Proofs of Green, parametrized Stokes and sliced divergence

Learning goals and scope

Prove integral theorems on precisely stated domain classes, derive the pullback identity behind Stokes, and track orientation under patch cancellation. The first calculations treat simple regions; the final extension proves the stated finite regular boundary classes by localization and explicit corner and edge estimates. No coverage of an entire institutional syllabus is inferred.

Green's theorem on a region simple in both coordinates

Suppose a bounded region $D$ can be written both as $a\le x\le b$, $g_1(x)\le y\le g_2(x)$ and as $c\le y\le d$, $h_1(y)\le x\le h_2(y)$, with ordered continuous bounds whose graphs admit the finite piecewise $C^1$ boundary parametrizations justified in the Green proof, and positively oriented boundary. Let $P,Q$ be $C^1$ on a neighborhood of its closure. Then

$$\oint_{\partial D}P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA.$$

Theorem

Green's theorem, circulation form

Let $D$ be a bounded planar region whose boundary $C$ consists of finitely many disjoint piecewise smooth simple closed curves, positively oriented relative to $D$. Suppose $P,Q$ are $C^1$ on an open neighbourhood of $\overline D$. Then

$$\oint_C P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA.$$

Positive orientation keeps the region to the left as one traverses the boundary: counterclockwise on an outer boundary and clockwise around a hole.

Proof

Complete proof on the double-simple class

Assume first that $D$ has both descriptions $a\le x\le b$, $g_1(x)\le y\le g_2(x)$ and $c\le y\le d$, $h_1(y)\le x\le h_2(y)$, with ordered continuous boundary functions whose graphs are supplied as finitely many piecewise $C^1$ arcs, and the indicated positive orientation. The fundamental theorem of calculus gives

$$-\iint_D P_y\,dA=\int_a^b[P(x,g_1(x))-P(x,g_2(x))]dx.$$

The lower boundary is traversed left to right and the upper boundary right to left, so this is precisely their $P\,dx$ contribution; vertical pieces have $dx=0$. Similarly,

$$\iint_D Q_x\,dA=\int_c^d[Q(h_2(y),y)-Q(h_1(y),y)]dy.$$

The right boundary is traversed upwards and the left boundary downwards; horizontal pieces have $dy=0$. Adding proves Green's formula in this class. If a specific finite decomposition into such pieces is given, with disjoint interiors and matched piecewise smooth boundary arcs, sum their identities. Each internal arc has opposite induced directions and cancels.

The boundary graphs need not have finite graph derivatives at projection endpoints, provided their continuous graphs are supplied as finitely many piecewise $C^1$ arcs. Here is the required parameter argument. For the lower graph put $H(x)=\int_a^xP(s,g_1(s))ds$. Its integrand is continuous. Along an oriented regular arc with increasing $x=x(t)$, the chain rule gives $dH(x(t))/dt=P(x(t),g_1(x(t)))x'(t)$. The fundamental theorem therefore identifies the arc's $P\,dx$ integral with $\int_a^bP(x,g_1(x))dx$, even if the inverse graph parameter has a vertical tangent at an endpoint. Reversed traversal changes the sign. The same reasoning applies to the upper graph and to the two $Q\,dy$ graphs. This supplies the boundary-integral identifications above without assuming a bounded derivative of each graph function at its endpoints.

In particular the disk $x^2+y^2\le R^2$, $R>0$, has continuous bounds $g_{1,2}(x)=\mp\sqrt{R^2-x^2}$ and the analogous horizontal bounds. Its semicircles have regular angle parametrizations, so this disk lies in the proved class. For the annulus $1\le x^2+y^2\le4$, split into four quadrant pieces: in the first quadrant the vertical bounds are $0\le x\le2$, $\sqrt{\max(0,1-x^2)}\le y\le\sqrt{4-x^2}$, and the horizontal description is obtained by exchanging coordinates. They are continuous and their boundary arcs are piecewise $C^1$. The reflected pieces have matching internal edges, so the stated finite-cancellation extension also covers this annulus.

