bounded Jordan measurable, let $N \geq 1$ and let $A_{1}, \dots, A_{N} \subseteq A$ be bounded Jordan sets with pairwise intersections of content zero and with $A ∖ ⋃_{i} A_{i}$ of content zero; if $f$ is bounded on $A$ and integrable over $A$ and over each $A_{i}$ , then $\int_{A} f = \sum_{i = 1}^{N} \int_{A_{i}} f$ (Additivity of the integral over finitely many Jordan pieces that fill a Jordan set up to content zero).

[L3]

For compact Jordan $D \subseteq R^{m}$ , continuous $α \leq β$ on $D$ and $K = {(u, t) : u \in D, α (u) \leq t \leq β (u)}$ , the solid $K$ is compact and Jordan measurable and every continuous $H : K \to R$ satisfies $\int_{K} H = \int_{D} (\int_{α (u)}^{β (u)} H (u, t) d t) d u$ (A solid between continuous graphs over a compact Jordan base is Jordan measurable and integrates by vertical sections).

[L4]

For a cyclic coordinate permutation $σ_{k}$ of $R^{3}$ and compact Jordan $E^{'}$ , the set $σ_{k} [E^{'}]$ is compact Jordan and $\int_{σ_{k} [E^{'}]} H = \int_{E^{'}} H \circ σ_{k}$ for bounded $H$ integrable on either side (A cyclic permutation of the coordinates of $R^{3}$ preserves Jordan measurability and integrals).

[L5]

If $G$ is differentiable at every point of $[a, b]$ with $a < b$ and $G^{'}$ is integrable on $[a, b]$ , then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[L6]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

[L7]

For a $C^{1}$ map $φ$ of two variables into $R^{3}$ , $(φ_{u} \times φ_{v})_{k} = \det D (π_{k} \circ φ)$ (Each coordinate of the oriented area vector is the Jacobian determinant of the matching cyclic projection).

Proof

technique · direct

1.1givenF2F3F5L7

Let $j \in Σ^{0}$ . By [F5] the flux integrand of $R e_{k}$ through $(D_{j}, φ_{j})$ is $R (φ_{j}) (φ_{j, u} \times φ_{j, v})_{k}$ , which is continuous on $D_{j}$ because $φ_{j}$ is $C^{1}$ there by [F3] and $R$ is continuous. By [F2] its second factor vanishes on $D_{j}^{\circ}$ , and $D_{j}$ is the closure of $D_{j}^{\circ}$ by [F3], so a continuous function vanishing on $D_{j}^{\circ}$ vanishes on $D_{j}$ . Hence that patch's flux is $\int_{D_{j}} 0 = 0$ .

1.2givenF4F5L3L4L6

By [F4] the set $E$ is compact and Jordan measurable and $σ_{k} [E] = K : = {(w, t) : w \in D, γ_{1} (w) \leq t \leq γ_{2} (w)}$ . The function $H (w, t) : = (\partial_{k} R) (σ_{k}^{- 1} (w, t))$ is continuous on $K$ , since $\partial_{k} R$ is continuous on $O$ by [F5] and $σ_{k}^{- 1}$ is linear, and $H \circ σ_{k} = \partial_{k} R$ on $E$ . So [L4] applied with $E^{'} = E$ gives $\int_{E} \partial_{k} R = \int_{K} H$ , both integrals existing by [L3] and [L6].

1.3givenF4F5L5

Fix $w \in D$ and put $G (t) : = R (σ_{k}^{- 1} (w, t))$ , defined and differentiable for every $t$ with $σ_{k}^{- 1} (w, t) \in O$ , with $G^{'} (t) = (\partial_{k} R) (σ_{k}^{- 1} (w, t)) = H (w, t)$ because varying $t$ moves only the $k$ th coordinate. If $γ_{1} (w) < γ_{2} (w)$ then $G$ is differentiable on $[γ_{1} (w), γ_{2} (w)]$ , whose points lie in $E \subseteq O$ by [F4], and $G^{'}$ is continuous there hence integrable, so [L5] gives $\int_{γ_{1} (w)}^{γ_{2} (w)} H (w, t) d t = G (γ_{2} (w)) - G (γ_{1} (w)) = R^{γ_{2}} (w) - R^{γ_{1}} (w)$ . If instead $γ_{1} (w) = γ_{2} (w)$ then the interval is degenerate, so the integral is $0$ , and the increment $R^{γ_{2}} (w) - R^{γ_{1}} (w)$ is also $0$ ; the identity holds in that case too.

1.4givenF2F6L1L2L6

The functions $R^{γ_{1}}$ and $R^{γ_{2}}$ are continuous on the compact Jordan base $D$ , hence bounded and integrable over $D$ by [L6] and [F6]. By [L1] each $V_{j}$ with $j \in Σ^{+}$ is a bounded Jordan subset of $D$ over which $R^{γ_{2}}$ is integrable, and by [F2] those sets are pairwise disjoint — so their pairwise intersections are empty and have content zero — and their union omits from $D$ only a set of content zero. So [L2] gives $\sum_{j \in Σ^{+}} \int_{V_{j}} R^{γ_{2}} = \int_{D} R^{γ_{2}}$ , and by [L1] the left side is the sum of the upper faces' fluxes.

1.5givenF2F6L1L2L6

The same argument applied to the lower sublist gives $\sum_{j \in Σ^{-}} \int_{V_{j}} R^{γ_{1}} = \int_{D} R^{γ_{1}}$ , and by [L1] each lower face's flux is $- \int_{V_{j}} R^{γ_{1}}$ , so the lower faces' fluxes sum to $- \int_{D} R^{γ_{1}}$ .

2.1step 1.2step 1.3L3L6

By step 1.3 the inner integral in [L3] is $R^{γ_{2}} (w) - R^{γ_{1}} (w)$ for every $w \in D$ , a continuous function of $w$ ; so [L3] applied to $H$ on $K$ and step 1.2 give $\int_{E} \partial_{k} R = \int_{K} H = \int_{D} (R^{γ_{2}} - R^{γ_{1}}) = \int_{D} R^{γ_{2}} - \int_{D} R^{γ_{1}}$ , the last step by linearity of the integral over $D$ .

2.2step 1.1step 1.4step 1.5F1

By [F1] the flux over the presentation is the sum of the $P$ patch fluxes, which splits along the three supplied sublists. Step 1.1 makes the lateral sum zero, step 1.4 makes the upper sum $\int_{D} R^{γ_{2}}$ and step 1.5 makes the lower sum $- \int_{D} R^{γ_{1}}$ , so the total is $\int_{D} R^{γ_{2}} - \int_{D} R^{γ_{1}}$ .

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 give the same number for the two sides of the asserted identity, so it holds.

Remarks

The degenerate slice needs its own line. Where the two graph functions agree, the second fundamental theorem is unavailable, since it requires a nondegenerate interval; step 1.3 handles that case separately, and both sides are zero there. Folding it into the main computation would apply [L5] with $a = b$ .
Nothing here says the normals point outward. The identity is proved from the sign conditions of the adapted presentation alone. Reading its right-hand side as an outward flux is At interior base points, the graph faces of an adapted presentation induce the outward unit normal and Every patch of an elementary solid region's presentation is a graph face in some direction, and at interior base points its normal is outward, and no step above depends on them.

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

At interior base points, the graph faces of an adapted presentation induce the outward unit normal

Statement

Let $(k, D, γ_{1}, γ_{2})$ be a simple description of a solid $E$ in the direction $k$ and let $Σ$ be a boundary presentation adapted to it. Let $j \in Σ^{+} \cup Σ^{-}$ , let $c$ be an interior point of the parameter region $D_{j}$ whose projection $w_{0} : = π_{k} (φ_{j} (c))$ lies in the interior of the base $D$ , and put $p : = φ_{j} (c)$ .

Then $p \in \partial E$ , the tangent plane $T : = span {φ_{j, u} (c), φ_{j, v} (c)}$ is defined, and the induced unit normal of an upper or lower face is the outward unit normal: the vector

$N : = \frac{φ_{j, u} (c) \times φ_{j, v} (c)}{∥ φ_{j, u} (c) \times φ_{j, v} (c) ∥_{2}}$

is outward at $p$ in the sense of The outward unit normal at a boundary point of a compact solid, while $- N$ is not; so $N$ is the outward unit normal to $T$ at $p$ .

The displayed interior condition makes explicit the part of the base on which the strict graph separation is used below.

Facts & Assumptions

Given: The simple description $(k, D, γ_{1}, γ_{2})$ of $E$ , the adapted presentation $Σ$ , the index $j \in Σ^{+} \cup Σ^{-}$ , the interior parameter point $c \in D_{j}^{\circ}$ with $w_{0} = π_{k} (φ_{j} (c)) \in D^{\circ}$ , and $p = φ_{j} (c)$ . Write $ψ_{j} = π_{k} \circ φ_{j}$ , write $γ$ for $γ_{2}$ when $j \in Σ^{+}$ and for $γ_{1}$ when $j \in Σ^{-}$ , and write $σ_{k}^{- 1} (w, t)$ for the point with $π_{k}$ -projection $w$ and $k$ th coordinate $t$ .

[F1]

$E = {q \in R^{3} : π_{k} (q) \in D, γ_{1} (π_{k} (q)) \leq q_{k} \leq γ_{2} (π_{k} (q))}$ , with $D$ compact Jordan of nonempty interior, $γ_{1}, γ_{2}$ continuous on $D$ , $γ_{1} \leq γ_{2}$ on $D$ and $γ_{1} < γ_{2}$ on the interior of $D$ (Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F2]

[F3]

A regular patch has a compact Jordan parameter region that is the closure of its nonempty interior, its parametrization is $C^{1}$ on an open neighbourhood of that region, and $φ_{u} \times φ_{v} \neq 0$ on the interior (Regular parametrized surface patches on compact Jordan parameter regions).

[F4]

At an interior parameter point the tangent plane of a regular patch is $span {φ_{u}, φ_{v}}$ , a two-dimensional subspace of $R^{3}$ (The tangent plane of a regular surface patch).

[F5]

The parametrization induces on the interior the unit normal $N_{φ} = (φ_{u} \times φ_{v}) / ∥ φ_{u} \times φ_{v} ∥_{2}$ , which is orthogonal to the tangent plane (Unit normal fields, orientations, and flux through a regular surface patch).

[F6]

A unit vector $ν$ is outward at $p \in \partial E$ when there is a real $ε > 0$ with $p + t ν \notin E$ and $p - t ν \in E$ for every $t$ with $0 < t < ε$ ; when a two-dimensional subspace $T$ is given and one of its two unit normals is outward at $p$ , the other is not, and the outward one is called the outward unit normal to $T$ at $p$ (The outward unit normal at a boundary point of a compact solid).

[F7]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ); the gradient of a scalar function is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case); a map is $C^{k}$ when each component is ( $C^{k}$ Euclidean maps and diffeomorphisms); and the Jacobian determinant of a square-dimensional $C^{1}$ map is $\det D g$ (The Jacobian determinant of a square-dimensional $C^{1}$ map is the determinant of its Jacobian matrix).

[F8]

The boundary of $A$ is $\partial A = \overline{A} ∖ int (A)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space), and $f^{'} (c)$ is the limit of the difference quotient at $c$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L1]

If $f : U \to R^{n}$ is $C^{1}$ on an open $U$ and $D f (a)$ is invertible, then there are open $V^{'}, W^{'}$ with $a \in V^{'} \subseteq U$ and $f (a) \in W^{'}$ such that $f ∣_{V^{'}} : V^{'} \to W^{'}$ is bijective with $C^{1}$ inverse (The Euclidean inverse function theorem).

[L2]

[L3]

If $f$ is totally differentiable at $a$ then $D_{v} f (a)$ exists for every $v$ and equals $D f (a) v$ , and the matrix of $D f (a)$ is $J f (a)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L4]

[L5]

[L6]

If $\lim_{x \to c} f (x) = L > 0$ then there is $δ > 0$ with $f (x) > L / 2 > 0$ for every $x$ in the domain with $0 < ∣ x - c ∣ < δ$ ; if $L < 0$ then $f (x) < L / 2 < 0$ there (If $\lim_{x \to c} f (x) = L \neq 0$ then $∣ f ∣ > ∣ L ∣ / 2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L / 2 > 0$ there).

Proof

technique · direct

1.1givenF2F3F7L1L2

By [F3] the map $φ_{j}$ is $C^{1}$ on an open neighbourhood of $D_{j}$ , so $ψ_{j} = π_{k} \circ φ_{j}$ is $C^{1}$ there by [F7], and by [F2] and [L2] its Jacobian determinant at $c$ is nonzero, hence $D ψ_{j} (c)$ is invertible. So [L1] supplies open sets $P ∋ c$ and $Q ∋ w_{0}$ with $ψ_{j} ∣_{P} : P \to Q$ bijective and with $C^{1}$ inverse $λ : Q \to P$ ; shrinking $P$ and $Q$ , which stays possible because $D_{j}^{\circ}$ and $D^{\circ}$ are open and contain $c$ and $w_{0}$ , we may take $P \subseteq D_{j}^{\circ}$ and $Q \subseteq D^{\circ}$ .

2.1step 1.1F2F7L3L4L5

Let $w \in Q$ . By [F2] the point $φ_{j} (λ (w))$ lies in the graph of $γ$ over $D$ , and its $π_{k}$ -projection is $ψ_{j} (λ (w)) = w$ , so $φ_{j} (λ (w)) = σ_{k}^{- 1} (w, γ (w))$ . Reading the $k$ th coordinate, $γ (w) = (φ_{j} (λ (w)))_{k}$ on $Q$ , a composite of $C^{1}$ maps and therefore $C^{1}$ on $Q$ by [F7] and [L4]; by [L5] it is totally differentiable at $w_{0}$ , and by [L3] and [F7] its total derivative there acts by $v \mapsto ⟨ g, v ⟩$ with $g : = \nabla γ (w_{0})$ .

3.1step 1.1step 2.1F4L3L4

The map $Φ (w) : = σ_{k}^{- 1} (w, γ (w))$ on $Q$ equals $φ_{j} \circ λ$ by step 2.1, and its two parameter derivatives at $w_{0}$ are $τ_{i} = σ_{k}^{- 1} (e_{i}, \partial_{i} γ (w_{0}))$ for $i = 0, 1$ . By [L4] the derivative $D Φ (w_{0}) = D φ_{j} (c) \circ D λ (w_{0})$ with $D λ (w_{0})$ invertible, so $span {τ_{0}, τ_{1}}$ and $span {φ_{j, u} (c), φ_{j, v} (c)}$ are the same subspace, namely the tangent plane $T$ of [F4].

4.1step 3.1F2F3F5F7L2

By [F3] and [F5] the vector $N$ is defined at $c$ , has norm $1$ and is orthogonal to $T$ . Write $N = σ_{k}^{- 1} (a, b)$ with $a \in R^{2}$ and $b = N_{k}$ ; since $σ_{k}^{- 1}$ merely permutes coordinates, [F7] gives $⟨ σ_{k}^{- 1} (a, b), σ_{k}^{- 1} (a^{'}, b^{'}) ⟩ = ⟨ a, a^{'} ⟩ + b b^{'}$ . Orthogonality to $τ_{i}$ of step 3.1 therefore reads $a_{i} + b \partial_{i} γ (w_{0}) = 0$ for $i = 0, 1$ , that is $a = - b g$ . If $b$ were $0$ then $a = 0$ and $N = 0$ , contradicting $∥ N ∥_{2} = 1$ ; so $b \neq 0$ , and by [F2] and [L2] the number $b$ has the sign of $\det D ψ_{j} (c)$ , hence $b > 0$ for $j \in Σ^{+}$ and $b < 0$ for $j \in Σ^{-}$ .

5.1step 2.1step 4.1F8L3L6

For real $t$ near $0$ the projection $π_{k} (p + t N) = w_{0} + t a$ lies in the open $Q$ , so $u (t) : = (p + t N)_{k} - γ (π_{k} (p + t N)) = γ (w_{0}) + t b - γ (w_{0} + t a)$ is defined there, using $p_{k} = γ (w_{0})$ from step 2.1. Then $u (0) = 0$ , and by step 2.1 and [L3] the function $u$ is differentiable at $0$ with $u^{'} (0) = b - ⟨ g, a ⟩ = b + b ∥ g ∥_{2}^{2} = b (1 + ∥ g ∥_{2}^{2})$ , using $a = - b g$ from step 4.1. Since $u (0) = 0$ , the difference quotient at $0$ is $u (t) / t$ , so [F8] and [L6] give $ε_{0} > 0$ such that $u (t) / t$ has the sign of $b$ for every $t$ with $0 < ∣ t ∣ < ε_{0}$ ; hence $u (t)$ has the sign of $t b$ there.

