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Integration of Forms and the General Stokes Theorem — Examples

1 · Prerequisites

2 · Summary

These calculations test chart weights, reflection signs, density integration on the Möbius band, both interval endpoint signs, and the agreement of forms with Green, surface Stokes, and Gauss flux. The angular period gives a nonbounding obstruction. Counterexamples isolate the need for convergence and the induced boundary orientation.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Partition weights in two overlapping charts

Example

Let fCc((1,1)). Use the increasing charts x and y=2x on overlapping domains containing its support. For smooth weights ρ,1ρ with 0ρ1 on a neighborhood of the support, the two weighted chart integrals sum to f(x)dx, independently of ρ.

Facts & Assumptions

[F1]

Independence of atlas, partition and refinement: The compact-support integral on an oriented manifold is independent of the chart cover, coordinate maps, subordinate partition, and refinement. If UM is open and contains suppω, with its restricted orientation, then UωU=Mω.

[F2]

Chart integral with its orientation sign: Let Mn be oriented and ω a smooth top form with compact support contained in a connected chart (U,ϕ). For n1 write (ϕ1)ω=fdx1dxn. Let σϕ{1,1} be the sign of its coordinate frame relative to the chosen orientation. Define the chart integral by Iϕ(ω)=σϕRnf~(x)dx. Here f~ is the Riemann-integrable zero extension, including across a genuine half-space face, as in lem-chart-supported-coefficients-have-well-defined-riemann-integrable-half-space-extensions. For n=0, a connected chart is a point p, and set Ip(ω)=ε(p)ω(p) using its determinant-line sign. Empty support gives zero. Negative charts are allowed: the upper-half-line chart u=bt at the right endpoint of an increasing interval has sign 1.

Verification

Given: The objects and hypotheses in the statement above.

1.1

The x-chart coefficient of the first term is ρ(x)f(x). Since dx=dy/2, the y-chart coefficient of the second is (1ρ(y/2))f(y/2)/2. Both have compact support; weights outside the support do not affect the products. The chart definition gives their ordinary Riemann integrals.

F2algebra
2.1

Substitute y=2x in the second integral and add: ρfdx+(1ρ)fdx=fdx. This is the finite product-partition identity underlying global independence. It includes f=0 and weights identically zero or one; no unweighted overlap is counted twice.

F1step 1.1
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Reflection reverses the signed form integral

Example

For fCc(R) and reflection r(x)=x, with the increasing orientation, Rr(fdx)=Rfdx,Rr(fdx)=Rfdx. Densities retain the sign of f; the absolute value here belongs to the coordinate density.

Facts & Assumptions

[F1]

Change of variables on oriented manifolds: Let F:MN be a diffeomorphism of oriented smooth n-manifolds and ωΩcn(N). If F preserves orientation everywhere, MFω=Nω; if it reverses orientation everywhere, MFω=Nω. If the sign varies between components, apply the appropriate signed equality on each component and add.

[F2]

Orientation-free density integration and its properties: Compactly supported smooth density integration is independent of charts and partition, linear, local, nonnegative on nonnegative densities and strictly positive for a nonzero nonnegative density. It is invariant under every diffeomorphism, without choosing an orientation. The finite-parametrization formula holds under the hypotheses of prop-integration-of-top-forms-by-finite-parametrizations, with orientation preservation omitted and absolute Jacobians used.

Verification

Given: The objects and hypotheses in the statement above.

1.1

The derivative of reflection is 1, so r(fdx)=f(x)dx. The map is a globally orientation-reversing diffeomorphism, and its compact pullback support is the reflected support. Oriented change of variables gives the first identity.

F1
2.1

For the density the Jacobian factor is 1=1, giving f(x)dx. Diffeomorphism invariance of density integration gives the second identity. Empty support, zero f, and signed f all satisfy the same formulas.