This proves the formula for the double-simple class and for those given finite decompositions. The general region class in the stated theorem requires an additional decomposition or approximation result; this calculation does not by itself prove that every such region admits the required finite decomposition.

Stokes for a regular parametrized patch

Let $D$ be compact and let $r$ be a $C^2$ parametrization on an open neighborhood of $D$, one-to-one on its interior, with $r_s\times r_t\ne0$. Assume $D$ has the Green property just proved and its piecewise smooth boundary maps to the surface boundary without overlap except endpoints. Let $F$ be $C^1$ near the image. Orient the surface by $r_s\times r_t$ and the boundary by the positive orientation of $D$.

Set $A(s,t)=F(r)\cdot r_s$, $B(s,t)=F(r)\cdot r_t$. The chain rule and equality of $r_{st},r_{ts}$ give

$$B_s-A_t=(DF(r)r_s)\cdot r_t-(DF(r)r_t)\cdot r_s =(\operatorname{curl}F)(r)\cdot(r_s\times r_t).$$

The last equality follows by expanding the three curl components and the cross product. The boundary line integral pulls back to $\oint_{\partial D}A\,ds+B\,dt$. Green converts it to the integral of $B_s-A_t$, which is exactly the oriented curl surface integral. This proves Stokes for the patch. For finitely many compatible patches, add their identities and cancel paired internal boundary curves. Compatibility of normals is necessary for those orientations to oppose each other.

The sliced-integration foundation used in three dimensions

Here is the precise integration fact needed by the divergence proof. Let $D\subset\mathbb R^2$ be compact and Jordan measurable, let $a\le b$ be continuous real functions on $D$, and set

$$E=\{(x,q):q\in D,\ a(q)\le x\le b(q)\},\qquad q=(y,z).$$

For $f$ continuous on a box containing $E$,

$$\iiint_E f(x,q)\,dx\,dq =\iint_D\left(\int_{a(q)}^{b(q)}f(x,q)\,dx\right)dq.$$

First prove the region is Jordan. Compactness of $D$ and continuity of the two bounds make $E$ compact. Its boundary is contained in the two graphs and the vertical wall over $\partial D$. To cover either graph, use a small square grid in the base. On a base square meeting $D$, uniform continuity bounds the height variation by a single $\omega(h)\to0$; enclosing the graph values in a slab of height at most $\omega(h)+2h$ gives total covering volume at most a fixed base-box area times this height. Thus both graphs are Jordan null. The wall is null because a finite rectangle cover of $\partial D$ with arbitrarily small total area can be multiplied by one fixed bounded $x$ interval. The Jordan boundary criterion therefore makes $E$ measurable.

Next prove both integrals exist. Write $g(q)=\int_{a(q)}^{b(q)}f(x,q)dx$. If $H$ is a common interval length and $M$ bounds $|f|$, then

$$|g(q)-g(q_0)|\le H\sup_x|f(x,q)-f(x,q_0)| +M\bigl(|a(q)-a(q_0)|+|b(q)-b(q_0)|\bigr).$$

Hence $g$ is continuous on $D$. Continuous functions on compact Jordan sets, extended by zero, are Riemann integrable by the boundary-cell argument. This applies both to $f$ on $E$ and to $g$ on $D$.

Finally identify the values by finite sums. Extend $f$ by zero outside $E$ in the box. For a rectangular grid with $x$ intervals $J_i$ and base cells $C_j$, let $m_{ij},M_{ij}$ be the infimum and supremum of that zero extension on $J_i\times C_j$. For every fixed $q\in C_j$, integrating the cell bounds in $x$ gives

$$\sum_i m_{ij}|J_i|\le g_0(q)\le\sum_i M_{ij}|J_i|,$$

where $g_0$ is zero outside $D$. Integrate these inequalities over the base cells and add. The iterated value lies between the same three-dimensional lower and upper sums as $\iiint_Ef$. Their gap can be made arbitrarily small by integrability. This proves equality without presupposing a three-dimensional Fubini interchange. The three-dimensional Darboux criterion itself uses the same finite common-refinement proof: refinement raises lower sums and lowers upper sums, and a common grid compares any pair; equality of their supremum and infimum is equivalent to arbitrarily small gaps.