6.1step 5.1F1F6

Suppose $j \in Σ^{+}$ , so $γ = γ_{2}$ and $b > 0$ by step 4.1. Shrink $ε_{0}$ so that $π_{k} (p \pm t N) \in Q \subseteq D$ for $0 < t < ε_{0}$ and so that, $γ_{1}$ and $γ_{2}$ being continuous with $γ_{1} (w_{0}) < γ_{2} (w_{0})$ by [F1], one also has $γ_{1} (π_{k} (p - t N)) < γ_{2} (w_{0}) - t b$ there. For $0 < t < ε_{0}$ , step 5.1 gives $u (t) > 0$ , that is $(p + t N)_{k} > γ_{2} (π_{k} (p + t N))$ , so $p + t N \notin E$ by [F1]; and $u (- t) < 0$ , that is $(p - t N)_{k} < γ_{2} (π_{k} (p - t N))$ , while $(p - t N)_{k} = γ_{2} (w_{0}) - t b > γ_{1} (π_{k} (p - t N))$ by the choice of $ε_{0}$ , so $p - t N \in E$ by [F1]. Hence $N$ is outward at $p$ by [F6].

6.2step 5.1F1F6

Suppose instead $j \in Σ^{-}$ , so $γ = γ_{1}$ and $b < 0$ by step 4.1. Shrink $ε_{0}$ so that $π_{k} (p \pm t N) \in Q \subseteq D$ for $0 < t < ε_{0}$ and so that $γ_{2} (π_{k} (p - t N)) > γ_{1} (w_{0}) - t b$ there, which is possible since $γ_{1} (w_{0}) < γ_{2} (w_{0})$ by [F1] and $- t b > 0$ tends to $0$ . For $0 < t < ε_{0}$ , step 5.1 gives $u (t) < 0$ , that is $(p + t N)_{k} < γ_{1} (π_{k} (p + t N))$ , so $p + t N \notin E$ by [F1]; and $u (- t) > 0$ , that is $(p - t N)_{k} > γ_{1} (π_{k} (p - t N))$ , while $(p - t N)_{k} = γ_{1} (w_{0}) - t b < γ_{2} (π_{k} (p - t N))$ by the choice of $ε_{0}$ , so $p - t N \in E$ by [F1]. Hence $N$ is outward at $p$ by [F6].

7.1step 6.1step 6.2F6F8∎

In both cases $p \in E$ while $p + t N \notin E$ for arbitrarily small $t > 0$ , so $p$ is not interior to $E$ and therefore $p \in \partial E$ by [F8]. Replacing $N$ by $- N$ exchanges the two conditions of [F6], which then fail, so $- N$ is not outward at $p$ ; since $\pm N$ are the only unit vectors orthogonal to the two-dimensional $T$ , the vector $N$ is the outward unit normal to $T$ at $p$ .

Remarks

Why the projection is interior here. The nonzero projected Jacobian at the interior parameter point makes $ψ_{j}$ a local diffeomorphism. Its local image is open and, because the patch image lies in the graph over $D$ , is contained in $D$ ; hence $w_{0}$ is automatically an interior point of $D$ . The Statement records the condition explicitly because steps 6.1 and 6.2 use the strict inequality $γ_{1} (w_{0}) < γ_{2} (w_{0})$ attached to it.
The excluded points are the seams and the edges. Nothing is claimed at a parameter-boundary point of a patch, nor at a point whose projection lies on $\partial D$ . Those points form a set of content zero in every parameter region, which is why no integral identity on this page is affected by them; but a pointwise claim about the normal there would be false in general and is not made.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Elementary solid regions: one boundary presentation adapted in all three coordinate directions

Definition

An elementary solid region is a compact set $E \subseteq R^{3}$ supplied with a simple description in each of the three coordinate directions (Simple solid regions in a coordinate direction and their cyclic coordinate projection) together with one compatible finite patch presentation of $\partial E$ that is adapted to a simple description of $E$ in each of the three coordinate directions (Boundary presentations adapted to a simple solid region in a coordinate direction, Finitely patched regular surfaces, their area, scalar integrals, and flux).

Explicitly, the data are: three simple descriptions $(x, D_{x}, γ_{1}^{x}, γ_{2}^{x})$ , $(y, D_{y}, γ_{1}^{y}, γ_{2}^{y})$ and $(z, D_{z}, γ_{1}^{z}, γ_{2}^{z})$ , each describing the same set $E$ ; one compatible finite patch presentation $Σ = ((D_{1}, φ_{1}), \dots, (D_{P}, φ_{P}))$ whose patch images cover $\partial E$ and are contained in $\partial E$ ; and, for each of the three directions $k$ , a partition of ${1, \dots, P}$ into sublists $Σ_{k}^{+}, Σ_{k}^{-}, Σ_{k}^{0}$ making $Σ$ adapted to the $k$ th description. The boundary is that of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space.

One presentation, three partitions. The patch list is the same in all three directions; only the sorting of its indices into upper, lower and lateral changes with $k$ . That is what makes the three single-direction flux identities statements about one and the same boundary integral, and it is the whole content of the word "elementary" here.

Remarks

The descriptions and the presentation are supplied data. Nothing above is inferred from the set $E$ : a compact set may admit several simple descriptions in a given direction, several compatible presentations of its boundary, and several sortings of a presentation, and no claim is made that any of these exists for an arbitrary compact set or that it is unique when it does. The convention matches the one Type I, Type II, and elementary regions for Green's theorem uses in the plane, where the decomposition is likewise part of the data.
A patch may be lateral in one direction and a face in another. The six faces of a box illustrate both: the top and bottom faces are the upper and lower sublists for $k = z$ and lateral for $k = x$ and $k = y$ . What cannot happen is a patch lateral in all three directions, and that is Every patch of an elementary solid region's presentation is a graph face in some direction, and at interior base points its normal is outward.
What is not assumed. The solid need not be convex, its boundary need not be connected, and no patch is required to be a graph over a coordinate plane. Conversely, nothing here asserts that every compact solid with a piecewise smooth boundary can be presented this way; What the classical divergence and Stokes theorems here do and do not cover states what is and is not covered.

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Every patch of an elementary solid region's presentation is a graph face in some direction, and at interior base points its normal is outward

Statement

Let $E$ be an elementary solid region with presentation $Σ = ((D_{1}, φ_{1}), \dots, (D_{P}, φ_{P}))$ and sublists $Σ_{k}^{+}, Σ_{k}^{-}, Σ_{k}^{0}$ for $k \in {x, y, z}$ (Elementary solid regions: one boundary presentation adapted in all three coordinate directions). Then every patch of the presentation is an upper or a lower face in at least one coordinate direction: for each $j$ there is $k$ with $j \in Σ_{k}^{+} \cup Σ_{k}^{-}$ .

Moreover, for such a $j$ and $k$ and for every interior parameter point $c \in D_{j}^{\circ}$ whose projection $π_{k} (φ_{j} (c))$ lies in the interior of the base $D_{k}$ of the $k$ th description, the induced unit normal $N_{φ_{j}} (c)$ is the outward unit normal to the tangent plane at $φ_{j} (c)$ .

Facts & Assumptions

Given: The elementary solid region $E$ with its presentation $Σ$ , its three simple descriptions and the three partitions of ${1, \dots, P}$ into sublists.

[F1]

For $j \in Σ_{k}^{0}$ the $k$ th coordinate of $φ_{j, u} \times φ_{j, v}$ vanishes on the interior of $D_{j}$ , and the three sublists $Σ_{k}^{+}, Σ_{k}^{-}, Σ_{k}^{0}$ partition ${1, \dots, P}$ (Boundary presentations adapted to a simple solid region in a coordinate direction, Elementary solid regions: one boundary presentation adapted in all three coordinate directions).

[F2]

A regular patch has $φ_{u} \times φ_{v} \neq 0$ at every point of the interior of its parameter region, and that interior is nonempty (Regular parametrized surface patches on compact Jordan parameter regions).

[F3]

For $v \in R^{3}$ , $⟨ v, e_{k} ⟩ = v_{k}$ , so $v = 0$ if and only if all three of $v_{x}, v_{y}, v_{z}$ are zero (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[F4]

The parametrization induces on the interior the unit normal $N_{φ} = (φ_{u} \times φ_{v}) / ∥ φ_{u} \times φ_{v} ∥_{2}$ (Unit normal fields, orientations, and flux through a regular surface patch).

[F5]

A unit vector $ν$ is outward at $p \in \partial E$ when for some $ε > 0$ one has $p + t ν \notin E$ and $p - t ν \in E$ for every $t$ with $0 < t < ε$ ; with a two-dimensional subspace $T$ supplied, the outward one of its two unit normals is the outward unit normal to $T$ at $p$ (The outward unit normal at a boundary point of a compact solid).

[L1]

Under the hypotheses of an adapted presentation, for $j \in Σ^{+} \cup Σ^{-}$ and an interior parameter point $c$ whose projection lies in the interior of the base, the induced unit normal at $c$ is the outward unit normal to the tangent plane at $φ_{j} (c)$ (At interior base points, the graph faces of an adapted presentation induce the outward unit normal).

Proof

technique · direct

1.1givenF2F3

Fix $j$ and, by [F2], a point $c \in D_{j}^{\circ}$ ; then $φ_{j, u} (c) \times φ_{j, v} (c) \neq 0$ , so by [F3] at least one of its three coordinates is nonzero at $c$ . Fix a direction $k$ for which the $k$ th coordinate is nonzero at $c$ .

2.1step 1.1F1F3F4

By [F1], if $j$ belonged to $Σ_{k}^{0}$ then that $k$ th coordinate would vanish at every point of $D_{j}^{\circ}$ , in particular at $c$ , which step 1.1 excludes. The three sublists partition the index set by [F1], so $j \in Σ_{k}^{+} \cup Σ_{k}^{-}$ . This is the first assertion; equivalently, by [F3] and [F4], a patch lateral in all three directions would have an induced unit normal orthogonal to $e_{x}$ , $e_{y}$ and $e_{z}$ and hence equal to $0$ , which no unit vector is.

3.1step 2.1F1F4F5L1∎

Let $j$ and $k$ be as in the second assertion and let $c \in D_{j}^{\circ}$ have $π_{k} (φ_{j} (c))$ in the interior of $D_{k}$ . The presentation is adapted to the $k$ th description by [F1] and $j \in Σ_{k}^{+} \cup Σ_{k}^{-}$ , so [L1] applies and gives that $N_{φ_{j}} (c)$ of [F4] is the outward unit normal to the tangent plane at $φ_{j} (c)$ in the sense of [F5].

Remarks

The claim is qualified, and the qualification is real. Outwardness is asserted only at interior parameter points whose projection lands in the interior of the relevant base. The excluded points are the parameter-boundary points of a patch and the points sitting over the boundary of the base — the seams and the edges — and at those a normal need not exist or need not be outward. That is not a defect of the presentation: no integral on this page sees a set of content zero in a parameter region.
Why one direction suffices. A patch may be a graph face in one direction and lateral in the other two, as the top face of a box is; the corollary asserts existence of one such direction for each patch, not the same direction for all patches.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The divergence theorem on an elementary solid region

Statement

Let $E$ be an elementary solid region with presentation $Σ = ((D_{1}, φ_{1}), \dots, (D_{P}, φ_{P}))$ (Elementary solid regions: one boundary presentation adapted in all three coordinate directions) and let $F$ be a $C^{1}$ vector field on an open set containing $E$ . Then

$∭_{E} div F = \iint_{\partial E} ⟨ F, n ⟩,$

where the left side is the integral of $div F$ over $E$ and the right side is the flux of $F$ over the presentation $Σ$ , that is $\sum_{j = 1}^{P} \int_{D_{j}} ⟨ F (φ_{j}), φ_{j, u} \times φ_{j, v} ⟩$ . At every interior parameter point whose projection lies in the interior of the relevant base, the orientation in which that flux is taken is the outward one, by Every patch of an elementary solid region's presentation is a graph face in some direction, and at interior base points its normal is outward.

Facts & Assumptions

Given: The elementary solid region $E$ with its three simple descriptions, its presentation $Σ$ and the three partitions of ${1, \dots, P}$ into sublists, and the $C^{1}$ field $F$ on an open $O \supseteq E$ .

[F1]

For a compatible finite patch presentation the oriented flux is the sum of the patch values, each being $\int_{D_{j}} (F \circ φ_{j}) \cdot (φ_{j, u} \times φ_{j, v})$ (Finitely patched regular surfaces, their area, scalar integrals, and flux, Unit normal fields, orientations, and flux through a regular surface patch).

[F2]

The divergence of a $C^{1}$ field $F$ on an open subset of $R^{n}$ is $div F = \sum_{i < n} \partial_{i} F_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ , and $e_{k}$ has $k$ th coordinate $1$ and the others $0$ ; so a vector $v \in R^{3}$ is $v_{x} e_{x} + v_{y} e_{y} + v_{z} e_{z}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[F4]

An elementary solid region carries one presentation adapted to a simple description of $E$ in each of the three coordinate directions (Elementary solid regions: one boundary presentation adapted in all three coordinate directions, Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F5]

Integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

Let $(k, D, γ_{1}, γ_{2})$ be a simple description of $E$ in the direction $k$ , let $Σ$ be adapted to it, and let $R$ be $C^{1}$ on an open set containing $E$ . Then the flux of $R e_{k}$ over the presentation equals the integral of the $k$ th partial derivative of $R$ over $E$ (The single-direction flux identity on a simple solid region).

[L2]

For integrable $f, g$ on a nondegenerate rectangle and scalars $α, β$ , the function $α f + β g$ is integrable with integral $α \int f + β \int g$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L3]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

Proof

technique · direct

1.1givenF1F3F5L2L3

By [F3] the field splits as $F = F_{x} e_{x} + F_{y} e_{y} + F_{z} e_{z}$ on $O$ , each $F_{k}$ being a $C^{1}$ real function there. For each patch, [F1] and [F3] make the flux integrand $⟨ F (φ_{j}), φ_{j, u} \times φ_{j, v} ⟩ = \sum_{k} F_{k} (φ_{j}) (φ_{j, u} \times φ_{j, v})_{k}$ , a sum of three continuous functions on the compact Jordan parameter region $D_{j}$ ; each is integrable by [L3], so [L2] and [F5] split that patch's flux into the three corresponding patch fluxes of the fields $F_{k} e_{k}$ . Summing over $j$ and using [F1] again, the flux of $F$ over $Σ$ is the sum over $k$ of the fluxes of $F_{k} e_{k}$ over $Σ$ .

1.2givenF4L1

Fix a direction $k$ . By [F4] the same presentation $Σ$ is adapted to the $k$ th simple description of $E$ , and $F_{k}$ is $C^{1}$ on the open $O \supseteq E$ , so [L1] applies and gives that the flux of $F_{k} e_{k}$ over $Σ$ equals $\int_{E} \partial_{k} F_{k}$ . This holds for each of the three directions, with the one presentation and the three descriptions supplied with $E$ .

2.1step 1.1step 1.2F2F5L2L3

Adding the three identities of step 1.2 and substituting into step 1.1, the flux of $F$ over $Σ$ equals $\int_{E} \partial_{x} F_{x} + \int_{E} \partial_{y} F_{y} + \int_{E} \partial_{z} F_{z}$ . Each $\partial_{k} F_{k}$ is continuous on the compact Jordan set $E$ , hence integrable over it by [L3], so [L2] and [F5] combine those three integrals into $\int_{E} (\partial_{x} F_{x} + \partial_{y} F_{y} + \partial_{z} F_{z})$ , which is $\int_{E} div F$ by [F2].

3.1step 2.1∎

Step 2.1 is the asserted identity. The requirement that one presentation be adapted in all three directions is used exactly once, in step 1.2, where the three applications of [L1] must be to the same boundary integral; and the outward reading of the normals is Every patch of an elementary solid region's presentation is a graph face in some direction, and at interior base points its normal is outward, on which no step above depends.

Remarks

The field must be $C^{1}$ on an open set containing all of $E$ , not only on $\partial E$ . Step 1.2 integrates $\partial_{k} F_{k}$ over the whole solid, so the partial derivatives must exist there. The companion examples page records the failure that quietly weakening this hypothesis produces.
Nothing is asserted for a solid presented without the data. The three descriptions, the presentation and the three sortings are hypotheses. A compact set with a piecewise smooth boundary may admit them, may admit them only after being cut into pieces — which is what The divergence theorem for finite gluings of elementary solid regions is for — or may not be shown to admit them by anything on this page.

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Finite gluings of elementary solid regions and their outward boundary presentation

Definition

A finite gluing of elementary solid regions consists of the following supplied data.

An integer $N \geq 1$ and elementary solid regions $E_{1}, \dots, E_{N}$ with pairwise disjoint interiors whose union is $E$ (Elementary solid regions: one boundary presentation adapted in all three coordinate directions, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space). Each $E_{i}$ carries its own three simple descriptions, its own presentation $Σ_{i}$ and its own three sortings.
For each $i$ , a designation of every patch of $Σ_{i}$ as internal or outer.
An involution without fixed points on the set of all internal patches of all the pieces, under which each internal patch is paired with an internal patch of a different piece that is an orientation-reversing regular reparametrization of it in the sense of Surface reparametrizations and their orientation sign: for paired patches $(D, φ)$ and $(D^{'}, φ^{'})$ there is a $C^{1}$ diffeomorphism $h$ between open neighbourhoods of $D^{'}$ and $D$ with $h [D^{'}] = D$ , $φ^{'} = φ \circ h$ and $\det D h < 0$ .
A requirement that the list of all outer patches of all the pieces, taken together, be a compatible finite patch presentation in the sense of Finitely patched regular surfaces, their area, scalar integrals, and flux whose patch images cover $\partial E$ and are contained in $\partial E$ . That list is the outer boundary presentation of the gluing, written $\partial E$ where an integral is taken over it.