F2step 1.1
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A density integral on the Mobius band

Example

On the compact Möbius band B=(R×[1,1])/T,T(s,t)=(s+1,t), the density dsdt descends to a smooth positive density δ, and Bδ=2.

Facts & Assumptions

[F1]

Orientation-free density integration and its properties: Compactly supported smooth density integration is independent of charts and partition, linear, local, nonnegative on nonnegative densities and strictly positive for a nonzero nonnegative density. It is invariant under every diffeomorphism, without choosing an orientation. The finite-parametrization formula holds under the hypotheses of prop-integration-of-top-forms-by-finite-parametrizations, with orientation preservation omitted and absolute Jacobians used.

[F2]

Pullback of densities by local diffeomorphisms: For a local diffeomorphism F:MnNn, pullback of smooth densities is smooth and in coordinates satisfies F(fdy)=(fF)detDFdx. It is real-linear, obeys F(aδ)=(aF)Fδ for smooth functions a on N, and (FG)=GF for composable local diffeomorphisms.

Verification

Given: The objects and hypotheses in the statement above.

1.1

The quotient map is open because the inverse image of an image-open set is the union of its translates. A rectangle with s-width less than one is disjoint from all its nontrivial translates, so maps homeomorphically onto its image; at t=1 or t=-1 use a half-rectangle. For two inequivalent points only finitely many translates of a bounded neighborhood of one can approach a bounded neighborhood of the other; shrink to separate these finitely many translates. Their saturated neighborhoods are disjoint, proving Hausdorffness. Images of rational rectangles form a countable base. The transition maps are restrictions of Tk, hence smooth, so these charts define a smooth manifold with boundary. It is compact as the image of [0,1]×[1,1].

construct
2.1

The seam transition has determinant 1, with absolute value one, so the local densities glue and are positive. Equivalently Tdsdt=dsdt by the pullback formula; the same holds for all integer powers.

F2step 1.1
3.1

Use the single finite parametrization from D=(0,1)×(1,1) to the quotient. It is a diffeomorphism onto the open complement of seam and boundary, extends continuously from the closed rectangle, and is smooth up to each edge in target coordinates. Its image closure is B and its pulled-back density coefficient is one. The density parametrization formula therefore gives Bδ=01111dtds=2. Seam and boundary are covered by that formula’s null-boundary control.

F1step 1.1step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Stokes on an interval with both endpoint chart signs

Example

For f(t)=t2 on the increasingly oriented interval [0,1], [0,1]df=1=f(1)f(0). The right endpoint chart u=1t is negative; its chart sign must be retained in the upper-half-line calculation.

Facts & Assumptions

[F1]

Stokes agrees with the fundamental theorem of calculus: For a<b, orient [a,b] increasingly. Every smooth f on this interval satisfies [a,b]df=f(b)f(a), where the boundary point signs are 1 at a and +1 at b. This agrees with the Riemann fundamental theorem of calculus.

[F2]

Compact-support Stokes on the upper half-space: Give Hn={xn0} the standard orientation, n1, and its face the outward-normal-first orientation. If ηΩcn1(Hn) and j:HnHn, then Hndη=Hnjη. With η=iaidx1dxi^dxn, both sides are (1)nRn1an(x,0)dx for n>1, and a1(0) for n=1.

Verification

Given: The objects and hypotheses in the statement above.

1.1

The interval formula gives df=012tdt=1 and boundary values (1)f(0)+(+1)f(1)=1. These are induced endpoint signs, not unsigned point counting.

F1
2.1

At the left endpoint u=t is positive and the half-line boundary sign is negative. At the right endpoint u=1t is negative: the half-line calculation contributes f(1), and the chart sign 1 changes it to +f(1). More explicitly apply that local calculation to partition-weighted f supported near the endpoint; the two signs multiply in exactly this way. Thus the local calculation reproduces both endpoint values, including the zero value at t=0.

F2step 1.1
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Green circulation and flux on a disk

Example

On the unit disk with orientation dxdy, let α=(ydx+xdy)/2 and F=(x/2,y/2). Then dα=dxdy and the boundary circulation and outward flux both equal π.

