False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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FALSE: Stokes' theorem requires the surface to be a graph over a coordinate plane

Statement

False claim: Stokes' theorem applies only when the surface is a graph of a function over one of the coordinate planes.

The actual hypothesis on the A page is about the parameter region and the smoothness of the parametrization. The image need not be a graph.

Facts & Assumptions

Given: The lateral cylinder patch $σ (θ, z) = (\cos θ, \sin θ, z)$ on $[0, 2 π] \times [0, 1]$ , and the field $F (x, y, z) = (- y z, x z, 0)$ .

[L1]

Stokes' theorem identifies circulation around the induced boundary chain with the curl flux in the induced orientation (The classical Stokes theorem for a $C^{2}$ patch over a finite elementary Green region).

[F1]

The induced boundary chain is obtained from the positive boundary chain of the parameter region (The induced boundary chain and circulation of a $C^{2}$ patch over a finite elementary Green region).

[F2]

A regular patch has no interior parameter point sharing its image with a distinct point of the parameter region (Regular parametrized surface patches on compact Jordan parameter regions).

[F3]

Flux is computed against the oriented area vector (Unit normal fields, orientations, and flux through a regular surface patch).

[F4]

The curl is that of Divergence and curl of a $C^{1}$ vector field.

[F5]

The cross product is that of The cross product in $R^{3}$ .

[L2]

$(\sin t)^{'} = \cos t$ and $(\cos t)^{'} = - \sin t$ (The derivatives of sine and cosine are cosine and minus sine).

[L3]

$\sin^{2} t + \cos^{2} t = 1$ (Parity and the Pythagorean identity for sine and cosine).

[L4]

Jordan Fubini computes a multiple integral by iterated section integrals (Fubini over a bounded Jordan set when all but a content-zero family of sections are integrable).

[L5]

If $G^{'} = f$ on $[a, b]$ , then $\int_{a}^{b} f = G (b) - G (a)$ (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]

Line integrals negate under path reversal (Line integrals under reversal and concatenation).

[F6]

Rectangles are elementary Green regions (Type I, Type II, and elementary regions for Green's theorem).

[F7]

The positive boundary of a rectangle runs along the bottom, right, top, and left edges in that order (Positive orientation of elementary-region boundaries).

[F8]

Vector line integrals are computed from $⟨ F (γ (t)), γ^{'} (t) ⟩$ (Scalar line integrals with respect to arc length and vector-field line integrals).

[F9]

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

Refutation

technique · direct

1.1

The cylinder patch is $C^{2}$ and regular over a rectangle, and [F5], [L2], and [L3] give $σ_{θ} \times σ_{z} = (\cos θ, \sin θ, 0)$ .

F2F5F6L2L3given

2.1

This surface is not a graph over any coordinate plane: over the $x y$ plane the same point $(x, y)$ on the unit circle carries all heights $z \in [0, 1]$ , while over the $x z$ or $y z$ plane the missing horizontal coordinate is two-valued.

step 1.1L3

2.2

By [F1], [F7], [L6], and [F8], the two seam edges of the parameter rectangle cancel in the induced boundary chain, leaving the bottom circle traversed counterclockwise and the top circle traversed clockwise.

step 1.1F1F7L6F8

2.3

Direct differentiation in [F4] gives $curl F = (- x, - y, 2 z)$ , and step 1.1 gives $⟨ curl F (σ (θ, z)), σ_{θ} \times σ_{z} ⟩ = - 1$ , so the curl flux is $\int_{0}^{1} \int_{0}^{2 π} (- 1) d θ d z = - 2 π$ by [L4] and [L5].

step 1.1F3F4F9L4L5

3.1

On the bottom circle the field vanishes, so that contribution is $0$ ; on the top circle, traversed clockwise, the circulation is $- 2 π$ . Thus the total circulation is $- 2 π$ , agreeing with step 2.3 and [L1].

step 2.2step 2.3F8F9L2L3L5L1

4.1

Stokes' theorem therefore holds on this surface even though step 2.1 shows it is not a graph over any coordinate plane, so the claim is false.

step 2.1step 3.1L1

5.1

What the theorem actually uses is that the parameter region is a finite elementary Green region and the parametrization is $C^{2}$ ; the image being a graph is irrelevant.

step 4.1L1F6∎

Remarks

The seam cancellation in step 2.1 is the same mechanism as in the hemisphere example, but here the image is genuinely cylindrical rather than a graph in disguise.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

67 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Feldman, A. Rechnitzer and E. Yeager, CLP-4 Vector Calculus, Example 4.4.6 (standard reference, not scraped)
G. Strang and E. Herman, Calculus Volume 3, Example 6.75 (standard reference, not scraped)