Each patch is a regular parametrized surface patch of Regular parametrized surface patches on compact Jordan parameter regions, and the flux of a continuous field over the outer boundary presentation is the sum of the flux over its patches.

Remarks

The pairing is a condition on parametrizations, not on images. Clause 3 asks for an orientation-reversing reparametrization, so the two paired patches have the same image and induced normals that are negatives of each other where both are defined. Two patches whose images merely coincide as sets do not satisfy it, and neither do two patches one of whose images is strictly larger: a reparametrization is a bijection between the parameter regions. A face of one piece that meets a smaller face of its neighbour must therefore be subdivided before it can be paired, and the companion examples page shows a case where that is unavoidable.
Everything is supplied. As with an elementary solid region, nothing here is inferred from the set $E$ : neither the decomposition, nor the internal-or-outer designation, nor the pairing, nor the fact that the outer patches present $\partial E$ . No claim is made that an arbitrary compact solid admits such data.
$N = 1$ is allowed and carries no internal patch. Then the involution of clause 3 is the empty map, the outer presentation is the piece's own presentation, and a finite gluing of one piece is the elementary solid region itself. The pieces themselves are indexed by a nonempty finite set: $N \geq 1$ is part of clause 1.

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Internal faces cancel and volume integrals add when elementary solid regions are glued

Statement

Let a finite gluing of elementary solid regions be given, with pieces $E_{1}, \dots, E_{N}$ , presentations $Σ_{1}, \dots, Σ_{N}$ , union $E$ and outer boundary presentation $Σ^{o u t}$ (Finite gluings of elementary solid regions and their outward boundary presentation). Then $E$ is compact and Jordan measurable, and the sum of the piece fluxes is the flux over the outer presentation, and the sum of the piece volume integrals is the integral over the union:

$\sum_{i = 1}^{N} \iint_{Σ_{i}} ⟨ G, n ⟩ = \iint_{Σ^{o u t}} ⟨ G, n ⟩ for every continuous vector field G on \partial E \cup ⋃_{i} \partial E_{i},$

$\sum_{i = 1}^{N} \int_{E_{i}} H = \int_{E} H for every continuous H : E \to R,$

both integrals in the second identity existing.

Facts & Assumptions

Given: The finite gluing with its pieces, presentations, internal-or-outer designations and pairing involution, together with the continuous $G$ and the continuous $H : E \to R$ .

[F1]

For a compatible finite patch presentation the oriented flux is the sum of the flux over its patches (Finitely patched regular surfaces, their area, scalar integrals, and flux, Unit normal fields, orientations, and flux through a regular surface patch).

[F2]

In a finite gluing the pieces are elementary solid regions with pairwise disjoint interiors whose union is $E$ ; every patch of every $Σ_{i}$ is designated internal or outer; each internal patch is paired with an internal patch of a different piece that is an orientation-reversing regular reparametrization of it; and the outer patches together form a compatible finite patch presentation of $\partial E$ (Finite gluings of elementary solid regions and their outward boundary presentation, Elementary solid regions: one boundary presentation adapted in all three coordinate directions).

[F3]

A regular reparametrization is orientation-reversing when its parameter Jacobian determinant is negative (Surface reparametrizations and their orientation sign).

[F4]

The simple solid region described by a simple description is compact and Jordan measurable (Simple solid regions in a coordinate direction and their cyclic coordinate projection).

[F5]

The boundary of $A$ is $\partial A = \overline{A} ∖ int (A)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space); a set has content zero when it admits finite cube covers of arbitrarily small total volume, and content zero passes to subsets (Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[F6]

Integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

An orientation-preserving reparametrization preserves flux and an orientation-reversing reparametrization negates it (Flux is invariant under orientation-preserving reparametrization and changes sign under reversal).

[L2]

[L3]

A metric-bounded set is Jordan measurable if and only if its boundary has content zero (A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero).

[L4]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

[L5]

A continuous real function on a nonempty compact metric space has bounded image (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L6]

A subset of $R^{m}$ is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct

1.1givenF1F2

By [F1] the sum $\sum_{i = 1}^{N} \iint_{Σ_{i}} ⟨ G, n ⟩$ is the sum of the flux of $G$ over every patch of every $Σ_{i}$ , a finite list of real numbers. By [F2] each entry of that list is designated internal or outer.

1.2givenF4F5L3L6

Each $E_{i}$ is compact and Jordan measurable by [F4], so $E = ⋃_{i} E_{i}$ is closed and bounded, hence compact by [L6], and each $\partial E_{i}$ has content zero by [L3]. If $p \in \partial E$ then $p \in E$ , so $p \in E_{i}$ for some $i$ , and $p$ cannot lie in $E_{i}^{\circ}$ , since $E_{i} \subseteq E$ would then put $p$ in $E^{\circ}$ ; so $p \in \partial E_{i}$ and $\partial E \subseteq ⋃_{i} \partial E_{i}$ . Concatenating the $N$ finite covers shows that union has content zero, so $E$ is Jordan measurable by [L3] and [F5].

2.1givenF2F3L1

Let $(D, φ)$ and $(D^{'}, φ^{'})$ be a paired internal pair, so by [F2] and [F3] there is a $C^{1}$ diffeomorphism $h$ between neighbourhoods of $D^{'}$ and $D$ with $h [D^{'}] = D$ , $φ^{'} = φ \circ h$ and $\det D h < 0$ ; that is an orientation-reversing regular reparametrization. So [L1] gives that the flux of $G$ over $(D^{'}, φ^{'})$ is the negative of its flux over $(D, φ)$ , and the two contributions to the sum of step 1.1 add to $0$ . The pairing of [F2] is an involution without fixed points, so the internal entries of the list are exhausted by such pairs.

2.2step 1.2F5F6L2L4L5

By [L5] the continuous $H$ is bounded on the nonempty compact $E$ , and by [L4] it is integrable over $E$ and over each compact Jordan $E_{i}$ . If $i \neq j$ and $p \in E_{i} \cap E_{j}$ then $p$ lies in at most one of the two interiors, so $E_{i} \cap E_{j} \subseteq \partial E_{i} \cup \partial E_{j}$ , which has content zero by step 1.2 and [F5]; and $E ∖ ⋃_{i} E_{i}$ is empty, hence of content zero. So [L2] applies with $A = E$ and $A_{i} = E_{i}$ and gives $\sum_{i} \int_{E_{i}} H = \int_{E} H$ , the integrals being those of [F6].

3.1step 1.1step 2.1F1F2

Deleting the cancelling internal pairs of step 2.1 from the finite sum of step 1.1 leaves exactly the outer entries, whose sum is $\iint_{Σ^{o u t}} ⟨ G, n ⟩$ by [F1] and [F2]. This is a rearrangement of finitely many reals, so it needs no connectedness of $E$ or of its boundary; for $N = 1$ there is no internal patch and the two lists coincide.

4.1step 3.1step 2.2step 1.2∎

Steps 3.1 and 2.2 are the two asserted identities, and step 1.2 is the assertion that $E$ is compact and Jordan measurable.

Remarks

The cancellation is between parametrizations, not between images. [L1] compares the flux of two patches related by a reparametrization; two patches with the same image but no such relation are not covered, and neither are two patches whose images overlap only partly. That is why the gluing data asks for the reparametrization explicitly, and why a face meeting a smaller neighbouring face has to be cut first.
The sign condition is pointwise and needs no connectedness argument. The gluing data requires $\det D h < 0$ everywhere, so the reparametrization is orientation-reversing in the sense of [F3] at every parameter point. A regular reparametrization of a connected parameter region has a constant orientation sign says that on a connected parameter region the sign cannot change, so the requirement costs nothing beyond one sign check per pair.

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The divergence theorem for finite gluings of elementary solid regions

Statement

Let a finite gluing of elementary solid regions be given, with pieces $E_{1}, \dots, E_{N}$ , union $E$ and outer boundary presentation $Σ^{o u t}$ (Finite gluings of elementary solid regions and their outward boundary presentation), and let $F$ be a $C^{1}$ vector field on an open set containing $E$ . Then

$∭_{E} div F = \iint_{\partial E} ⟨ F, n ⟩,$

the right-hand side being the flux of $F$ over $Σ^{o u t}$ .

The decomposition into pieces, the internal-or-outer designation of the patches and the pairing of the internal patches are hypotheses supplied with the gluing; nothing is asserted about a solid presented without them.

Facts & Assumptions

Given: The finite gluing with its pieces $E_{1}, \dots, E_{N}$ , their presentations $Σ_{i}$ , the internal-or-outer designations, the pairing involution, the union $E$ , the outer presentation $Σ^{o u t}$ , and the $C^{1}$ field $F$ on an open $O \supseteq E$ .

[F1]

In a finite gluing the pieces are elementary solid regions with pairwise disjoint interiors whose union is $E$ , and the outer patches together form a compatible finite patch presentation of $\partial E$ (Finite gluings of elementary solid regions and their outward boundary presentation).

[F2]

The divergence of a $C^{1}$ field $F$ on an open subset of $R^{n}$ is $div F = \sum_{i < n} \partial_{i} F_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

For a compatible finite patch presentation the oriented flux is the sum of the flux over its patches (Finitely patched regular surfaces, their area, scalar integrals, and flux), and $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L1]

For an elementary solid region $E^{'}$ with presentation $Σ^{'}$ and a $C^{1}$ field $F$ on an open set containing $E^{'}$ , $∭_{E^{'}} div F = \iint_{Σ^{'}} ⟨ F, n ⟩$ (The divergence theorem on an elementary solid region).

[L2]

For a finite gluing, $E$ is compact and Jordan measurable; for a continuous vector field on the union of the piece boundaries, the sum of the piece fluxes is the flux over the outer presentation; and for a continuous scalar function on $E$ , the sum of the piece volume integrals is the integral over the union (Internal faces cancel and volume integrals add when elementary solid regions are glued).

Proof

technique · direct

1.1givenF1L1

By [F1] each $E_{i}$ is an elementary solid region contained in $E$ , so $O$ is an open set containing $E_{i}$ and $F$ is $C^{1}$ on it; hence [L1] applies to each piece and gives $∭_{E_{i}} div F = \iint_{Σ_{i}} ⟨ F, n ⟩$ for $i = 1, \dots, N$ .

1.2givenF2

The function $div F$ is continuous on $O$ by [F2], since a $C^{1}$ field has continuous first partial derivatives, and in particular continuous on $E$ .

2.1step 1.1step 1.2F1F3L2

Summing the $N$ identities of step 1.1 over $i$ and applying [L2] to each side — the volume clause with $H = div F$ , continuous on $E$ by step 1.2, and the flux clause with $G = F$ , continuous on $\partial E$ and on every $\partial E_{i}$ — turns the left sum into $∭_{E} div F$ and the right sum into $\iint_{Σ^{o u t}} ⟨ F, n ⟩$ , which by [F1] and [F3] is the flux over the outer boundary presentation of $E$ .

3.1step 2.1∎

Step 2.1 is the asserted identity. The field is required to be $C^{1}$ on an open set containing the whole union, because step 1.1 applies the piecewise identity with that same field on each piece and step 2.1 integrates $div F$ over $E$ .

Remarks

What the gluing clause buys. A solid need not be simple in every coordinate direction: a U-shaped prism has sections in one direction that are unions of two disjoint intervals, so it admits no simple description there, and yet it is a gluing of three boxes. The companion examples page carries that computation.
No connectedness is used. The pieces need not touch and the boundary need not be connected: step 2.1 rearranges finitely many real numbers and integrates over a finite union.

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Vector forms: the boundary integrals of $f n$ and of $n \times F$

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t} = ((D_{1}, φ_{1}), \dots, (D_{M}, φ_{M}))$ . Vector-valued integrals below are taken coordinatewise, so that for a continuous $R^{3}$ -valued $W$ on $E$ the symbol $∭_{E} W$ denotes the vector whose $k$ th coordinate is $\int_{E} W_{k}$ , and for a continuous $R^{3}$ -valued $Z$ on the boundary the symbol $\iint_{\partial E} Z$ denotes the vector whose $k$ th coordinate is $\sum_{j = 1}^{M} \int_{D_{j}} Z_{k} (φ_{j})$ , where $n$ inside such an integrand is read as the oriented area vector $φ_{j, u} \times φ_{j, v}$ of the patch, exactly as in the scalar flux.

Then, for $f$ of class $C^{1}$ on an open set containing $E$ and $F$ of class $C^{1}$ on an open set containing $E$ ,

$∭_{E} \nabla f = \iint_{\partial E} f n, ∭_{E} curl F = \iint_{\partial E} n \times F .$

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , the $C^{1}$ scalar $f$ and the $C^{1}$ field $F$ , both on open sets containing $E$ , and the coordinatewise reading of the vector integrals fixed in the Statement.

[F1]

The divergence of a $C^{1}$ field is $div G = \sum_{i < n} \partial_{i} G_{i}$ and the curl of a $C^{1}$ field on an open subset of $R^{3}$ is $curl G = (\partial_{y} G_{z} - \partial_{z} G_{y}, \partial_{z} G_{x} - \partial_{x} G_{z}, \partial_{x} G_{y} - \partial_{y} G_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

[F3]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ , and $e_{k}$ has $k$ th coordinate $1$ and the others $0$ , so $⟨ x, e_{k} ⟩ = x_{k}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[F4]

For scalar-valued $f$ the gradient is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F5]

In a finite gluing the outer patches form a compatible finite patch presentation of $\partial E$ , over which flux is the sum of the patch values (Finite gluings of elementary solid regions and their outward boundary presentation, Finitely patched regular surfaces, their area, scalar integrals, and flux).

[L1]

For $C^{1}$ fields and a $C^{1}$ scalar on an open subset of $R^{n}$ , $div (g G) = ⟨ \nabla g, G ⟩ + g div G$ (Divergence and curl are linear and satisfy the scalar product rules).

[L2]

For $C^{1}$ fields $G, H$ on an open subset of $R^{3}$ , $div (G \times H) = ⟨ curl G, H ⟩ - ⟨ G, curl H ⟩$ (The divergence and curl of a cross product).

[L3]

For a finite gluing with union $E$ and outer presentation $Σ^{o u t}$ and a $C^{1}$ field $G$ on an open set containing $E$ , $∭_{E} div G = \iint_{\partial E} ⟨ G, n ⟩$ (The divergence theorem for finite gluings of elementary solid regions).

Proof

technique · direct

1.1givenF1F3L1

Let $c \in R^{3}$ and let $\underline{c}$ be the constant field with value $c$ on the open set where $f$ is $C^{1}$ . Its partial derivatives all vanish, so it is $C^{1}$ with $div \underline{c} = 0$ and $curl \underline{c} = 0$ by [F1]. The field $f \underline{c}$ is $C^{1}$ and [L1] gives $div (f \underline{c}) = ⟨ \nabla f, \underline{c} ⟩ + f div \underline{c} = ⟨ \nabla f, c ⟩$ , while its flux integrand against a vector $ν$ is $⟨ f c, ν ⟩ = f ⟨ c, ν ⟩$ by [F3].

1.2F2F3algebra

For all $a, b, d \in R^{3}$ , expanding both sides by [F2] and [F3] gives $⟨ a \times b, d ⟩ = (a_{y} b_{z} - a_{z} b_{y}) d_{x} + (a_{z} b_{x} - a_{x} b_{z}) d_{y} + (a_{x} b_{y} - a_{y} b_{x}) d_{z},$ $⟨ d \times a, b ⟩ = (d_{y} a_{z} - d_{z} a_{y}) b_{x} + (d_{z} a_{x} - d_{x} a_{z}) b_{y} + (d_{x} a_{y} - d_{y} a_{x}) b_{z},$ and the six monomials of the first list are the six of the second with the same signs, matched as $a_{y} b_{z} d_{x}$ with $d_{x} a_{y} b_{z}$ , $- a_{z} b_{y} d_{x}$ with $- d_{x} a_{z} b_{y}$ , $a_{z} b_{x} d_{y}$ with $d_{y} a_{z} b_{x}$ , $- a_{x} b_{z} d_{y}$ with $- d_{y} a_{x} b_{z}$ , $a_{x} b_{y} d_{z}$ with $d_{z} a_{x} b_{y}$ and $- a_{y} b_{x} d_{z}$ with $- d_{z} a_{y} b_{x}$ . Hence $⟨ a \times b, d ⟩ = ⟨ d \times a, b ⟩$ .