Facts & Assumptions

[F1]

General Stokes agrees with both planar Green formulas: For a compact smooth planar region D oriented by dxdy and smooth P,Q on a neighborhood, general Stokes gives D(Pdx+Qdy)=D(QxPy)dxdy, D(PdyQdx)=D(Px+Qy)dxdy. When D also has the supplied finite elementary Green decomposition required by the classical results, these are exactly their circulation and outward-flux formulas. Outer boundary curves run counterclockwise and holes clockwise.

[F2]

Computing form integrals by finite parametrizations: Let n1, let Mn be oriented, and let ωΩcn(M). For 1im let DiRn be bounded open Jordan domains and Fi:DiM continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose FiDi is an orientation-preserving diffeomorphism onto an open WiM, the Wi are pairwise disjoint, and suppωiWi. Then Mω=i=1mDiFiω. An empty family is allowed when the support is empty. No nonsingularity of DFi on Di, and no M-valued extension across a genuine target boundary, is assumed.

Verification

Given: The objects and hypotheses in the statement above.

1.1

Differentiation gives dα=dxdy; also FxdyFydx=α and divF=1. The Green agreement identifies these as the circulation and flux integrands.

F1algebra
1.2

The polar parametrization (r,θ)(rcosθ,rsinθ) on (0,1)×(0,2π) has positive determinant r, extends smoothly in coordinates from its closure, and covers the disk except its cut, center, and boundary. Hence the area integral is 02π01rdrdθ=π. Singularities at r=0 are permitted at parameter boundary.

F2
2.1

For the counterclockwise boundary c(θ)=(cosθ,sinθ), cα=12dθ, so the boundary integral is π by the interval parametrization with its cut point. The outward-first boundary orientation is increasing angle since the ordered pair of radial outward normal and this tangent has positive determinant.

F2step 1.1step 1.2
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Surface Stokes on a graph disk

Example

Let S be the graph z=x2+y2 over the closed unit disk, with upward orientation, and let F=(y/2,x/2,0). Then curlF=(0,0,1), its upward curl flux over S is π, and its induced boundary circulation is also π.

Facts & Assumptions

[F1]

Agreement of general and classical surface Stokes: Let SR3 be a compact oriented smooth embedded surface with boundary, and let F be smooth on an open neighborhood of S. Set α=Fxdx+Fydy+Fzdz and μ=dxdydz. Then dα=ιcurlFμ,Sα=SιcurlFμ. On an oriented parametrization r(u,v) the latter integrand is (curlF)(r)(ru×rv)dudv; on a boundary curve it is F(r)rdt. On the common smooth patch scope this is the published classical Stokes theorem, using the standard Euclidean metric identification.

[F2]

Computing form integrals by finite parametrizations: Let n1, let Mn be oriented, and let ωΩcn(M). For 1im let DiRn be bounded open Jordan domains and Fi:DiM continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose FiDi is an orientation-preserving diffeomorphism onto an open WiM, the Wi are pairwise disjoint, and suppωiWi. Then Mω=i=1mDiFiω. An empty family is allowed when the support is empty. No nonsingularity of DFi on Di, and no M-valued extension across a genuine target boundary, is assumed.

Verification

Given: The objects and hypotheses in the statement above.

1.1

The graph parametrization is r(x,y)=(x,y,x2+y2) with rx×ry=(2x,2y,1), whose last component is positive. The curl is (0,0,1), so its scalar product with this cross product is one. The surface is a smooth compact embedded disk and F is smooth on all of Euclidean space.

F1algebra
2.1

The flux is the area of the unit parameter disk. Using polar parameters it is 02π01rdrdθ=π; finite parametrizations allow their boundary degeneracy.