The field in the divergence theorem is initially $C^1$ only on an open neighborhood $U$ of $\overline E$, not necessarily on the whole containing box. This causes no hidden extension assumption. Cover $\overline E$ by finitely many open half-cubes whose full closed cubes lie in $U$. On each cube take a $C^1$ product cutoff $\chi_i$, equal to one on the half-cube and zero outside the full cube. Such a one-variable cutoff is obtained by joining the constants zero and one with $3t^2-2t^3$, whose derivatives vanish at both ends; coordinate products give the cube cutoff. Then

$$\chi=1-\prod_i(1-\chi_i)$$

equals one near $\overline E$ and has compact support inside $U$. Extend $\chi F$ by zero outside $U$. It is globally $C^1$, because its support stays away from the boundary of $U$, and it and all first derivatives agree with $F$ near $\overline E$. Its component derivatives are continuous on the containing box, so the proved sliced formula applies to $P_x,Q_y,R_z$ as used below. No extension of a merely arbitrary discontinuous integrand is being asserted.

Divergence for a solid simple in each coordinate

Let $E$ be a compact solid describable as the interval between two ordered continuous graphs, piecewise $C^1$ on their regular boundary pieces, in the $x$ direction over a compact Jordan planar domain, and likewise in the $y$ and $z$ directions. Assume the associated boundary graph formulas apply outside finitely many edges, and let $F=(P,Q,R)$ be $C^1$ near its closure. The boundary is oriented outward.

Require, in each coordinate description, that $a<b$ off a Jordan-null part of the projection and that the two graphs represent distinct actual boundary faces away from the exceptional edges. This excludes positive-area zero-thickness pieces with coincident faces; the slab integration lemma itself permits $a=b$, but such a degenerate set does not supply the two outward faces used by this flux proof.

Write the $x$ description as $a(y,z)\le x\le b(y,z)$ over $D_{yz}$. The fundamental theorem and iterated integration give

$$\iiint_E P_x\,dV=\iint_{D_{yz}}[P(b(y,z),y,z)-P(a(y,z),y,z)]\,dy\,dz.$$

On the right graph, the outward vector area has $x$ component $+dy\,dz$; on the left it has $-dy\,dz$. Side pieces generated parallel to the $x$ direction have normal $x$ component zero. Thus the right side is exactly the flux of $(P,0,0)$ over the boundary. Repeat using the $y$ and $z$ descriptions for the other components. Addition proves

$$\iiint_E\operatorname{div}F\,dV=\iint_{\partial E}F\cdot n\,dS.$$

Finite unions again follow by cancellation of matched interior faces. General approximation or decomposition beyond this class needs an additional theorem; it is not hidden in the word “cancellation.”

Worked example: checking the signs on a box

For $F=(x,y,z)$ on $[0,a]\times[0,b]\times[0,c]$ with $a,b,c>0$, divergence is three and the volume integral is $3abc$. On the three coordinate-zero faces, flux is zero. On $x=a$, flux is $a$ times area $bc$, and similarly the other two opposite faces each contribute $abc$. The total is $3abc$. If the zero faces had nonzero components, their outward normals would carry negative coordinate signs.

Graded practice

1. Foundation. Prove Green's formula for $P=0,Q=x$ on $[0,2]\times[0,3]$ by both sides.
2. Core. Use $r(s,t)=(s,t,s+t)$ on $[0,1]^2$ to compute $r_s\times r_t$ and its compatible boundary direction.
3. Core. For $F=(x^2,0,0)$ on the unit cube, check divergence and direct outward flux.
4. Proof. Explain why opposite orientations are necessary when adding patch formulas. What remains if an internal edge is traversed in the same direction twice?