2.1givenF1F2L2

With $\underline{c}$ as in step 1.1 on the open set where $F$ is $C^{1}$ , the field $F \times \underline{c}$ is $C^{1}$ and [L2] gives $div (F \times \underline{c}) = ⟨ curl F, c ⟩ - ⟨ F, curl \underline{c} ⟩ = ⟨ curl F, c ⟩$ .

2.2step 1.1F3F4F5L3

Apply [L3] to the field $f \underline{c}$ of step 1.1: $\int_{E} ⟨ \nabla f, c ⟩ = \sum_{j = 1}^{M} \int_{D_{j}} f (φ_{j}) ⟨ c, φ_{j, u} \times φ_{j, v} ⟩$ , using [F5] to read the right side patch by patch. Take $c = e_{k}$ : by [F3] and [F4] the left side becomes $\int_{E} \partial_{k} f$ , the $k$ th coordinate of $∭_{E} \nabla f$ , and the right side becomes $\sum_{j} \int_{D_{j}} f (φ_{j}) (φ_{j, u} \times φ_{j, v})_{k}$ , the $k$ th coordinate of $\iint_{\partial E} f n$ . As $k$ ranges over the three directions this is the first identity.

3.1step 2.1step 1.2F3F5L3

Apply [L3] to the field $F \times \underline{c}$ of step 2.1: $\int_{E} ⟨ curl F, c ⟩ = \sum_{j = 1}^{M} \int_{D_{j}} ⟨ F (φ_{j}) \times c, φ_{j, u} \times φ_{j, v} ⟩$ . Step 1.2 with $a = F (φ_{j})$ , $b = c$ and $d = φ_{j, u} \times φ_{j, v}$ rewrites each integrand as $⟨ (φ_{j, u} \times φ_{j, v}) \times F (φ_{j}), c ⟩$ . Take $c = e_{k}$ : by [F3] the left side becomes $\int_{E} (curl F)_{k}$ and the right side becomes $\sum_{j} \int_{D_{j}} ((φ_{j, u} \times φ_{j, v}) \times F (φ_{j}))_{k}$ , so as $k$ ranges over the three directions this is the second identity.

4.1step 2.2step 3.1∎

Steps 2.2 and 3.1 are the two asserted identities.

Remarks

Why a constant vector is the right device. Both clauses assert an equality of vectors, and the divergence theorem produces only scalars. Pairing with a fixed $c$ turns each vector identity into a scalar one; running $c$ over the standard basis recovers the vector identity coordinate by coordinate, and nothing else about $c$ is used.
The triple-product identity of step 1.2 is the determinant identity in disguise. By The cross product is bilinear, alternating, and orthogonal to both factors each of $⟨ a \times b, d ⟩$ and $⟨ d \times a, b ⟩$ is the determinant of the matrix with the three vectors as columns, in the orders $a, b, d$ and $d, a, b$ ; those two orders differ by a cyclic permutation of three columns. The coordinate expansion above is the same fact written out, and it is what the proof uses.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The volume of a glued elementary solid is a third of the outward flux of the position field

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t}$ , and let $P$ be the position field $P (p) = p$ on $R^{3}$ . Then the content of the solid is a third of the outward flux of the position field through its boundary:

$cont (E) = \frac{1}{3} \iint_{\partial E} ⟨ P, n ⟩ .$

Moreover each of the three single-coordinate fields $p \mapsto p_{x} e_{x}$ , $p \mapsto p_{y} e_{y}$ and $p \mapsto p_{z} e_{z}$ satisfies

$\iint_{\partial E} ⟨ p_{k} e_{k}, n ⟩ = cont (E) (k \in {x, y, z}) .$

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , and the position field $P$ .

[F1]

The divergence of a $C^{1}$ field is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

For $f = 1$ the zero extension is $1_{E}$ , so $\int_{E} 1 = cont (E)$ (The Riemann integral of a bounded function over a bounded Jordan measurable set, A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content).

[F3]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ , and $e_{k}$ has $k$ th coordinate $1$ and the others $0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L1]

[L2]

For a finite gluing, $E$ is compact and Jordan measurable (Internal faces cancel and volume integrals add when elementary solid regions are glued, Finite gluings of elementary solid regions and their outward boundary presentation).

[L3]

Proof

technique · direct

1.1givenF1F2L2

The position field has $P_{k} (p) = p_{k}$ , so $\partial_{i} P_{k}$ is $1$ when $i = k$ and $0$ otherwise; these are continuous on $R^{3}$ , so $P$ is $C^{1}$ there and [F1] gives $div P = 1 + 1 + 1 = 3$ at every point. By [L2] the set $E$ is compact and Jordan measurable, so $\int_{E} 1 = cont (E)$ by [F2].

1.2givenF1F3

For each $k$ the field $p \mapsto p_{k} e_{k}$ has $k$ th coordinate $p_{k}$ and the other two coordinates $0$ by [F3], so its only nonvanishing first partial derivative is $\partial_{k} p_{k} = 1$ ; it is therefore $C^{1}$ on $R^{3}$ with divergence $1$ by [F1].

2.1step 1.1L1L3

Applying [L1] with $G = P$ , which is $C^{1}$ on the open set $R^{3} \supseteq E$ , gives $\iint_{\partial E} ⟨ P, n ⟩ = \int_{E} div P = \int_{E} 3$ , and by [L3] with $α = 3$ , $β = 0$ and step 1.1 this is $3 \int_{E} 1 = 3 cont (E)$ . Dividing by $3$ gives the first identity.

2.2step 1.1step 1.2L1

Applying [L1] with $G$ the field $p \mapsto p_{k} e_{k}$ of step 1.2 gives $\iint_{\partial E} ⟨ p_{k} e_{k}, n ⟩ = \int_{E} 1 = cont (E)$ by step 1.1, for each of the three directions $k$ .

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 are the asserted identities.

Remarks

The solid is Jordan measurable because its pieces are, not by assumption. Step 1.1 takes that from [L2]; without it the symbol $cont (E)$ would not denote anything and $\int_{E} 1$ would not exist.
The orientation is what fixes the sign. Reversing every patch of the presentation negates the right-hand sides and would give a negative content. That the presentation of a glued elementary solid carries the outward normal is Every patch of an elementary solid region's presentation is a graph face in some direction, and at interior base points its normal is outward applied to each piece, and the companion examples page checks the sign against the known volume of a ball.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A field with vanishing divergence has zero outward flux through the boundary of a glued elementary solid

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t}$ , and let $F$ be a $C^{1}$ vector field on an open set $O$ containing $E$ . If the divergence vanishes on an open set containing the solid then the outward boundary flux is zero: if $div F = 0$ at every point of $O$ , then

$\iint_{\partial E} ⟨ F, n ⟩ = 0.$

The hypothesis is that $F$ is $C^{1}$ with vanishing divergence on an open set containing the whole of $E$ , not merely on $\partial E$ and not merely wherever $F$ happens to be defined.

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , the open set $O \supseteq E$ , and the $C^{1}$ field $F$ on $O$ with $div F = 0$ throughout $O$ .

[F1]

The divergence of a $C^{1}$ field is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

Integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

[L2]

For a finite gluing, $E$ is compact and Jordan measurable (Internal faces cancel and volume integrals add when elementary solid regions are glued).

[L3]

Proof

technique · direct

1.1givenF1F2L2L3

By [L2] the set $E$ is compact and Jordan measurable, and by hypothesis and [F1] the function $div F$ is identically zero on $E$ . Its zero extension to a bounding rectangle is the zero function, which by [L3] with $α = β = 0$ is integrable with integral $0$ ; so $\int_{E} div F = 0$ by [F2].

2.1step 1.1L1∎

The field $F$ is $C^{1}$ on the open $O \supseteq E$ , so [L1] applies and gives $\iint_{\partial E} ⟨ F, n ⟩ = ∭_{E} div F$ , which is $0$ by step 1.1.

Remarks

The hypothesis is about an open set containing $E$ , and that is exactly what fails in the standard counterexample. The inverse-square field has vanishing divergence at every point where it is defined, yet its outward flux through the unit sphere is $4 π$ ; the field is not defined at the origin, so no open set containing the closed unit ball carries it. The companion examples page states the false weakening and carries the computation.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The flux of a curl through the boundary of a glued elementary solid vanishes

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t}$ , and let $F$ be a vector field of class $C^{2}$ on an open set $O \subseteq R^{3}$ containing $E$ . Then

$\iint_{\partial E} ⟨ curl F, n ⟩ = 0.$

The hypothesis is $C^{2}$ , not $C^{1}$ : with $F$ only $C^{1}$ the field $curl F$ need not have a divergence at any point, so neither the degree-two identity nor the divergence theorem has a hypothesis to consume.

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , the open $O \supseteq E$ , and the $C^{2}$ field $F$ on $O$ .

[F1]

The curl of a $C^{1}$ field on an open subset of $R^{3}$ is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ , and the divergence of a $C^{1}$ field is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

[F3]

In a finite gluing the outer patches form a compatible finite patch presentation of $\partial E$ , over which flux is the sum of the patch values, with $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (Finite gluings of elementary solid regions and their outward boundary presentation, Finitely patched regular surfaces, their area, scalar integrals, and flux, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L1]

For $U \subseteq R^{3}$ open and $F : U \to R^{3}$ of class $C^{2}$ , the field $curl F$ is $C^{1}$ on $U$ and $div (curl F) = 0$ on $U$ (The divergence of the curl of a $C^{2}$ field vanishes).

[L2]

For a finite gluing with union $E$ and a $C^{1}$ field $G$ on an open set containing $E$ whose divergence vanishes there, $\iint_{\partial E} ⟨ G, n ⟩ = 0$ (A field with vanishing divergence has zero outward flux through the boundary of a glued elementary solid).

Proof

technique · direct

1.1givenF1F2L1

Each coordinate of $curl F$ is a difference of two first partial derivatives of components of $F$ by [F1]. Since $F$ is $C^{2}$ on $O$ , [F2] makes every iterated derivative $\partial_{i} \partial_{j} F_{a}$ exist and be continuous on $O$ , so each coordinate of $curl F$ has continuous first partial derivatives; hence $curl F$ is a $C^{1}$ field on $O$ , which is what [L1] asserts and which is exactly the regularity [L2] requires of the field it is applied to.

2.1step 1.1F3L1L2∎

By [L1] the divergence of $curl F$ vanishes at every point of $O$ , and $O$ is an open set containing $E$ . So [L2] applied to $G = curl F$ , a $C^{1}$ field on $O$ by step 1.1 with vanishing divergence there, gives $\iint_{\partial E} ⟨ curl F, n ⟩ = 0$ , the flux being read over the outer presentation as in [F3].

Remarks

Where $C^{2}$ is spent. It is used once, in step 1.1, to make $curl F$ a $C^{1}$ field. Everything after that is the divergence-free corollary applied to that field. The identity $div curl F = 0$ is itself a $C^{2}$ statement, so the hypothesis cannot be weakened by rearranging the argument.
The converse is false. A field with zero outward flux through the boundary of every glued elementary solid need not be a curl on the whole of $O$ : the divergence-free field is a curl on a star-shaped open set by A divergence-free $C^{1}$ field on a star-shaped open subset of $R^{3}$ has a vector potential, and on a general open set that theorem's hypothesis is unavailable. Nothing here asserts otherwise.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The divergence at a point is the limit of outward flux per unit volume

Statement

Let $O \subseteq R^{3}$ be open, let $F : O \to R^{3}$ be $C^{1}$ and let $p \in O$ . For each $m \in N$ let a finite gluing of elementary solid regions be given whose union $E^{(m)}$ satisfies $E^{(m)} \subseteq O$ , $p \in E^{(m)}$ and $cont (E^{(m)}) > 0$ , and suppose $diam (E^{(m)}) \to 0$ . Then

$\lim_{m \to \infty} \frac{1}{cont (E^{(m)})} \iint_{\partial E^{(m)}} ⟨ F, n ⟩ = div F (p),$

that is: for every rational $ε > 0$ there is $M$ such that every $m \geq M$ satisfies

$∣ \frac{1}{cont (E^{(m)})} \iint_{\partial E^{(m)}} ⟨ F, n ⟩ - div F (p) ∣ < ε .$

Positive content is required only so that the quotient is defined; no relation between the content and the diameter is assumed.

Facts & Assumptions

Given: The open $O \subseteq R^{3}$ , the $C^{1}$ field $F$ on $O$ , the point $p \in O$ , and for each $m$ the finite gluing with union $E^{(m)} \subseteq O$ containing $p$ , of positive content, with $diam (E^{(m)}) \to 0$ .

[F1]

The divergence of a $C^{1}$ field is $div G = \sum_{i < n} \partial_{i} G_{i}$ ; a $C^{1}$ map has continuous first partial derivatives, so $div G$ is continuous (Divergence and curl of a $C^{1}$ vector field, $C^{k}$ Euclidean maps and diffeomorphisms).

[F2]

[F3]

For a nonempty bounded $A$ in a metric space, $diam (A) = \sup {d (a, b) : a, b \in A}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), the metric on $R^{3}$ being $d (a, b) = ∥ a - b ∥_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

A map between metric spaces is continuous at a point when for every real $ε > 0$ there is a real $δ > 0$ such that points within $δ$ of it have images within $ε$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

[F5]

A sequence of reals converges to $x$ when for every rational $ε > 0$ there is $K$ with $∣ x_{k} - x ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[L1]

[L2]

[L3]

For integrable $f, g$ on a nondegenerate rectangle and scalars $α, β$ : $α f + β g$ is integrable with integral $α \int f + β \int g$ ; if $f \leq g$ then $\int f \leq \int g$ ; and $∣ f ∣$ is integrable with $∣ \int f ∣ \leq \int ∣ f ∣$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L4]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

Proof

technique · direct

1.1givenF1L1L2L4

For each $m$ the set $E^{(m)}$ is compact and Jordan measurable by [L2], and $div F$ is continuous on $O$ by [F1], hence integrable over $E^{(m)}$ by [L4]. Since $F$ is $C^{1}$ on the open $O \supseteq E^{(m)}$ , [L1] gives $\iint_{\partial E^{(m)}} ⟨ F, n ⟩ = \int_{E^{(m)}} div F$ .

1.2givenF1F3F4F5

Let $ε > 0$ be rational. The function $div F$ is continuous at $p$ by [F1], so [F4] with the real number $ε / 2$ supplies $δ > 0$ such that every $q \in O$ with $∥ q - p ∥_{2} < δ$ satisfies $∣ div F (q) - div F (p) ∣ < ε / 2$ . Since $diam (E^{(m)}) \to 0$ , there is $M$ with $diam (E^{(m)}) < δ$ for every $m \geq M$ .

2.1step 1.1step 1.2F2F3L3

Fix $m \geq M$ . Since $p \in E^{(m)}$ , every $q \in E^{(m)}$ has $∥ q - p ∥_{2} \leq diam (E^{(m)}) < δ$ by [F3], so step 1.2 bounds $∣ div F - div F (p) ∣$ by $ε / 2$ on $E^{(m)}$ . By [L3] and [F2], $\int_{E^{(m)}} div F (p) = div F (p) cont (E^{(m)})$ , and $∣ \int_{E^{(m)}} div F - div F (p) cont (E^{(m)}) ∣ = ∣ \int_{E^{(m)}} (div F - div F (p)) ∣ \leq \int_{E^{(m)}} \frac{ε}{2} = \frac{ε}{2} cont (E^{(m)}) .$

3.1step 2.1F5∎

Dividing the estimate of step 2.1 by the positive number $cont (E^{(m)})$ and substituting step 1.1 gives $∣ \frac{1}{cont (E^{(m)})} \iint_{\partial E^{(m)}} ⟨ F, n ⟩ - div F (p) ∣ \leq \frac{ε}{2} < ε$ for every $m \geq M$ . As $ε$ was an arbitrary positive rational, [F5] gives the asserted limit.

Remarks

No shape hypothesis is needed. The content cancels between the estimate and the quotient, so nothing forces the solids to be balls, cubes or comparable to their diameters. What is needed is that each carries the gluing data, that each contains $p$ , and that the diameters vanish.
Positive content is a hypothesis about the quotient, not about the estimate. Step 2.1 holds whatever $cont (E^{(m)})$ is; step 3.1 divides by it. A solid of content zero would make the left-hand side undefined rather than make the estimate fail.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

Green's first identity on a glued elementary solid region

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t}$ , let $O$ be an open set containing $E$ , let $u : O \to R$ be $C^{1}$ and let $v : O \to R$ be $C^{2}$ . Then

$∭_{E} (⟨ \nabla u, \nabla v ⟩ + u Δ v) = \iint_{\partial E} u ⟨ \nabla v, n ⟩,$

the right-hand side being the flux of the field $u \nabla v$ over $Σ^{o u t}$ .

No symmetry between $u$ and $v$ is claimed: the hypotheses on them differ.

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , the open $O \supseteq E$ , the $C^{1}$ function $u$ and the $C^{2}$ function $v$ on $O$ .