F2step 1.1
3.1

The induced boundary is c(t)=(cost,sint,1) with increasing t. Along it F(c)c=1/2, so circulation is 02π(1/2)dt=π. The graph orientation gives the same increasing-angle boundary orientation as its disk parametrization, so these are the two sides of Stokes with matching signs.

F1F2step 1.1step 2.1
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Volume-form divergence on the Euclidean ball

Example

For the closed unit ball B3 oriented by dxdydz, the field F=(x,y,z) has divergence 3 and outward flux 4π. The field G=(x,0,0) has divergence 1 and outward flux 4π/3. In each case the volume integral equals the flux.

Facts & Assumptions

[F1]

Agreement with classical Gauss flux in Euclidean space: For μ=dxdydz and a smooth Euclidean field F, the volume-form divergence is xFx+yFy+zFz. For a surface parametrization r(u,v), r(ιFμ)=(F(r)(ru×rv))dudv. Consequently the volume-form divergence theorem agrees with the classical Gauss flux theorem on compact smooth regions that also admit the supplied elementary-solid presentation required by that classical theorem.

[F2]

Computing form integrals by finite parametrizations: Let n1, let Mn be oriented, and let ωΩcn(M). For 1im let DiRn be bounded open Jordan domains and Fi:DiM continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose FiDi is an orientation-preserving diffeomorphism onto an open WiM, the Wi are pairwise disjoint, and suppωiWi. Then Mω=i=1mDiFiω. An empty family is allowed when the support is empty. No nonsingularity of DFi on Di, and no M-valued extension across a genuine target boundary, is assumed.

Verification

Given: The objects and hypotheses in the statement above.

1.1

Use R(r,ϕ,θ)=(rsinϕcosθ,rsinϕsinθ,rcosϕ) on (0,1)×(0,π)×(0,2π). Its determinant is r2sinϕ>0, it is a diffeomorphism onto its open image and extends smoothly in coordinates from the closed box. The missing radial cut and axes lie in the image of its parameter boundary. Finite parametrizations give VolB3=01r2dr0πsinϕdϕ02πdθ=4π/3.

F2
2.1

For the sphere parametrization q=R(1,ϕ,θ), qϕ×qθ=sinϕq, which points outward for 0<ϕ<π. Thus F has flux integrand sinϕ, whose double integral is 4π; its volume divergence integral is 3(4π/3)=4π.

F1F2step 1.1
3.1

For G the flux integrand is qx2sinϕ=sin3ϕcos2θ. The two factors integrate to 4/3 and π, giving 4π/3. The first value follows from u=cosϕ, and the second from cos2θ=(1+cos2θ)/2. Divergence is one, so its volume integral agrees. The poles and seam are parameter-boundary images handled by the finite-parametrization formula.

F1F2step 1.1step 2.1
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The angular period and the obstruction to bounding

Example

On R2{0} let ω=ydx+xdyx2+y2. Its integral over the counterclockwise unit circle is 2π. Therefore that circle cannot be the induced oriented boundary of a compact oriented embedded smooth surface contained in the punctured plane; the form is also not exact there.

Facts & Assumptions

[F1]

A nonzero period obstructs exactness and bounding: Let SM be an oriented compact boundaryless embedded k-submanifold, k1, and let ω be a closed smooth k-form on M. If Sω0, then ω is not exact on M, and S cannot be the induced oriented boundary of a compact embedded (k+1)-submanifold of M.

[F2]

Computing form integrals by finite parametrizations: Let n1, let Mn be oriented, and let ωΩcn(M). For 1im let DiRn be bounded open Jordan domains and Fi:DiM continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose FiDi is an orientation-preserving diffeomorphism onto an open WiM, the Wi are pairwise disjoint, and suppωiWi. Then Mω=i=1mDiFiω. An empty family is allowed when the support is empty. No nonsingularity of DFi on Di, and no M-valued extension across a genuine target boundary, is assumed.

Verification

Given: The objects and hypotheses in the statement above.