Full solutions

1. The area integral of $Q_x=1$ is six. The only nonzero $x\,dy$ boundary integral is on the right edge, where $x=2$ and $y$ increases from zero to three, giving six. The left edge has $x=0$.
2. The tangent vectors are $(1,0,1)$ and $(0,1,1)$; the cross product is $(-1,-1,1)$. Positive boundary order in parameter space is $(0,0)\to(1,0)\to(1,1)\to(0,1)\to(0,0)$. Mapping these vertices gives the compatible space-curve direction.
3. Divergence is $2x$, whose unit-cube integral is one. Only the face $x=1$ contributes direct flux, with value one; the face $x=0$ has zero field component and all other normals are perpendicular to the field.
4. Reversing a line-integral orientation changes its sign, so paired internal contributions sum to zero only with opposite traversal. Same-direction traversal gives twice that edge integral instead, an extra term not belonging to the external boundary. Orientation consistency is an algebraic requirement, not cosmetic notation.

Summary and term definitions

A pullback expresses a spatial integral in parameter coordinates. A regular patch has nonzero tangent cross product. A vector area element includes both normal direction and area scaling. The fundamental theorem proves sliced flux identities; Green plus a chain-rule identity proves parametrized Stokes. Extensions to more complicated regions need justified localization or decomposition; any shared internal boundaries must cancel with the correct orientations.

Beyond simple regions: the precise regular classes

The preceding calculations establish the local identities. We now supply the missing passage from local identities to general regular boundaries. The argument uses overlapping local charts and a finite partition of unity; it does not assume that an arbitrary region has a finite subdivision into coordinate-simple pieces.

For Green, let $D$ be a bounded open planar set whose boundary is a finite union of disjoint simple closed curves, each made of finitely many regular $C^1$ arcs. Each arc has a regular $C^1$ extension through its endpoints. Away from their finitely many junctions, the boundary is locally a $C^1$ graph and $D$ occupies exactly one side. Take the positive boundary direction, with $D$ on the left. For Stokes, first let $\Sigma$ be a compact oriented embedded $C^2$ surface with $C^2$ boundary; boundary charts extend regularly across the boundary. More generally, allow regular $C^1$ boundary arcs in $C^2$ surface parameter charts extending regularly across those arcs, finitely many boundary corners, and finite unions of such surface pieces with disjoint interiors and matched edges, each internal edge occurring exactly twice with opposite induced directions. For divergence, let $E$ be a bounded open solid whose compact boundary consists of finitely many regular compact $C^1$ faces with regular extensions, meeting only along finitely many finite-length regular $C^1$ edge arcs and vertices. Away from these edges and vertices, the boundary is locally a $C^1$ graph with $E$ on exactly one side. Faces cover the boundary once, apart from their edges. Smooth boundaries are the case with no edges.

In all three cases the field is $C^1$ on an open neighbourhood of the entire closed set used in the theorem: $\overline D$, $\Sigma$, or $\overline E$, respectively. Under these hypotheses all three displayed integral theorems hold. Holes, cavities, disconnected components and surfaces that need several charts are allowed. These are sufficient classical regularity hypotheses; arbitrary nonregular Jordan curves, cusps without regular endpoint extensions, self-intersecting surfaces and measure-theoretic weak boundaries are outside the claim.

The prerequisites are the ordinary chain and product rules, the previously proved proper Riemann change-of-variables and sliced-integration results, and the local graph descriptions just specified. The next steps construct the localization and estimate the exceptional sets explicitly.