[F1]

For scalar-valued $f$ the gradient is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[F2]

For a $C^{2}$ function $f$ on an open subset of $R^{n}$ , $Δ f = div \nabla f = \sum_{i < n} \partial_{i} \partial_{i} f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F3]

A scalar $f$ is of class $C^{k}$ on $U$ when every iterated derivative of length at most $k$ exists and is continuous on $U$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space), and a map is $C^{k}$ when each component is ( $C^{k}$ Euclidean maps and diffeomorphisms).

[F4]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ), and the divergence of a $C^{1}$ field is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[L1]

Let $U \subseteq R^{n}$ be open, let $G : U \to R^{n}$ be $C^{1}$ and let $f : U \to R$ be $C^{1}$ . Then $f G$ is $C^{1}$ on $U$ and $div (f G) = ⟨ \nabla f, G ⟩ + f div G$ (Divergence and curl are linear and satisfy the scalar product rules).

[L2]

Proof

technique · direct

1.1givenF1F3

Since $v$ is $C^{2}$ on $O$ , [F1] and [F3] make each component $\partial_{i} v$ of $\nabla v$ a function with continuous first partial derivatives, so $\nabla v$ is a $C^{1}$ field on $O$ .

2.1step 1.1F2F4L1

The function $u$ is $C^{1}$ on $O$ and $\nabla v$ is a $C^{1}$ field there by step 1.1, so [L1] with $f = u$ and $G = \nabla v$ makes $u \nabla v$ a $C^{1}$ field on $O$ with $div (u \nabla v) = ⟨ \nabla u, \nabla v ⟩ + u div \nabla v = ⟨ \nabla u, \nabla v ⟩ + u Δ v,$ the last equality by [F2] and [F4].

3.1step 2.1F4L2∎

Applying [L2] to the $C^{1}$ field $u \nabla v$ on the open $O \supseteq E$ and substituting step 2.1 on the left gives $∭_{E} (⟨ \nabla u, \nabla v ⟩ + u Δ v) = \iint_{\partial E} ⟨ u \nabla v, n ⟩$ , and by [F4] the boundary integrand is $u ⟨ \nabla v, n ⟩$ . That is the asserted identity.

Remarks

The regularity is asymmetric because the identity is. The left-hand side applies $Δ$ to $v$ and only $\nabla$ to $u$ , so $v$ must be $C^{2}$ and $u$ need only be $C^{1}$ . Interchanging them is a different statement and needs $u$ to be $C^{2}$ as well; that is Green's second identity on a glued elementary solid region.
The boundary integrand is the normal derivative of $v$ , weighted by $u$ . The quantity $⟨ \nabla v, n ⟩$ is the derivative of $v$ in the direction of the boundary normal, and the identity says that its $u$ -weighted boundary integral is controlled by $Δ v$ and by the pairing of the two gradients inside the solid. Taking $u$ identically $1$ makes the first volume term vanish, which is the form used on the companion examples page.

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Green's second identity on a glued elementary solid region

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t}$ , let $O$ be an open set containing $E$ , and let $u, v : O \to R$ both be $C^{2}$ . Then

$∭_{E} (u Δ v - v Δ u) = \iint_{\partial E} (u ⟨ \nabla v, n ⟩ - v ⟨ \nabla u, n ⟩) .$

Both functions are required to be $C^{2}$ , which is a stronger hypothesis than the first identity places on either of them.

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , the open $O \supseteq E$ , and the $C^{2}$ functions $u, v$ on $O$ .

[F1]

For scalar-valued $f$ the gradient is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case), and $Δ f = div \nabla f$ for $C^{2}$ $f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F2]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ ; in particular $⟨ x, y ⟩ = ⟨ y, x ⟩$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

A scalar $f$ is of class $C^{k}$ on $U$ when every iterated derivative of length at most $k$ exists and is continuous on $U$ ; in particular a $C^{2}$ function is $C^{1}$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F4]

The flux over a finite patch presentation is a finite sum of parameter integrals of continuous integrands (The divergence theorem for finite gluings of elementary solid regions).

[L1]

Under the hypotheses above with $u$ of class $C^{1}$ and $v$ of class $C^{2}$ , $∭_{E} (⟨ \nabla u, \nabla v ⟩ + u Δ v) = \iint_{\partial E} u ⟨ \nabla v, n ⟩$ (Green's first identity on a glued elementary solid region).

[L2]

[L3]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set), and for a finite gluing $E$ is compact and Jordan measurable (Internal faces cancel and volume integrals add when elementary solid regions are glued).

Proof

technique · direct

1.1givenF3L1

Both $u$ and $v$ are $C^{2}$ on $O$ , hence also $C^{1}$ there by [F3]. So [L1] applies as it stands and gives $∭_{E} (⟨ \nabla u, \nabla v ⟩ + u Δ v) = \iint_{\partial E} u ⟨ \nabla v, n ⟩$ ; and it applies again with the roles of the two functions exchanged, which is legitimate exactly because both are $C^{2}$ , giving $∭_{E} (⟨ \nabla v, \nabla u ⟩ + v Δ u) = \iint_{\partial E} v ⟨ \nabla u, n ⟩$ .

2.1givenF1F3F4L2L3

All the integrands appearing in step 1.1 are continuous: $\nabla u$ and $\nabla v$ have continuous components by [F1] and [F3], $Δ u$ and $Δ v$ are continuous by [F1] and [F3], and each boundary integrand is a continuous function on a compact Jordan parameter region by [F4]. So every one of them is integrable over the relevant set by [L3], and differences of them may be taken inside the integrals by [L2].

3.1step 1.1step 2.1F2L2∎

Subtract the second identity of step 1.1 from the first, using step 2.1 to combine the integrals. By the symmetry of the inner product in [F2] the two terms $⟨ \nabla u, \nabla v ⟩$ and $⟨ \nabla v, \nabla u ⟩$ are equal and cancel, leaving $∭_{E} (u Δ v - v Δ u)$ on the left and $\iint_{\partial E} (u ⟨ \nabla v, n ⟩ - v ⟨ \nabla u, n ⟩)$ on the right.

Remarks

What the extra hypothesis buys. The first identity needs only one of the two functions to be $C^{2}$ ; using it twice with the roles exchanged needs both. That is the whole difference between the two identities, and it is why the second is stated separately rather than as a rearrangement of the first.
The cancellation is the symmetry of the inner product, nothing more. No integration by parts and no mixed-partials theorem enters here: the term that cancels is literally the same function written two ways.

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The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region

Definition

A $C^{2}$ patch over a finite elementary Green region is a regular parametrized surface patch $(D, φ)$ in the sense of Regular parametrized surface patches on compact Jordan parameter regions whose parameter region $D$ is supplied, in addition, with a decomposition making it a finite elementary Green region in the sense of Type I, Type II, and elementary regions for Green's theorem, and whose parametrization $φ$ is of class $C^{2}$ on an open neighbourhood of $D$ ( $C^{k}$ Euclidean maps and diffeomorphisms). Both requirements on $D$ are part of the data: it is a compact Jordan parameter region, so it is the closure of its nonempty connected interior, and it carries a supplied elementary decomposition.

Let $\partial D = (σ_{1}, \dots, σ_{m})$ be the positive boundary chain of that decomposition, the finite list of oriented piecewise- $C^{1}$ arcs of Positive orientation of elementary-region boundaries. Then the induced boundary chain is the list of arcs obtained by composing the positive boundary chain of the parameter region with the parametrization, namely

$φ (\partial D) : = (φ \circ σ_{1}, \dots, φ \circ σ_{m}),$

each entry a piecewise- $C^{1}$ path in $R^{3}$ in the sense of Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations. For a continuous vector field $F$ on a set containing $φ [D]$ , the circulation of $F$ around the induced boundary chain is the finite sum

$\oint_{φ (\partial D)} F \cdot d r : = \sum_{l = 1}^{m} \int_{φ \circ σ_{l}} F \cdot d r,$

with the vector line integrals of Scalar line integrals with respect to arc length and vector-field line integrals. The value does not depend on the order of the list, a finite sum of reals being independent of its order.

If instead $F$ is defined on an open set $U$ containing $φ [D]$ , continuity of $φ$ and compactness of $D$ give an open neighbourhood $V$ of $D$ in the domain of $φ$ with $φ [V] \subseteq U$ . On $V$ , the pulled-back functions are the inner products of the field along the parametrization with the two parameter derivatives:

$P^{*} : = ⟨ F \circ φ, φ_{u} ⟩, Q^{*} : = ⟨ F \circ φ, φ_{v} ⟩,$

with the inner product of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ . The oriented area vector $φ_{u} \times φ_{v}$ and the flux it computes are those of Unit normal fields, orientations, and flux through a regular surface patch. A merely continuous field on an arbitrary set containing $φ [D]$ is enough for circulation, but not for these neighbourhood-defined pullbacks.

Remarks

The orientation of the boundary is defined mechanically, not by a hand rule. Which way the induced boundary chain runs is decided entirely by Positive orientation of elementary-region boundaries in the parameter plane and then transported by $φ$ . The informal descriptions in the literature — walking along the curve with the head pointing along the normal and the surface on the left, or the right-hand rule — agree with this, but none of them is used here as a definition, and none of them is quoted as one. What makes the sign agreement a fact rather than a convention is that Green's theorem is proved on the parameter region.
A closed disc is not a legal parameter region here. An elementary Green region is bounded by continuous piecewise- $C^{1}$ graphs over a nondegenerate interval, and the two semicircular graphs of a disc are not piecewise $C^{1}$ at the endpoints. Every parameter region used with this definition on this page is a rectangle; a disc-shaped patch image is obtained instead by a polar parametrization over a rectangle, whose induced boundary chain then has two radial edges that cancel and one degenerate edge.
Why $C^{2}$ and where the elementary decomposition is spent. When $F$ is $C^{1}$ on the open set $U$ , the class $C^{2}$ makes the parameter derivatives $φ_{u}, φ_{v}$ of class $C^{1}$ , so the pullback coefficients $P^{*}, Q^{*}$ are differentiable on a neighbourhood of $D$ ; then The curl flux integrand of a $C^{2}$ patch is a two-dimensional curl of the pulled-back field uses $C^{2}$ once more to exchange the mixed second parameter derivatives of $φ$ . The elementary decomposition of $D$ is what lets Green's theorem be applied on the parameter region, and the positive boundary chain it carries is what the induced chain is the image of.

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A vector line integral along an image arc is the parameter line integral of the pulled-back field

Statement

Let $O \subseteq R^{2}$ be open, let $φ : O \to R^{3}$ be $C^{1}$ , let $σ = (σ_{1}, σ_{2}) : [a, b] \to O$ be a piecewise- $C^{1}$ path, and let $F$ be a continuous vector field on a set containing $φ (σ ([a, b]))$ . Put $P^{*} = ⟨ F \circ φ, φ_{u} ⟩$ and $Q^{*} = ⟨ F \circ φ, φ_{v} ⟩$ where these are defined. Then $φ \circ σ$ is a piecewise- $C^{1}$ path in $R^{3}$ and the vector line integral of the field along the image arc equals the parameter line integral of the pulled-back pair:

$\int_{φ \circ σ} F \cdot d r = \int_{σ} (P^{*}, Q^{*}) \cdot d r .$

Facts & Assumptions

Given: The open $O \subseteq R^{2}$ , the $C^{1}$ map $φ : O \to R^{3}$ , the piecewise- $C^{1}$ path $σ : [a, b] \to O$ , and the continuous field $F$ on a set containing the image of the trace of $σ$ under $φ$ .

[F1]

For a piecewise- $C^{1}$ path $γ : [a, b] \to R^{n}$ with $a < b$ , an admissible partition $a = t_{0} < \dots < t_{m} = b$ and continuous derivative extensions $v_{i}$ on the pieces, $\int_{γ} F \cdot d r = \sum_{i < m} \int_{t_{i}}^{t_{i + 1}} ⟨ F (γ (t)), v_{i} (t) ⟩ d t$ ; if $a = b$ the integral is $0$ (Scalar line integrals with respect to arc length and vector-field line integrals).

[F2]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

A piecewise- $C^{1}$ path admits a partition on whose pieces its derivative has a continuous extension, and constant paths are allowed (Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[F4]

The pulled-back functions of a patch and a field are $P^{*} = ⟨ F \circ φ, φ_{u} ⟩$ and $Q^{*} = ⟨ F \circ φ, φ_{v} ⟩$ (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region).

[F5]

A map is $C^{k}$ when each component is ( $C^{k}$ Euclidean maps and diffeomorphisms), and the Jacobian matrix of $φ$ has columns $φ_{u}, φ_{v}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case); a regular patch's parametrization is $C^{1}$ on an open neighbourhood of its parameter region (Regular parametrized surface patches on compact Jordan parameter regions).

[L1]

[L2]

If $f$ is totally differentiable at $a$ then $D_{w} f (a) = D f (a) w$ for every $w$ , and the matrix of $D f (a)$ is $J f (a)$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L3]

[L4]

A continuous function on a closed bounded interval is Riemann integrable (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

Proof

technique · direct

1.1givenF1F3

If $a = b$ then both line integrals are $0$ by [F1] and the identity holds. Assume $a < b$ , and by [F3] fix an admissible partition $a = t_{0} < \dots < t_{m} = b$ and continuous extensions $v_{i} = (v_{i, 1}, v_{i, 2})$ of $σ^{'}$ on the pieces $[t_{i}, t_{i + 1}]$ .

2.1step 1.1F5L1L2L3

Fix $i$ and let $t$ be interior to $[t_{i}, t_{i + 1}]$ . By [F5] and [L3] the map $φ$ is totally differentiable at $σ (t)$ , so [L1] and [L2] give that $φ \circ σ$ is differentiable at $t$ with $(φ \circ σ)^{'} (t) = D φ (σ (t)) σ^{'} (t) = φ_{u} (σ (t)) v_{i, 1} (t) + φ_{v} (σ (t)) v_{i, 2} (t),$ the second equality because by [L2] and [F5] the matrix of $D φ$ has columns $φ_{u}$ and $φ_{v}$ . The right-hand side is continuous in $t$ on the whole of $[t_{i}, t_{i + 1}]$ , since $φ_{u}, φ_{v}$ are continuous by [F5] and $v_{i}$ is continuous; so it is a continuous extension of $(φ \circ σ)^{'}$ on that piece, and $φ \circ σ$ is a piecewise- $C^{1}$ path with that admissible partition.

3.1step 2.1F2F4L4

On each piece, pairing the extension of step 2.1 with $F (φ (σ (t)))$ and using [F2] gives $⟨ F (φ (σ (t))), (φ \circ σ)^{'} (t) ⟩ = ⟨ F (φ (σ (t))), φ_{u} (σ (t)) ⟩ v_{i, 1} (t) + ⟨ F (φ (σ (t))), φ_{v} (σ (t)) ⟩ v_{i, 2} (t),$ which by [F4] is $P^{*} (σ (t)) v_{i, 1} (t) + Q^{*} (σ (t)) v_{i, 2} (t) = ⟨ (P^{*}, Q^{*}) (σ (t)), v_{i} (t) ⟩$ . Both sides are continuous on the piece, hence integrable by [L4].

4.1step 3.1F1∎

Summing the integrals of step 3.1 over the $m$ pieces and reading each side by [F1] — the left as the vector line integral of $F$ along $φ \circ σ$ with the partition of step 2.1, the right as the vector line integral of $(P^{*}, Q^{*})$ along $σ$ with the partition of step 1.1 — gives the asserted identity. A piece on which $σ$ is constant has $v_{i} = 0$ and contributes $0$ to both sides.

Remarks

No regularity of the patch is used. The parametrization need only be $C^{1}$ near the trace of $σ$ ; nothing here asks that $φ_{u} \times φ_{v}$ be nonzero, and nothing asks $σ$ to be injective or the trace to avoid the parameter boundary. That matters because the arcs of a positive boundary chain lie exactly on the parameter boundary, where a patch is allowed to be irregular.
The identity is an equality of two integrals, not a reparametrization statement. The path $φ \circ σ$ traverses a curve in $R^{3}$ and $σ$ traverses one in the parameter plane; what is being compared is the integral of $F$ along the first with the integral of a different field, $(P^{*}, Q^{*})$ , along the second.

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The curl flux integrand of a $C^{2}$ patch is a two-dimensional curl of the pulled-back field

Statement

Let $O \subseteq R^{2}$ be open, let $φ : O \to R^{3}$ be $C^{2}$ , let $U \subseteq R^{3}$ be open with $φ [O] \subseteq U$ and let $F : U \to R^{3}$ be $C^{1}$ . Put $P^{*} = ⟨ F \circ φ, φ_{u} ⟩$ and $Q^{*} = ⟨ F \circ φ, φ_{v} ⟩$ on $O$ . Then $P^{*}$ and $Q^{*}$ are $C^{1}$ on $O$ and, at every point of $O$ , the difference of the two pulled-back partial derivatives equals the curl flux integrand:

$\partial_{u} Q^{*} - \partial_{v} P^{*} = ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩ .$

No regularity of the patch is used: the identity holds also at parameter points where $φ_{u} \times φ_{v} = 0$ .