1.1

Write P=y/(x2+y2) and Q=x/(x2+y2). Direct differentiation gives Qx=(y2x2)/(x2+y2)2=Py, hence dω=0 on the punctured plane. The origin is excluded from its domain.

algebra
1.2

For c(t)=(cost,sint), cω=dt. The interval (0,2π) maps diffeomorphically to the circle minus one point and extends smoothly to its closure. The finite-parametrization formula gives S1ω=2π.

F2
2.1

The circle is compact, embedded, oriented, and boundaryless, with dimension one. Its nonzero period and the closedness calculation meet all hypotheses of the period obstruction, giving both nonexactness and the stated nonbounding conclusion.

F1step 1.1step 1.2
ExampleConstruction: AI-adaptedVerification: AI-adaptedaudited 2026-09-07Open item page →

An exact top form with nonzero integral on a disk

Example

Assume ACω. On the closed unit disk D with its standard orientation dxdy, dxdy=d(xdy),Ddxdy=π=Dxdy. Thus an exact top form can have nonzero integral on a manifold with boundary. Its primitive xdy is not itself an exact one-form.

Facts & Assumptions

[F1]

The general Stokes theorem: Assume ACω. Let M be an oriented smooth n-manifold with boundary, n1, and let ηΩcn1(M). With j:MM and the outward-normal-first orientation, Mdη=Mjη. An empty boundary contributes zero; in dimension one its integral is a finite signed sum of point values.

[F2]

Computing form integrals by finite parametrizations: Let n1, let Mn be oriented, and let ωΩcn(M). For 1im let DiRn be bounded open Jordan domains and Fi:DiM continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose FiDi is an orientation-preserving diffeomorphism onto an open WiM, the Wi are pairwise disjoint, and suppωiWi. Then Mω=i=1mDiFiω. An empty family is allowed when the support is empty. No nonsingularity of DFi on Di, and no M-valued extension across a genuine target boundary, is assumed.

Verification

Given: The objects and hypotheses in the statement above.

1.1

Differentiation gives d(xdy)=dxdy. Polar parametrization evaluates its disk integral as 02π01rdrdt=π. The coordinate extensions are smooth on the closed parameter rectangle, so the finite-parametrization formula applies.

F2algebra
2.1

The counterclockwise circle c(t)=(cost,sint) pulls xdy back to cos2tdt, with integral π. General Stokes equates these two integrals on compact D with outward-first orientation.

F1F2step 1.1
3.1

If xdy=dh on D for a smooth h, then in coordinates hx=0 and hy=x. Equality of smooth mixed partials would give 0=yhx=xhy=1, impossible in the disk interior. Thus the primitive is not exact, even though its derivative is an exact top form.

step 1.1algebra
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A noncompactly supported form whose integral diverges

Statement refuted

False assertion: smoothness alone guarantees a finite integral of a top form on an oriented manifold, without a compact-support or convergence condition.

Facts & Assumptions

[F1]

Chart integral with its orientation sign: Let Mn be oriented and ω a smooth top form with compact support contained in a connected chart (U,ϕ). For n1 write (ϕ1)ω=fdx1dxn. Let σϕ{1,1} be the sign of its coordinate frame relative to the chosen orientation. Define the chart integral by Iϕ(ω)=σϕRnf~(x)dx. Here f~ is the Riemann-integrable zero extension, including across a genuine half-space face, as in lem-chart-supported-coefficients-have-well-defined-riemann-integrable-half-space-extensions. For n=0, a connected chart is a point p, and set Ip(ω)=ε(p)ω(p) using its determinant-line sign. Empty support gives zero. Negative charts are allowed: the upper-half-line chart u=bt at the right endpoint of an increasing interval has sign 1.

Counterexample

Given: The proposed assertion; use the data constructed below.

1.1

On the increasingly oriented line the form dx is smooth, but its support is all of R, which is not compact. Its restriction to every compact interval [R,R], R>0, has the ordinary integral RR1dx=2R, computed using chart integration or a finite endpoint chart partition.