A finite partition of unity, constructed rather than assumed

Cover a compact set $K$ by open chart neighbourhoods $U_i$ on which the desired local calculation is valid. For each point choose two concentric cubes, with the closure of the larger cube in one $U_i$ and with the smaller cube containing that point. Compactness gives finitely many smaller cubes covering $K$. Choose a nonnegative $C^1$ product cutoff $\chi_i$, equal to one on the smaller cube and zero outside the larger cube. The cubic transition $3t^2-2t^3$ used above supplies these cutoffs. Put

$$S=\sum_{i=1}^N\chi_i,\qquad \rho_i=\frac{\chi_i}{S}.$$

The quotients are defined on the open set $S>0$, which contains $K$. If ambient extensions are needed, use another finite cube cutoff $\lambda$, compactly supported in $S>0$ and equal to one near $K$, replace each quotient by $\lambda\chi_i/S$, and extend by zero outside $S>0$. Such a $\lambda$ is constructed as $1-\prod_j(1-\psi_j)$ from smaller cube cutoffs covering $K$. Thus the extended weights are $C^1$, each has support compactly contained in its assigned chart, and near $K$

$$\sum_i\rho_i=1,\qquad \sum_i\nabla\rho_i=0.$$

Here support on the set being integrated is meant; a chart term can be extended by zero across its artificial chart edges because its support stays away from them. On a surface restrict the ambient cutoffs to the surface. For a field $F$, linearity and the product rule give

$$\sum_i\operatorname{curl}(\rho_iF)=\operatorname{curl}F, \qquad \sum_i\operatorname{div}(\rho_iF)=\operatorname{div}F.$$

The same cancellation applies to $Q_x-P_y$ in the plane. Thus it suffices to prove each identity for a field supported in a single chart. No internal boundary of an overlapping cover has been added; the derivatives of the weights cancel algebraically. A finite sum can be interchanged with every integral here.

Green on a smooth boundary chart

After a translation and an orientation-preserving rotation, a boundary chart has $D$ above $y=g(x)$, with $g$ of class $C^1$. Let $G=(p,q)$ be a localized field, zero near the artificial sides and top of a chart rectangle. For either component $h$, define $H_h(x)=\int_{g(x)}^b h(x,y)\,dy$, where the top $b$ lies above the support. The fundamental theorem and the chain rule give

$$H_h'(x)=\int_{g(x)}^b h_x(x,y)\,dy-g'(x)h(x,g(x)).$$

The differentiation in $x$ is justified directly: on a compact rectangle, the fundamental theorem writes each difference quotient of $h$ as an average of $h_x$, and uniform continuity of $h_x$ makes these quotients converge uniformly to $h_x$. Integration therefore commutes with this limit. Combining this fixed-limit derivative with the ordinary chain rule at $y=g(x)$ proves the displayed formula. The same argument, with the other parameter fixed, justifies both variable-limit derivatives in the divergence calculation below.

Integrating $H_q'$ over $x$ gives zero because $H_q$ vanishes at the artificial endpoints. Integrating $-p_y$ vertically gives $p(x,g(x))$. Consequently

$$\iint_D(q_x-p_y)\,dA =\int\bigl[p(x,g(x))+g'(x)q(x,g(x))\bigr]\,dx.$$

The right side is exactly the boundary integral on $(x,g(x))$ with $x$ increasing: the region above is on the left. If $D$ is below the graph, the vertical integration and positive traversal both reverse the sign. A chart wholly inside $D$ has zero area integral, since each integrated derivative has zero endpoint values, and has no boundary contribution. Rigid orientation-preserving coordinate changes preserve the circulation integral, area and scalar curl by the chain rule. Summing the localized identities proves Green for every boundary covered by these graph charts. The corner step below handles the finitely many remaining points. In particular, infinitely many tangencies to a fixed coordinate direction do not force an unproved finite slicing decomposition.