Facts & Assumptions

Given: The open sets $O \subseteq R^{2}$ and $U \subseteq R^{3}$ , the $C^{2}$ map $φ : O \to R^{3}$ with $φ [O] \subseteq U$ , and the $C^{1}$ field $F : U \to R^{3}$ .

[F1]

In the present local setting, define the pulled-back functions directly by $P^{*} = ⟨ F \circ φ, φ_{u} ⟩$ and $Q^{*} = ⟨ F \circ φ, φ_{v} ⟩$ on $O$ . For a regular patch over a finite elementary Green region these agree with the notation of The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region.

[F2]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

For $u, v \in R^{3}$ , $u \times v = (u_{y} v_{z} - u_{z} v_{y}, u_{z} v_{x} - u_{x} v_{z}, u_{x} v_{y} - u_{y} v_{x})$ (The cross product in $R^{3}$ ), and the curl of a $C^{1}$ field is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F4]

A map is of class $C^{k}$ when each component is ( $C^{k}$ Euclidean maps and diffeomorphisms), and a scalar is $C^{k}$ when every iterated derivative of length at most $k$ exists and is continuous ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F5]

[L1]

For a $C^{1}$ field $F$ on an open $U \subseteq R^{3}$ , a point $p \in U$ and $u, v \in R^{3}$ , $⟨ D F (p) u, v ⟩ - ⟨ D F (p) v, u ⟩ = ⟨ curl F (p), u \times v ⟩$ (The curl measures the antisymmetric part of the total derivative).

[L2]

[L3]

[L4]

[L5]

[L6]

For real functions of one real variable differentiable at a point, $f + g$ is differentiable there with $(f + g)^{'} = f^{'} + g^{'}$ and $f g$ is differentiable there with $(f g)^{'} = f^{'} g + f g^{'}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · direct

1.1givenF1F2F4L3L4L5L6

Since $φ$ is $C^{2}$ on $O$ , [F4] makes each $\partial_{u} φ_{i}$ and $\partial_{v} φ_{i}$ a $C^{1}$ function on $O$ ; and since $F$ is $C^{1}$ on $U$ with $φ [O] \subseteq U$ , [L3], [L4] and [L5] make each $F_{i} \circ φ$ differentiable in each parameter with $\partial_{u} (F_{i} \circ φ) = \sum_{j} (\partial_{j} F_{i}) (φ) \partial_{u} φ_{j}, \partial_{v} (F_{i} \circ φ) = \sum_{j} (\partial_{j} F_{i}) (φ) \partial_{v} φ_{j},$ both continuous on $O$ , so $F \circ φ$ is $C^{1}$ there. By [F1], [F2] and [L6], $P^{*} = \sum_{i} F_{i} (φ) \partial_{u} φ_{i}$ and $Q^{*} = \sum_{i} F_{i} (φ) \partial_{v} φ_{i}$ are then $C^{1}$ on $O$ .

2.1step 1.1F2F5L4L6

Differentiating $Q^{*}$ with respect to $u$ by [L6] and substituting step 1.1, $\partial_{u} Q^{*} = \sum_{i} (\sum_{j} (\partial_{j} F_{i}) (φ) \partial_{u} φ_{j}) \partial_{v} φ_{i} + \sum_{i} F_{i} (φ) \partial_{u} \partial_{v} φ_{i},$ and by [F2], [F5] and [L4] the first double sum is $⟨ D F (φ) φ_{u}, φ_{v} ⟩$ while the second is $⟨ F (φ), \partial_{u} \partial_{v} φ ⟩$ .

2.2step 1.1F2F5L4L6

The same computation for $P^{*}$ with respect to $v$ gives $\partial_{v} P^{*} = \sum_{i} (\sum_{j} (\partial_{j} F_{i}) (φ) \partial_{v} φ_{j}) \partial_{u} φ_{i} + \sum_{i} F_{i} (φ) \partial_{v} \partial_{u} φ_{i} = ⟨ D F (φ) φ_{v}, φ_{u} ⟩ + ⟨ F (φ), \partial_{v} \partial_{u} φ ⟩ .$

3.1step 2.1step 2.2F4L2

Each component $φ_{i}$ is $C^{2}$ on $O$ by [F4], so [L2] gives $\partial_{u} \partial_{v} φ_{i} = \partial_{v} \partial_{u} φ_{i}$ for every $i$ ; hence the two terms $⟨ F (φ), \partial_{u} \partial_{v} φ ⟩$ and $⟨ F (φ), \partial_{v} \partial_{u} φ ⟩$ of steps 2.1 and 2.2 are equal. This is the only place where $φ$ being $C^{2}$ rather than $C^{1}$ is used.

4.1step 2.1step 2.2step 3.1F3L1∎

Subtracting step 2.2 from step 2.1 and cancelling by step 3.1 leaves $\partial_{u} Q^{*} - \partial_{v} P^{*} = ⟨ D F (φ) φ_{u}, φ_{v} ⟩ - ⟨ D F (φ) φ_{v}, φ_{u} ⟩$ , which by [L1] applied at the point $φ$ with the vectors $φ_{u}$ and $φ_{v}$ is $⟨ curl F (φ), φ_{u} \times φ_{v} ⟩$ , the coordinates being those of [F3]. No step used $φ_{u} \times φ_{v} \neq 0$ .

Remarks

The right-hand side is a flux integrand, but the identity is not about flux. It is a pointwise equality of two continuous functions on $O$ . Reading its right side as the flux integrand of $curl F$ through the patch requires the patch to be regular; the identity itself does not, which is why it also holds along the parameter boundary, where a regular patch is allowed to degenerate.
What each hypothesis is for. $F$ being $C^{1}$ makes $curl F$ exist and makes the chain rule of step 1.1 available; $φ$ being $C^{2}$ makes $φ_{u}$ and $φ_{v}$ differentiable, so that steps 2.1 and 2.2 can be written at all, and makes the two mixed second derivatives equal in step 3.1.

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The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region

Statement

Let $(D, φ)$ be a $C^{2}$ patch over a finite elementary Green region (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region), with positive boundary chain $\partial D = (σ_{1}, \dots, σ_{m})$ and induced boundary chain $φ (\partial D)$ , and let $F$ be a $C^{1}$ vector field on an open set $U \subseteq R^{3}$ containing $φ [D]$ . Then the circulation around the induced boundary chain equals the flux of the curl in the induced orientation:

$\oint_{φ (\partial D)} F \cdot d r = \int_{D} ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩ .$

The right-hand side is the flux of $curl F$ through the patch in the orientation induced by $φ$ , in the sense of Unit normal fields, orientations, and flux through a regular surface patch.

Facts & Assumptions

Given: The $C^{2}$ patch $(D, φ)$ over a finite elementary Green region with its supplied decomposition and positive boundary chain, and the $C^{1}$ field $F$ on the open $U \supseteq φ [D]$ .

[F1]

A $C^{2}$ patch over a finite elementary Green region is a regular patch whose parameter region carries a supplied elementary decomposition and whose parametrization is $C^{2}$ on an open neighbourhood of that region; the induced boundary chain is the list of arcs obtained by composing the positive boundary chain of the parameter region with the parametrization, and the circulation around it is the finite sum of the vector line integrals along those arcs; the pulled-back functions are $P^{*} = ⟨ F \circ φ, φ_{u} ⟩$ and $Q^{*} = ⟨ F \circ φ, φ_{v} ⟩$ (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region).

[F2]

For a finite elementary Green region the boundary integral over the positive boundary chain is the finite sum $\int_{\partial D} G \cdot d r = \sum_{k} \int_{σ_{k}} G \cdot d r$ , and likewise $\int_{\partial D} P d u + Q d v$ for the field $(P, Q)$ (Positive orientation of elementary-region boundaries, Scalar line integrals with respect to arc length and vector-field line integrals).

[F3]

A finite elementary Green region is a nonempty finite union of elementary Green regions with pairwise disjoint interiors and the stated shared-arc conditions, supplied as data (Type I, Type II, and elementary regions for Green's theorem).

[F4]

For a regular patch $(D, φ)$ and a continuous field $G$ , the flux in the orientation induced by $φ$ is $\int_{D} (G \circ φ) \cdot (φ_{u} \times φ_{v})$ , with the inner product of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ (Unit normal fields, orientations, and flux through a regular surface patch).

[F5]

A regular patch's parametrization is defined and $C^{1}$ on an open neighbourhood of its compact Jordan parameter region (Regular parametrized surface patches on compact Jordan parameter regions), and a map is $C^{k}$ when each component is ( $C^{k}$ Euclidean maps and diffeomorphisms); the curl of a $C^{1}$ field is that of Divergence and curl of a $C^{1}$ vector field.

[L1]

Let $O \subseteq R^{2}$ be open, $φ : O \to R^{3}$ be $C^{1}$ , $σ$ a piecewise- $C^{1}$ path in $O$ , and $F$ continuous on a set containing the image of its trace. Then $\int_{φ \circ σ} F \cdot d r = \int_{σ} (P^{*}, Q^{*}) \cdot d r$ (A vector line integral along an image arc is the parameter line integral of the pulled-back field).

[L2]

Let $O \subseteq R^{2}$ be open, $φ : O \to R^{3}$ be $C^{2}$ with $φ [O] \subseteq U$ and $F : U \to R^{3}$ be $C^{1}$ . Then $P^{*}, Q^{*}$ are $C^{1}$ on $O$ and $\partial_{u} Q^{*} - \partial_{v} P^{*} = ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩$ (The curl flux integrand of a $C^{2}$ patch is a two-dimensional curl of the pulled-back field).

[L3]

Let $D = D_{1} \cup \dots \cup D_{N}$ be a finite elementary Green region with its supplied decomposition, oriented positively, and let $P, Q$ be $C^{1}$ on an open neighbourhood of $D$ . Then $\int_{\partial D} P d x + Q d y = \iint_{D} (\partial_{x} Q - \partial_{y} P) d A$ (Green's theorem for finite unions of elementary regions).

Proof

technique · direct

1.1givenF1F5

By [F1] and [F5] there is an open $O_{0} \supseteq D$ on which $φ$ is defined and $C^{2}$ . The set $φ^{- 1} [U] \cap O_{0}$ is open, since $φ$ is continuous and $U$ is open, and it contains $D$ because $φ [D] \subseteq U$ ; call it $O$ . Then $O$ is an open neighbourhood of $D$ with $φ$ of class $C^{2}$ on $O$ and $φ [O] \subseteq U$ .

2.1step 1.1F1L2

By [L2] applied on $O$ , the pulled-back functions $P^{*}$ and $Q^{*}$ of [F1] are $C^{1}$ on $O$ , an open neighbourhood of $D$ , and satisfy $\partial_{u} Q^{*} - \partial_{v} P^{*} = ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩$ there.

3.1step 2.1F1F3L3

The region $D$ is a finite elementary Green region with its supplied decomposition by [F1] and [F3], and $P^{*}, Q^{*}$ are $C^{1}$ on the open neighbourhood $O$ of $D$ by step 2.1. So [L3] applies with the parameter names $u, v$ in place of $x, y$ and gives $\int_{\partial D} P^{*} d u + Q^{*} d v = \iint_{D} (\partial_{u} Q^{*} - \partial_{v} P^{*}) d A .$

4.1step 1.1step 3.1F1F2L1

By [F2] the left-hand side of step 3.1 is $\sum_{k = 1}^{m} \int_{σ_{k}} (P^{*}, Q^{*}) \cdot d r$ . Each $σ_{k}$ is a piecewise- $C^{1}$ path with trace in $\partial D \subseteq D \subseteq O$ , and $F$ is continuous on $U \supseteq φ [O]$ , so [L1] rewrites each summand as $\int_{φ \circ σ_{k}} F \cdot d r$ ; summing and using [F1] identifies the left-hand side with $\oint_{φ (\partial D)} F \cdot d r$ .

5.1step 2.1step 3.1step 4.1F4F5∎

By step 2.1 the right-hand side of step 3.1 is $\int_{D} ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩$ , which by [F4] and [F5] is the flux of the $C^{1}$ field $curl F$ through $(D, φ)$ in the orientation induced by $φ$ . With step 4.1 this is the asserted identity.

Remarks

The identity needs no regularity of the patch; the flux reading does. Steps 3.1 and 4.1 use only that $φ$ is $C^{2}$ near $D$ and that $D$ carries an elementary decomposition. What the regularity of the patch supplies is the right to call $\int_{D} ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩$ a flux in an orientation, which is [F4]; at parameter points where the oriented area vector vanishes there is no orientation to speak of and the equality still holds.
What the surface is allowed to be. Nothing requires the patch image to be a graph over a coordinate plane, and nothing requires it to be embedded: the companion examples page checks the theorem on a lateral cylinder, which is a graph over no coordinate plane. What is required is that the parameter region be a finite elementary Green region, a hypothesis about the parameter plane and not about the image.
The two sides depend on the parametrization in the same way. Replacing $φ$ by a reparametrization that reverses orientation negates the oriented area vector and reverses the positive boundary chain's image, so both sides change sign together; nothing here asserts independence of the presentation, which is why the theorem is stated for a patch with its parametrization rather than for a surface.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A curl-free field has zero circulation around the induced boundary chain of a $C^{2}$ patch

Statement

Let $(D, φ)$ be a $C^{2}$ patch over a finite elementary Green region and let $F$ be a $C^{1}$ vector field on an open set $U \subseteq R^{3}$ containing $φ [D]$ , with $curl F = 0$ at every point of $U$ . Then

$\oint_{φ (\partial D)} F \cdot d r = 0.$

The curl must vanish on an open set containing the whole patch image, not merely along the induced boundary chain. Equivalently, by A $C^{1}$ field on an open subset of $R^{3}$ is closed exactly when its curl vanishes, the hypothesis is that $F$ be closed on $U$ .

Facts & Assumptions

Given: The $C^{2}$ patch $(D, φ)$ over a finite elementary Green region, the open $U \supseteq φ [D]$ , and the $C^{1}$ field $F$ on $U$ with $curl F = 0$ throughout $U$ .

[F1]

[F2]

The circulation of $F$ around the induced boundary chain is the finite sum of the vector line integrals along the arcs $φ \circ σ_{l}$ (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region), and integration over a bounded Jordan measurable set is integration of the zero extension over a bounding rectangle (The Riemann integral of a bounded function over a bounded Jordan measurable set).

[L1]

For a $C^{2}$ patch over a finite elementary Green region and a $C^{1}$ field $F$ on an open set containing the patch image, the circulation around the induced boundary chain equals the flux of the curl in the induced orientation, $\oint_{φ (\partial D)} F \cdot d r = \int_{D} ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩$ (The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region).

[L2]

A $C^{1}$ field on an open subset of $R^{3}$ is closed if and only if its curl vanishes identically (A $C^{1}$ field on an open subset of $R^{3}$ is closed exactly when its curl vanishes).

[L3]

Proof

technique · direct

1.1givenF1F2L3

Since $φ [D] \subseteq U$ and $curl F$ vanishes at every point of $U$ by hypothesis and [F1], the integrand $⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩$ is identically zero on $D$ . Its zero extension to a bounding rectangle is the zero function, which by [L3] with $α = β = 0$ is integrable with integral $0$ , so $\int_{D} ⟨ (curl F) \circ φ, φ_{u} \times φ_{v} ⟩ = 0$ by [F2].

2.1step 1.1F2L1L2∎

By [L1] the circulation around the induced boundary chain equals that integral, hence is $0$ . By [L2] the hypothesis $curl F = 0$ on $U$ is the same as $F$ being closed on $U$ , so the corollary may be read either way.

Remarks

A closed field can still have nonzero circulation around a loop. What this corollary rules out is a nonzero circulation around the induced boundary chain of a $C^{2}$ patch whose whole image lies where the curl vanishes. A closed field on a domain that carries no such patch spanning the loop may circulate: the companion examples page gives a field with circulation $2 π$ around a circle encircling the excluded axis, and no patch over a finite elementary Green region has image inside that domain and that circle as its induced boundary.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The normal component of the curl is the limiting circulation per unit area of shrinking discs

Statement

Let $O \subseteq R^{3}$ be open, let $F : O \to R^{3}$ be $C^{1}$ , let $p \in O$ and let $n \in R^{3}$ have $∥ n ∥_{2} = 1$ . Then there are $a, b \in R^{3}$ with

$∥ a ∥_{2} = ∥ b ∥_{2} = 1, ⟨ a, b ⟩ = ⟨ a, n ⟩ = ⟨ b, n ⟩ = 0, a \times b = n,$

and a real $r_{0} > 0$ such that for every $r$ with $0 < r \leq r_{0}$ the map

$φ_{r} (ρ, θ) : = p + ρ \cos θ a + ρ \sin θ b ((ρ, θ) \in D_{r} : = [0, r] \times [0, 2 π])$

is a $C^{2}$ patch over a finite elementary Green region whose image lies in $O$ , with $φ_{r, ρ} \times φ_{r, θ} = ρ n$ ; the circulation of $F$ around its induced boundary chain is the vector line integral of $F$ along the circle $C_{r} (t) : = p + r \cos t a + r \sin t b$ on $[0, 2 π]$ ; and

$\lim_{r \to 0^{+}} \frac{1}{π r^{2}} \int_{C_{r}} F \cdot d r = ⟨ curl F (p), n ⟩,$

meaning: for every real $ε > 0$ there is a real $δ > 0$ such that every $r$ with $0 < r \leq r_{0}$ and $r < δ$ satisfies $∣ \frac{1}{π r^{2}} \int_{C_{r}} F \cdot d r - ⟨ curl F (p), n ⟩ ∣ \leq ε$ .