F1algebra
2.1

The values 2R are unbounded as R increases. Hence even the elementary symmetric improper-integral attempt fails to give a finite value. The compact-support integral defined on this page is simply inapplicable to dx on the whole line; the calculation does not introduce a general improper manifold integral.

step 1.1algebra
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The wrong boundary sign in the half-space computation

Statement refuted

False assertion: Stokes on the standard oriented half-line remains valid if its boundary point is assigned the positive sign instead of its induced negative sign.

Facts & Assumptions

[F1]

Compact-support Stokes on the upper half-space: Give Hn={xn0} the standard orientation, n1, and its face the outward-normal-first orientation. If ηΩcn1(Hn) and j:HnHn, then Hndη=Hnjη. With η=iaidx1dxi^dxn, both sides are (1)nRn1an(x,0)dx for n>1, and a1(0) for n=1.

Counterexample

Given: The proposed assertion; use the data constructed below.

1.1

Choose a smooth compactly supported f on [0,) with f=1 near zero. Explicitly, put b(u)=e1/u for u>0 and zero otherwise, and f(t)=b(2t)/(b(2t)+b(t1)). The denominator is positive for all t, and all derivatives of b vanish at zero, so f is smooth, equals one for t at most one, and zero for t at least two.

construct
2.1

The half-space Stokes formula in dimension one gives 0f(t)dt=f(0)=1, equivalently the FTC difference f(2)f(0). Giving the endpoint positive sign instead yields +f(0)=1, so the proposed convention breaks the identity.

F1step 1.1
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Change of variables on an oriented circle

Example

Fix aR with a<1. The map F(eit)=ei(t+asint) is an orientation-preserving smooth circle diffeomorphism. For the standard angular form ω on the unit circle, Fω=(1+acost)dt,S1Fω=2π=S1ω, where t denotes the angle on each cut chart.

Facts & Assumptions

[F1]

Change of variables on oriented manifolds: Let F:MN be a diffeomorphism of oriented smooth n-manifolds and ωΩcn(N). If F preserves orientation everywhere, MFω=Nω; if it reverses orientation everywhere, MFω=Nω. If the sign varies between components, apply the appropriate signed equality on each component and add.

[F2]

Computing form integrals by finite parametrizations: Let n1, let Mn be oriented, and let ωΩcn(M). For 1im let DiRn be bounded open Jordan domains and Fi:DiM continuous and smooth up to the boundary in target coordinates: near each parameter point, a target coordinate representative extends smoothly to a Euclidean neighborhood. Suppose FiDi is an orientation-preserving diffeomorphism onto an open WiM, the Wi are pairwise disjoint, and suppωiWi. Then Mω=i=1mDiFiω. An empty family is allowed when the support is empty. No nonsingularity of DFi on Di, and no M-valued extension across a genuine target boundary, is assumed.

Verification

Given: The objects and hypotheses in the statement above.

1.1

The lift h(t)=t+asint satisfies h(t+2π)=h(t)+2π and h(t)=1+acost1a>0. It is strictly increasing; since h(t)ta, its limits are the two infinities, so it is onto R. Its inverse is smooth by the one-dimensional inverse theorem, and both lift maps commute with shifts by 2π. They descend to inverse smooth circle maps, preserving increasing-angle orientation.

algebra
2.1

Pulling back the angular form in cut charts gives dh=(1+acost)dt. The cut-circle parametrization yields 02π(1+acost)dt=2π, while 02πdt=2π. Both cut parametrizations extend smoothly from the closed interval.

F2step 1.1
3.1

The circle is compact, so the form is compactly supported; oriented change of variables therefore predicts the same equality and all its hypotheses have just been checked. For a=0 the map is the identity. Values a=1 are excluded because the derivative can vanish, so the proof does not assert a diffeomorphism at those endpoints.

F1step 1.1step 2.1

Sources