Stokes without a single global parametrization

Use an oriented regular $C^2$ chart $r(s,t)$; an interior chart has a full-plane parameter neighbourhood, and a smooth boundary chart has $t\ge0$. Such a boundary chart follows by straightening its regular $C^2$ parameter boundary. For the localized field put $A=G(r)\cdot r_s$ and $B=G(r)\cdot r_t$. The earlier pullback calculation gives

$$B_s-A_t=(\operatorname{curl}G)(r)\cdot(r_s\times r_t).$$

The coefficients vanish near artificial chart edges. For an interior chart, integrating both derivatives gives zero. For a boundary chart, extend the compactly supported coefficients by zero in the tangential direction and above their support. Direct iterated integration on the half-plane gives

$$\int_{-\infty}^{\infty}\int_0^{\infty}(B_s-A_t)\,dt\,ds =\int_{-\infty}^{\infty}A(s,0)\,ds.$$

The infinite integration limits here are notation for a fixed bounded rectangle containing the support; no improper-integral theorem is needed. To integrate $B_s$ first, the continuous compactly supported integrand permits the previously proved rectangular Fubini interchange. The boundary direction is increasing $s$: the parameter region $t>0$ is on its left, consistent with $r_s\times r_t$. Hence the right side is the required localized boundary circulation. Sum the chart identities using the weights above. This proves Stokes on the whole smooth oriented surface, including a closed surface, where there are no boundary-chart contributions. A sphere needs several charts and is covered by this proof without a global nonsingular parametrization.

The same conclusion holds for a regular $C^1$ parameter boundary in a $C^2$ surface chart. Retain the original $C^2$ parametrization and apply the Green graph calculation to $A\,ds+B\,dt$ on its localized parameter region. The coefficients $A,B$ are $C^1$, and the pullback identity above is unchanged. This proves the boundary-chart identity without needing second derivatives of a $C^1$ boundary-straightening map. Finitely many boundary corners are handled below.

For finitely many surface pieces, apply this result, together with the corner argument below, to each piece. Every shared edge occurs exactly twice with opposite induced line-integral directions, so its two integrals sum to zero. The unpaired edges are precisely the external boundary. This requires the stated matched-edge structure; no existence of a triangulation of an arbitrary singular surface is being asserted.

Divergence on an arbitrary smooth boundary graph

In a boundary chart choose rigid coordinates so that $E$ lies above $z=g(x,y)$. Write the localized field as $G=(p,q,r)$ and choose a top $b$ above its support. Differentiating the variable-limit integrals in $x$ and $y$, and using their zero values near the artificial sides, yields

$$\iiint_E p_x\,dV=\iint g_xp(x,y,g)\,dx\,dy, \qquad \iiint_E q_y\,dV=\iint g_yq(x,y,g)\,dx\,dy.$$

The vertical fundamental theorem gives

$$\iiint_E r_z\,dV=-\iint r(x,y,g)\,dx\,dy.$$

Adding gives the surface integral of $G\cdot(g_x,g_y,-1)$. This is exactly the outward flux, since the solid is above the graph and the outward vector area is $(g_x,g_y,-1)\,dx\,dy$. For a solid below the graph the three signs reverse and the outward vector area is $(-g_x,-g_y,1)\,dx\,dy$. Interior-chart divergence integrates to zero. Under a rigid rotation the derivative matrix is conjugated by the orthogonal rotation matrix, so its trace, divergence, is unchanged; outward flux and volume are likewise unchanged. Thus these chart calculations apply in any spatial direction. Sum with the finite weights to prove divergence for a compact $C^1$ boundary, including every cavity wall with its own outward normal. The solid need not be simple in any fixed coordinate direction.

Why corners and edges do not leave an unproved limit

We now justify the exceptional-set step. For Green the exceptional set $Z$ is finite; for one Stokes surface piece it is its finitely many boundary corners. For divergence $Z$ is a finite union of finite-length edge arcs and vertices. Construct a $C^1$ function $\eta_\varepsilon$ with values in $[0,1]$, zero on a neighbourhood of $Z$, equal to one outside a distance $C\varepsilon$ from $Z$, and with $\lVert\nabla\eta_\varepsilon\rVert\le C/\varepsilon$. Constants are independent of $\varepsilon$.