Facts & Assumptions

Given: The open $O \subseteq R^{3}$ , the $C^{1}$ field $F$ on $O$ , the point $p \in O$ , the unit vector $n$ , and the notation $D_{r} = [0, r] \times [0, 2 π]$ of the Statement.

[F1]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ and $∥ x ∥_{2} = \sqrt{⟨ x, x ⟩}$ ; the inner product is symmetric and bilinear, and $⟨ x, x ⟩ = 0$ only for $x = 0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). The standard unit vector $e_{k}$ has $k$ th coordinate $1$ and the others $0$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[F2]

For $u, v \in R^{3}$ , $u \times v = (u_{y} v_{z} - u_{z} v_{y}, u_{z} v_{x} - u_{x} v_{z}, u_{x} v_{y} - u_{y} v_{x})$ (The cross product in $R^{3}$ ); the curl of a $C^{1}$ field is that of Divergence and curl of a $C^{1}$ vector field.

[F3]

A compact Type I region is ${(s, t) : a \leq s \leq b, α (s) \leq t \leq β (s)}$ with $a < b$ and continuous piecewise- $C^{1}$ $α \leq β$ , strict on $(a, b)$ ; it is compact and Jordan measurable, an elementary Green region admits both descriptions, and a finite elementary Green region is a nonempty finite union of them with the stated conditions (Type I, Type II, and elementary regions for Green's theorem).

[F4]

The positive boundary of a Type I region traverses the lower graph from left to right, the right endpoint arc upward, the upper graph from right to left, and the left endpoint arc downward, omitting zero-length arcs; the boundary integral over the resulting chain is the finite sum over its arcs (Positive orientation of elementary-region boundaries).

[F5]

A regular parametrized surface patch has a compact Jordan parameter region that is the closure of its nonempty connected interior, a parametrization $C^{1}$ on an open neighbourhood of it, nonvanishing parameter cross product on the interior, and no interior parameter point sharing its image with a distinct point of the region (Regular parametrized surface patches on compact Jordan parameter regions); a $C^{2}$ patch over a finite elementary Green region adds the supplied elementary decomposition and the class $C^{2}$ (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region).

[F6]

A vector line integral along a piecewise- $C^{1}$ path is $\sum_{i} \int_{t_{i}}^{t_{i + 1}} ⟨ F (γ (t)), v_{i} (t) ⟩ d t$ , and is $0$ on a degenerate parameter interval (Scalar line integrals with respect to arc length and vector-field line integrals); reversal of a path is $γ^{-} (t) = γ (a + b - t)$ and constant paths are allowed (Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[F7]

A set is open in a metric space when each of its points has a ball around it inside the set (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement); a map is continuous at a point when every $ε > 0$ admits a $δ > 0$ carrying the $δ$ -ball into the $ε$ -ball (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form); and $\lim_{x \to c} f (x) = L$ has the usual meaning (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). Integration over a bounded Jordan set is that of The Riemann integral of a bounded function over a bounded Jordan measurable set, and $C^{k}$ is the componentwise class of $C^{k}$ Euclidean maps and diffeomorphisms.

[L1]

The cross product is bilinear and alternating, $⟨ u \times v, w ⟩ = \det [u v w]$ , and $u \times v$ is orthogonal to both $u$ and $v$ (The cross product is bilinear, alternating, and orthogonal to both factors).

[L2]

For $u, v \in R^{3}$ , $∥ u \times v ∥_{2}^{2} = ∥ u ∥_{2}^{2} ∥ v ∥_{2}^{2} - ⟨ u, v ⟩^{2}$ , and this is positive exactly when $u$ and $v$ are linearly independent (The squared cross-product norm is the Gram determinant of two vectors).

[L3]

$(\sin t)^{'} = \cos t$ , $(\cos t)^{'} = - \sin t$ , $\sin 0 = 0$ and $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine); $\sin^{2} t + \cos^{2} t = 1$ and $\sin (- t) = - \sin t$ , $\cos (- t) = \cos t$ (Parity and the Pythagorean identity for sine and cosine).

[L4]

$\sin (s + t) = \sin s \cos t + \cos s \sin t$ and $\cos (s + t) = \cos s \cos t - \sin s \sin t$ (The addition formulas for sine and cosine).

[L5]

$\sin x = 0$ if and only if $x = m π$ for some integer $m$ , and both sine and cosine have period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi); $\sin π = 0$ and $\cos π = - 1$ (Quarter-turn values and shifts by pi/2 and pi).

[L6]

For a bounded Jordan set $E \subseteq R^{p + q}$ and integrable $g$ whose sections are integrable outside a content-zero set, $\int_{E} g = \int h (x) d x$ with $h (x) = \int_{E_{x}} g_{x}$ (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable).

[L7]

If $G$ is differentiable at every point of $[a, b]$ with $a < b$ and $G^{'}$ is integrable there, then $\int_{a}^{b} G^{'} = G (b) - G (a)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ); sums, scalar multiples and products of differentiable functions differentiate by the usual rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L8]

[L9]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set).

[L10]

Vector line integrals negate under reversal (Line integrals under reversal and concatenation).

[L11]

For a $C^{2}$ patch over a finite elementary Green region and a $C^{1}$ field on an open set containing the patch image, the circulation around the induced boundary chain equals the flux of the curl in the induced orientation (The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region).

Proof

technique · constructive

1.1givenF1construct

Since $\sum_{k} n_{k}^{2} = ∥ n ∥_{2}^{2} = 1$ by [F1], not all three coordinates can have $n_{k}^{2} > 1 / 3$ ; fix $k$ with $n_{k}^{2} \leq 1 / 3$ . Then $n$ and $e_{k}$ are linearly independent: a relation $e_{k} = λ n$ would force $∣ λ ∣ = 1$ by comparing norms, hence $n = \pm e_{k}$ and $n_{k}^{2} = 1$ , contradicting $n_{k}^{2} \leq 1 / 3$ ; and $n \neq 0$ .

1.2F1F2algebra

For all $u, v, w \in R^{3}$ , expanding by [F1] and [F2] gives $⟨ u \times v, w ⟩ = (u_{y} v_{z} - u_{z} v_{y}) w_{x} + (u_{z} v_{x} - u_{x} v_{z}) w_{y} + (u_{x} v_{y} - u_{y} v_{x}) w_{z},$ $⟨ w \times u, v ⟩ = (w_{y} u_{z} - w_{z} u_{y}) v_{x} + (w_{z} u_{x} - w_{x} u_{z}) v_{y} + (w_{x} u_{y} - w_{y} u_{x}) v_{z},$ and the six signed monomials of the first expression are those of the second, matched as $u_{y} v_{z} w_{x}$ with $w_{x} u_{y} v_{z}$ , $- u_{z} v_{y} w_{x}$ with $- w_{x} u_{z} v_{y}$ , $u_{z} v_{x} w_{y}$ with $w_{y} u_{z} v_{x}$ , $- u_{x} v_{z} w_{y}$ with $- w_{y} u_{x} v_{z}$ , $u_{x} v_{y} w_{z}$ with $w_{z} u_{x} v_{y}$ and $- u_{y} v_{x} w_{z}$ with $- w_{z} u_{y} v_{x}$ . Hence $⟨ u \times v, w ⟩ = ⟨ w \times u, v ⟩$ .

1.3givenF7choose

The set $O$ is open and $p \in O$ , so by [F7] there is a real $r_{0} > 0$ with every $q$ satisfying $∥ q - p ∥_{2} \leq r_{0}$ lying in $O$ ; take such an $r_{0}$ .

1.4givenF3L6L7L9

Fix $r$ with $0 < r \leq r_{0}$ . The function $(ρ, θ) \mapsto ρ$ is continuous on the compact Jordan rectangle $D_{r}$ , hence integrable by [L9] and [F3]. Its sections in $θ$ are the continuous functions $ρ \mapsto ρ$ on $[0, r]$ , so [L6] gives $\int_{D_{r}} ρ = \int_{0}^{2 π} (\int_{0}^{r} ρ d ρ) d θ$ ; by [L7] with $G (ρ) = ρ^{2} / 2$ the inner integral is $r^{2} / 2$ , and again by [L7] with $G (θ) = r^{2} θ / 2$ the outer integral is $π r^{2}$ . So $\int_{D_{r}} ρ = π r^{2}$ .

2.1step 1.1F1L1L2construct

By step 1.1 and [L2] the number $∥ n \times e_{k} ∥_{2}^{2}$ is positive, so $n \times e_{k} \neq 0$ ; put $a : = \frac{n \times e_{k}}{∥ n \times e_{k} ∥_{2}} .$ Then $∥ a ∥_{2} = 1$ by [F1], and $⟨ a, n ⟩ = 0$ because $n \times e_{k}$ is orthogonal to $n$ by [L1].

3.1step 2.1F1L1L2construct

Put $b : = n \times a$ . By [L1] it is orthogonal to $n$ and to $a$ , so $⟨ b, n ⟩ = ⟨ b, a ⟩ = 0$ ; and by [L2] with step 2.1, $∥ b ∥_{2}^{2} = ∥ n ∥_{2}^{2} ∥ a ∥_{2}^{2} - ⟨ n, a ⟩^{2} = 1$ , so $∥ b ∥_{2} = 1$ .

4.1step 1.2step 2.1step 3.1F1L2

By step 1.2 with $u = a$ , $v = b$ and $w = n$ , and then step 3.1, $⟨ a \times b, n ⟩ = ⟨ n \times a, b ⟩ = ⟨ b, b ⟩ = 1$ . By [L2] and steps 2.1 and 3.1, $∥ a \times b ∥_{2}^{2} = ∥ a ∥_{2}^{2} ∥ b ∥_{2}^{2} - ⟨ a, b ⟩^{2} = 1$ . Hence $∥ a \times b - n ∥_{2}^{2} = ∥ a \times b ∥_{2}^{2} - 2 ⟨ a \times b, n ⟩ + ∥ n ∥_{2}^{2} = 1 - 2 + 1 = 0$ by [F1], and positive definiteness in [F1] gives $a \times b = n$ .

4.2step 2.1step 3.1F7L1L3L7

By [L3] and [L7] the map $φ_{r}$ is differentiable in each parameter with $φ_{r, ρ} = \cos θ a + \sin θ b$ and $φ_{r, θ} = ρ (- \sin θ a + \cos θ b)$ , and all its iterated parameter derivatives of order at most $2$ exist and are continuous, so $φ_{r}$ is $C^{2}$ on the whole plane by [F7]. Expanding by bilinearity and the alternating law in [L1], $φ_{r, ρ} \times φ_{r, θ} = ρ (\cos^{2} θ (a \times b) - \sin^{2} θ (b \times a)) = ρ (\cos^{2} θ + \sin^{2} θ) (a \times b),$ which is $ρ (a \times b)$ by [L3].

5.1step 4.1step 4.2

Combining steps 4.1 and 4.2, $φ_{r, ρ} \times φ_{r, θ} = ρ n$ , which is nonzero exactly when $ρ > 0$ .

6.1step 1.3step 2.1step 3.1step 5.1F1F3F5L3L4L5

The rectangle $D_{r}$ is a Type I and a Type II region with $0 < r$ and constant graphs $0 < 2 π$ , hence an elementary Green region and a nonempty finite elementary Green region with the one-piece decomposition, compact and Jordan measurable, and it is the closure of its nonempty convex, hence connected, interior $(0, r) \times (0, 2 π)$ ([F3], [F5]). The cross product of step 5.1 is nonzero on that interior. For injectivity, let $(ρ, θ)$ be interior and $(ρ^{'}, θ^{'}) \in D_{r}$ have the same image; pairing $ρ \cos θ a + ρ \sin θ b = ρ^{'} \cos θ^{'} a + ρ^{'} \sin θ^{'} b$ with $a$ and with $b$ and using steps 2.1 and 3.1 gives $ρ \cos θ = ρ^{'} \cos θ^{'}$ and $ρ \sin θ = ρ^{'} \sin θ^{'}$ ; squaring and adding with [L3] gives $ρ^{2} = ρ^{' 2}$ , so $ρ^{'} = ρ > 0$ and $\cos θ = \cos θ^{'}$ , $\sin θ = \sin θ^{'}$ . Then [L4] and [L3] give $\cos (θ - θ^{'}) = \cos θ \cos θ^{'} + \sin θ \sin θ^{'} = \cos^{2} θ^{'} + \sin^{2} θ^{'} = 1$ , so $\sin^{2} (θ - θ^{'}) = 0$ and $θ - θ^{'} = m π$ for an integer $m$ by [L3] and [L5]; since $θ \in (0, 2 π)$ and $θ^{'} \in [0, 2 π]$ we have $∣ θ - θ^{'} ∣ < 2 π$ , so $m \in {- 1, 0, 1}$ , and $\cos (\pm π) = - 1 \neq 1$ by [L5], leaving $θ = θ^{'}$ . Finally $∥ φ_{r} (ρ, θ) - p ∥_{2} = ρ \leq r \leq r_{0}$ by [F1], steps 2.1 and 3.1 and [L3], so the image lies in $O$ by step 1.3. Hence $(D_{r}, φ_{r})$ is a $C^{2}$ patch over a finite elementary Green region with image in $O$ .

7.1step 6.1F4F6L3L5L10

By [F4] the positive boundary chain of $D_{r}$ in its Type I description, with $ρ$ horizontal, is the four arcs $σ_{1} (t) = (t, 0)$ on $[0, r]$ , $σ_{2} (t) = (r, t)$ on $[0, 2 π]$ , $σ_{3} (t) = (r - t, 2 π)$ on $[0, r]$ and $σ_{4} (t) = (0, 2 π - t)$ on $[0, 2 π]$ . Composing with $φ_{r}$ and using $\cos 0 = \cos 2 π = 1$ , $\sin 0 = \sin 2 π = 0$ from [L3] and [L5]: $φ_{r} \circ σ_{1} (t) = p + t a$ , $φ_{r} \circ σ_{2} (t) = C_{r} (t)$ , $φ_{r} \circ σ_{3} (t) = p + (r - t) a$ and $φ_{r} \circ σ_{4} (t) = p$ . The third is the reversal of the first in the sense of [F6], so [L10] makes their integrals cancel; the fourth is constant, so its derivative extension is $0$ and its integral is $0$ by [F6]. Hence the circulation of $F$ around the induced boundary chain of $(D_{r}, φ_{r})$ is $\int_{C_{r}} F \cdot d r$ .

8.1step 5.1step 6.1step 7.1F1L9L11

By step 6.1 the pair $(D_{r}, φ_{r})$ satisfies the hypotheses of [L11], and $F$ is $C^{1}$ on the open $O$ containing $φ_{r} [D_{r}]$ . So [L11] and step 7.1 give $\int_{C_{r}} F \cdot d r = \int_{D_{r}} ⟨ (curl F) \circ φ_{r}, φ_{r, ρ} \times φ_{r, θ} ⟩ = \int_{D_{r}} ρ g_{r}, g_{r} (ρ, θ) : = ⟨ (curl F) (φ_{r} (ρ, θ)), n ⟩,$ using step 5.1 and the bilinearity of the inner product in [F1]; the integrand is continuous on the compact Jordan $D_{r}$ , hence integrable by [L9].

9.1step 1.4step 4.1step 6.1step 7.1step 8.1F7L8discharge-construct: the polar patch∎

Let $ε > 0$ be real. The field $curl F$ is continuous on $O$ and $⟨ \cdot, n ⟩$ is continuous, so [F7] gives $δ_{0} > 0$ such that $∣ ⟨ curl F (q), n ⟩ - ⟨ curl F (p), n ⟩ ∣ \leq ε$ for every $q \in O$ with $∥ q - p ∥_{2} < δ_{0}$ ; put $δ : = δ_{0}$ . Let $0 < r \leq r_{0}$ with $r < δ$ . Every point of $φ_{r} [D_{r}]$ is within $r < δ_{0}$ of $p$ by step 6.1, so $∣ g_{r} - ⟨ curl F (p), n ⟩ ∣ \leq ε$ on $D_{r}$ ; hence by [L8] and step 1.4 $∣ \int_{D_{r}} ρ g_{r} - ⟨ curl F (p), n ⟩ π r^{2} ∣ = ∣ \int_{D_{r}} ρ (g_{r} - ⟨ curl F (p), n ⟩) ∣ \leq ε \int_{D_{r}} ρ = ε π r^{2} .$ Dividing by $π r^{2} > 0$ and substituting step 8.1 gives $∣ \frac{1}{π r^{2}} \int_{C_{r}} F \cdot d r - ⟨ curl F (p), n ⟩ ∣ \leq ε$ , which by [F7] is the asserted limit; with steps 4.1, 6.1 and 7.1 every clause of the Statement is established.