Here is a construction and the necessary size estimate. Take a grid of cubes of side $\varepsilon$ meeting $Z$; around each choose a product cutoff equal to one on a slightly enlarged cube and zero outside a fixed further enlargement. Define $\eta_\varepsilon$ as the finite product of one minus these cutoffs. Enlarged grid cubes have uniformly bounded overlap, so the product rule bounds its gradient by $C/\varepsilon$. A finite point set needs only a bounded number of cubes. A rectifiable arc of length $L$ is covered by at most $1+L/\varepsilon$ balls of radius $\varepsilon$, by subdivision in arc length; each ball meets a bounded number of grid cubes. Thus the number of edge cubes is $O(1/\varepsilon)$, and their total spatial volume is $O(\varepsilon^2)$. For finitely many points the planar area is $O(\varepsilon^2)$. On a regular surface chart both the map and its local inverse have bounded Lipschitz constants on sufficiently small compact subcharts, and the area density is bounded. This gives surface area $O(\varepsilon^2)$ near finitely many points, surface area $O(\varepsilon)$ near finitely many regular edge arcs lying on the faces, and boundary length $O(\varepsilon)$ near boundary corners. Finite chart covers make the constants uniform. These facts also show that the solids and planar regions are Jordan measurable and that edges have zero surface area: bounded regular graph pieces have zero ambient volume by the graph-cover argument used earlier, and the lower-dimensional exceptional sets admit the cube covers just given.

Apply the localized proof to $G_\varepsilon=\eta_\varepsilon F$. It vanishes near $Z$, so only interior and smooth boundary charts meeting its support are needed. Every added derivative is explicitly controlled. In the plane, its extra scalar-curl term is $(\eta_\varepsilon)_xQ-(\eta_\varepsilon)_yP$. On a Stokes surface it is $\nabla\eta_\varepsilon\times F$. The corresponding area error, for a bound $M$ on the field, is at most $CM\varepsilon^{-1}\varepsilon^2=CM\varepsilon$. For divergence,

$$\operatorname{div}(\eta_\varepsilon F) =\eta_\varepsilon\operatorname{div}F+\nabla\eta_\varepsilon\cdot F, \qquad \left|\iiint_E\nabla\eta_\varepsilon\cdot F\,dV\right| \le CM\varepsilon.$$

The terms with $\eta_\varepsilon$ converge to the original volume or surface integrals because the bounded integrands differ only on sets of volume or area tending to zero. Boundary circulation loses at most $CM\varepsilon$ near finitely many corners; boundary flux loses at most $CM\varepsilon$ near finitely many edges. Therefore every error tends to zero. This proves Green with the stated finite corners, Stokes on each stated piece with corners followed by edge cancellation, and divergence with the stated finite face-and-edge boundary. There is no appeal to convergence of unspecified polyhedral approximations.

A check beyond coordinate-simple solids

Let $E$ be the solid torus with major radius $R$ and tube radius $a$, where $R>a>0$, and take $F=(x,y,z)$. It has a smooth compact boundary but is not simple in all three coordinate directions. The theorem now proved applies. Its volume is $2\pi^2Ra^2$, so divergence three gives flux $6\pi^2Ra^2$. This volume follows directly from cylindrical coordinates: integrate $\rho$ over the disk $(\rho-R)^2+z^2\le a^2$ and multiply by $2\pi$; the odd part $\rho-R$ integrates to zero. We can also check the flux independently. With

$$r(\theta,\phi)=((R+a\cos\phi)\cos\theta,(R+a\cos\phi)\sin\theta,a\sin\phi), \qquad 0\le\theta,\phi\le2\pi,$$

the outward vector area is $r_\theta\times r_\phi=a(R+a\cos\phi)(\cos\phi\cos\theta,\cos\phi\sin\theta,\sin\phi)$. Direct integration gives

$$\iint_{\partial E}F\cdot n\,dS =2\pi a\int_0^{2\pi}(R+a\cos\phi)(R\cos\phi+a)\,d\phi =6\pi^2Ra^2.$$

The parameter seam has zero surface area and causes no double-counting in this direct check. More generally, the proof just supplied handles all the regular classes stated at the start of this extension. It does not establish any broader weak-boundary theorem or certify coverage of an institution's entire MATH2020 syllabus.