Remarks

The orthonormal pair is built, not chosen by an extension theorem. Steps 1.1, 2.1 and 3.1 write $a$ and $b$ down from $n$ and one standard basis vector, and step 4.1 fixes the sign of $a \times b$ by a computation rather than by replacing $b$ with $- b$ after the fact. No choice principle and no basis-extension theorem is used, which matters because the general extension of an independent set to a basis in this library assumes the Axiom of Choice and would be a disproportionate hypothesis for a statement about $R^{3}$ .
The two radial edges are what make the chain a circle. The induced boundary chain of a polar patch has four arcs, and only one of them is the circle: the two radial ones are reverses of each other and the fourth is the constant path at the centre. That is why a disc-shaped patch may be used at all, since a closed disc is not an elementary Green region and cannot be a parameter region here.
No area comparison between the disc and its diameter is needed. The factor $π r^{2}$ appears on both sides of the estimate in step 9.1 and cancels; what drives the limit is the continuity of $curl F$ at $p$ alone.

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Green's theorem is the curl statement for a planar field lifted to $R^{3}$

Statement

Let $D$ be a finite elementary Green region with its supplied decomposition, positively oriented, and let $P, Q$ be $C^{1}$ on an open $U \subseteq R^{2}$ containing $D$ . Define the lift

$\tilde{F} (x, y, z) : = (P (x, y), Q (x, y), 0) ((x, y, z) \in U \times R),$

a field on the open set $U \times R \subseteq R^{3}$ . Then $\tilde{F}$ is $C^{1}$ , its curl has first and second coordinates identically $0$ and third coordinate $\partial_{x} Q - \partial_{y} P$ at every point, independent of $z$ , and the circulation of the planar field around the positive boundary chain equals the integral of the third coordinate of the curl of the lift:

$\int_{\partial D} (P, Q) \cdot d r = \iint_{D} (curl \tilde{F})_{z} (x, y, 0) d A .$

Facts & Assumptions

Given: The finite elementary Green region $D$ with its supplied decomposition and positive orientation, the $C^{1}$ functions $P, Q$ on the open $U \supseteq D$ , and the lift $\tilde{F}$ of the Statement.

[F1]

The curl of a $C^{1}$ field on an open subset of $R^{3}$ is $curl G = (\partial_{y} G_{z} - \partial_{z} G_{y}, \partial_{z} G_{x} - \partial_{x} G_{z}, \partial_{x} G_{y} - \partial_{y} G_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

A map is of class $C^{k}$ when each component is, a scalar component being $C^{1}$ when its first partial derivatives exist and are continuous ( $C^{k}$ Euclidean maps and diffeomorphisms).

[F3]

For a finite elementary Green region the positive boundary integral is the finite sum over the surviving oriented arcs, and $\int_{\partial D} G \cdot d r$ and $\int_{\partial D} P d x + Q d y$ denote that sum for the field $(P, Q)$ (Positive orientation of elementary-region boundaries, Scalar line integrals with respect to arc length and vector-field line integrals).

[F4]

[F5]

[L1]

Proof

technique · direct

1.1givenF2F5

The three components of $\tilde{F}$ are $(x, y, z) \mapsto P (x, y)$ , $(x, y, z) \mapsto Q (x, y)$ and the constant $0$ . Their first partial derivatives are $\partial_{x} {\tilde{F}}_{x} = \partial_{x} P$ , $\partial_{y} {\tilde{F}}_{x} = \partial_{y} P$ , $\partial_{z} {\tilde{F}}_{x} = 0$ ; $\partial_{x} {\tilde{F}}_{y} = \partial_{x} Q$ , $\partial_{y} {\tilde{F}}_{y} = \partial_{y} Q$ , $\partial_{z} {\tilde{F}}_{y} = 0$ ; and all three of $\partial_{x} {\tilde{F}}_{z}$ , $\partial_{y} {\tilde{F}}_{z}$ , $\partial_{z} {\tilde{F}}_{z}$ are $0$ . Each of these exists and is continuous on $U \times R$ because $P$ and $Q$ are $C^{1}$ on $U$ , so $\tilde{F}$ is $C^{1}$ there by [F2].

2.1step 1.1F1

By [F1] and step 1.1 the three coordinates of $curl \tilde{F}$ are $\partial_{y} {\tilde{F}}_{z} - \partial_{z} {\tilde{F}}_{y} = 0 - 0 = 0$ , then $\partial_{z} {\tilde{F}}_{x} - \partial_{x} {\tilde{F}}_{z} = 0 - 0 = 0$ , and then $\partial_{x} {\tilde{F}}_{y} - \partial_{y} {\tilde{F}}_{x} = \partial_{x} Q - \partial_{y} P$ . All three are computed, and the third depends only on $(x, y)$ , so its value at $(x, y, z)$ is its value at $(x, y, 0)$ .

3.1step 2.1F3F4L1∎

By [F4] the region $D$ carries its supplied decomposition and $P, Q$ are $C^{1}$ on the open neighbourhood $U$ of $D$ , so [L1] gives $\int_{\partial D} P d x + Q d y = \iint_{D} (\partial_{x} Q - \partial_{y} P) d A$ ; by [F3] the left side is $\int_{\partial D} (P, Q) \cdot d r$ , and by step 2.1 the integrand on the right is $(curl \tilde{F})_{z} (x, y, 0)$ . That is the asserted identity, and step 2.1 is the assertion about the three curl coordinates.

Remarks

This is a dictionary, not a new theorem. Both sides are the two sides of Green's theorem, rewritten. What the corollary records is that the planar integrand $\partial_{x} Q - \partial_{y} P$ is a curl, so that the planar and the spatial developments on this page speak about one operator rather than two unrelated ones.
The route is deliberately one-way. The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region is proved from Green's theorem, so re-deriving Green's theorem from it would be circular. Nothing above uses Stokes' theorem.

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The planar divergence theorem: the flux form of Green's theorem

Statement

Let $D$ be a finite elementary Green region with its supplied decomposition, positively oriented, and let $F = (F_{x}, F_{y})$ be $C^{1}$ on an open $U \subseteq R^{2}$ containing $D$ . Then

$\int_{\partial D} (- F_{y}) d x + F_{x} d y = \iint_{D} (\partial_{x} F_{x} + \partial_{y} F_{y}) d A,$

the right-hand integrand being the divergence of $F$ as a field on an open subset of $R^{2}$ .

Moreover, if $σ : [α, β] \to R^{2}$ is one of the arcs of the positive boundary chain and its derivative is nowhere zero on a piece with continuous derivative extension $v$ , then on that piece

$\int_{σ} (- F_{y}) d x + F_{x} d y = \int_{σ} ⟨ F, ν ⟩ d s, ν : = \frac{(v_{2}, - v_{1})}{∥ v ∥_{2}},$

where $ν$ is the unit vector obtained from the tangent $v$ by a quarter turn clockwise.

Facts & Assumptions

Given: The finite elementary Green region $D$ with its supplied decomposition and positive orientation, and the $C^{1}$ field $F = (F_{x}, F_{y})$ on the open $U \supseteq D$ .

[F1]

The divergence of a $C^{1}$ field on an open subset of $R^{n}$ is $div G = \sum_{i < n} \partial_{i} G_{i}$ ; for $n = 2$ and coordinates named $x, y$ this is $\partial_{x} G_{x} + \partial_{y} G_{y}$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

For a finite elementary Green region the positive boundary integral is the finite sum over the surviving oriented arcs, written $\int_{\partial D} P d x + Q d y$ for the field $(P, Q)$ (Positive orientation of elementary-region boundaries).

[F3]

For a piecewise- $C^{1}$ path with admissible partition and continuous derivative extensions $v_{i}$ , $\int_{γ} G \cdot d r = \sum_{i} \int_{t_{i}}^{t_{i + 1}} ⟨ G (γ (t)), v_{i} (t) ⟩ d t$ and $\int_{γ} h d s = \sum_{i} \int_{t_{i}}^{t_{i + 1}} h (γ (t)) ∥ v_{i} (t) ∥_{2} d t$ (Scalar line integrals with respect to arc length and vector-field line integrals, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[F4]

[F5]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ and $∥ x ∥_{2} = \sqrt{⟨ x, x ⟩}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ); a map is $C^{k}$ when each component is ( $C^{k}$ Euclidean maps and diffeomorphisms).

[L1]

Proof

technique · direct

1.1givenF4F5L1

Put $P : = - F_{y}$ and $Q : = F_{x}$ on $U$ . These are $C^{1}$ on $U$ by [F5], since the components of $F$ are, so [L1] applies with the supplied decomposition of [F4] and gives $\int_{\partial D} P d x + Q d y = \iint_{D} (\partial_{x} Q - \partial_{y} P) d A$ .

1.2givenF3F5

Fix such an arc $σ$ of the positive boundary chain, a piece of it carrying a continuous derivative extension $v = (v_{1}, v_{2})$ with $v$ nowhere zero there, and set $ν : = (v_{2}, - v_{1}) / ∥ v ∥_{2}$ . By [F5] the vector $ν$ has norm $1$ , since $v_{2}^{2} + v_{1}^{2} = ∥ v ∥_{2}^{2}$ , and $⟨ ν, v ⟩ = (v_{2} v_{1} - v_{1} v_{2}) / ∥ v ∥_{2} = 0$ . Writing $v = ∥ v ∥_{2} (\cos τ, \sin τ)$ for the direction of $v$ is not needed: the map $(s, t) \mapsto (t, - s)$ is the quarter turn clockwise, as it carries $(1, 0)$ to $(0, - 1)$ and $(0, 1)$ to $(1, 0)$ .

2.1step 1.1F1F2

By step 1.1 and [F1], $\partial_{x} Q - \partial_{y} P = \partial_{x} F_{x} - \partial_{y} (- F_{y}) = \partial_{x} F_{x} + \partial_{y} F_{y} = div F$ on $U$ ; substituting into step 1.1 and reading the left side by [F2] gives the first asserted identity.

3.1step 1.1step 1.2F3F5∎

By [F3] the integral of $P d x + Q d y$ over that piece of $σ$ is $\int (P (σ (t)) v_{1} (t) + Q (σ (t)) v_{2} (t)) d t = \int (- F_{y} (σ (t)) v_{1} (t) + F_{x} (σ (t)) v_{2} (t)) d t$ , while by step 1.2 and [F5] $⟨ F (σ (t)), ν (t) ⟩ ∥ v (t) ∥_{2} = F_{x} (σ (t)) v_{2} (t) - F_{y} (σ (t)) v_{1} (t)$ . The two integrands are equal, so by [F3] the two integrals over that piece agree, which is the second asserted identity.

Remarks

The first identity needs no regularity of the boundary arcs; the second does. The positive boundary chain of an elementary Green region is built from continuous piecewise- $C^{1}$ graphs, whose derivative may vanish, and where it vanishes there is no unit tangent and hence no $ν$ . That is why the normal reading is a separate clause under an extra hypothesis, and why the identity that Green's theorem actually delivers is stated in the $d x, d y$ form.
Outwardness of $ν$ is not claimed here. For a positively oriented boundary the quarter turn clockwise of the tangent does point out of the region, but establishing that at a boundary point requires the same kind of local analysis that At interior base points, the graph faces of an adapted presentation induce the outward unit normal carries out in space, and it is not carried out for plane regions on this page. What is proved is the equality of the two integrals for the stated $ν$ .
Why this is called a divergence theorem. The right-hand integrand is the divergence of a field on an open subset of $R^{2}$ , and the left-hand side is the boundary integral of the normal component. The three-dimensional statement of The divergence theorem for finite gluings of elementary solid regions has the same shape; neither is derived from the other on this page.

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What the classical divergence and Stokes theorems here do and do not cover

The decomposition is a hypothesis, not a conclusion. The divergence theorem for finite gluings of elementary solid regions applies to a solid supplied with its three simple descriptions per piece, its boundary presentation, its internal-or-outer designation and its pairing of internal patches (Finite gluings of elementary solid regions and their outward boundary presentation). It does not say that a compact set with a piecewise smooth boundary admits such data, and it does not construct the interior of a given closed surface. The same convention governs The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region, whose hypothesis is that the parameter region be a finite elementary Green region with a supplied decomposition; the plane case is stated the same way, and Limitation: arbitrary Jordan domains are not covered by the elementary Green theorem records the corresponding limitation there.

What that excludes. Two kinds of statement are outside the reach of these theorems as proved.

A theorem of the form "every closed surface bounds a solid to which the divergence theorem applies" would need a separation result for surfaces in space, which is not among this page's declared prerequisites. Nothing here proves that a given closed surface bounds anything.
A theorem of the form "the flux and the volume integral do not depend on the presentation" would need a comparison of two different presentations of the same boundary. Flux over a finite patch presentation is defined as a sum over the supplied list (Finitely patched regular surfaces, their area, scalar integrals, and flux), and no independence-of-presentation result is asserted or used.

The surface side is a single patch. The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region is a statement about one $C^{2}$ patch and the boundary chain its parametrization induces (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region). It says nothing about a surface presented by several patches whose induced boundary arcs are meant to cancel in pairs: that pairing is exactly the gluing data the divergence theorem receives explicitly, and no analogue of it is supplied for surfaces here.

No differential form appears among this page's declared prerequisites. The general statement that unifies the gradient theorem, Green's theorem, the divergence theorem and the classical Stokes theorem is an identity between the integral of a differential form over the boundary of a chain and the integral of its exterior derivative over the chain. No differential form, no exterior derivative and no manifold is available among the prerequisites this page declares, so no such unification is stated or used; every theorem above is proved from Jordan content, Fubini, change of variables, line integrals, patch flux and Green's theorem, and each is stated in the vector-field language those tools supply.

What is genuinely established. The divergence theorem holds for every finite gluing of elementary solid regions and every $C^{1}$ field on an open set containing it, and the classical Stokes theorem holds for every $C^{2}$ patch over a finite elementary Green region and every $C^{1}$ field on an open set containing the patch image. Those classes are wide enough to contain boxes, balls, right circular cylinders and finite gluings of boxes, and wide enough for the flat disc, the hemisphere and the lateral surface of a cylinder on the Stokes side; the companion page carries each of those as a worked case.

5 · Examples, counterexamples and false statements

None yet.

The Divergence Theorem and Classical Stokes

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Divergence and curl of a C1C1 vector field

Definition

Remarks

Divergence and curl are linear and satisfy the scalar product rules

Statement

Facts & Assumptions

Proof

Remarks

The divergence and curl of a cross product

Statement

Facts & Assumptions

Proof

Remarks

A C1C1 field on an open subset of R3R3 is closed exactly when its curl vanishes

Statement

Facts & Assumptions

Proof

Remarks

The curl of the gradient of a C2C2 function vanishes

Statement

Facts & Assumptions

Proof

Remarks

The divergence of the curl of a C2C2 field vanishes

Statement

Facts & Assumptions

Proof

Remarks

The Laplacian of a C2C2 function and of a C2C2 vector field

Definition

Remarks

The curl of a curl is the gradient of the divergence minus the Laplacian

Statement

Facts & Assumptions

Proof

Remarks

A C1C1 field with vanishing curl on a star-shaped open subset of R3R3 is conservative

Statement

Facts & Assumptions

Proof

Remarks

Vector potentials of a continuous field on an open subset of R3R3

Definition

Remarks

A divergence-free C1C1 field on a star-shaped open subset of R3R3 has a vector potential

Statement

Facts & Assumptions

Proof

Remarks

The curl measures the antisymmetric part of the total derivative

Statement

Facts & Assumptions

Proof

Remarks

A C1C1 map sends a compact set of content zero to a set of content zero

Statement

Facts & Assumptions

Proof

Remarks

Additivity of the integral over finitely many Jordan pieces that fill a Jordan set up to content zero

Statement

Facts & Assumptions

Proof

Remarks

Change of variables for a C1C1 map injective and regular only on the interior of a compact Jordan set

Statement

Facts & Assumptions

Proof

Remarks

A cyclic permutation of the coordinates of R3R3 preserves Jordan measurability and integrals

Statement

Facts & Assumptions

Proof

Remarks

Simple solid regions in a coordinate direction and their cyclic coordinate projection

Divergence and curl of a $C^{1}$ vector field

A $C^{1}$ field on an open subset of $R^{3}$ is closed exactly when its curl vanishes

The curl of the gradient of a $C^{2}$ function vanishes

The divergence of the curl of a $C^{2}$ field vanishes

The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field

A $C^{1}$ field with vanishing curl on a star-shaped open subset of $R^{3}$ is conservative

Vector potentials of a continuous field on an open subset of $R^{3}$

A divergence-free $C^{1}$ field on a star-shaped open subset of $R^{3}$ has a vector potential

A $C^{1}$ map sends a compact set of content zero to a set of content zero

Change of variables for a $C^{1}$ map injective and regular only on the interior of a compact Jordan set

A cyclic permutation of the coordinates of $R^{3}$ preserves Jordan measurability and integrals

Vector forms: the boundary integrals of $f n$ and of $n \times F$