Stokes' theorem and the divergence theorem

The localization proof and corner/edge estimates extend the elementary calculations below to the precisely stated finite regular boundary classes, including holes and cavities. The local proofs retained here do not assume an unproved decomposition.

Circulation and curl

Theorem

Stokes' theorem

Let $\Sigma$ be a compact oriented piecewise $C^2$ regular surface, made of finitely many regular patches meeting along their edges, with piecewise smooth boundary $C$, and let $\mathbf F$ have continuous first partial derivatives on an open neighbourhood of the surface. With $C$ oriented compatibly by the right-hand rule,

$$\oint_C\mathbf F\cdot d\mathbf r=\iint_{\Sigma}(\nabla\times\mathbf F)\cdot\mathbf n\,dS.$$

The thumb of the right hand points along the chosen normal while curled fingers indicate the positive boundary direction.

The complete patch proof in the integral-theorem supplement assumes a $C^2$ regular parametrization on a neighborhood of its compact parameter region, interior once-only coverage and boundary coverage without overlap except endpoints. Its parameter region must have the proved Green property, or an explicitly given finite decomposition with matched edges. Finite compatible patches can then be added with internal boundaries canceled. This does not silently supply a decomposition theorem for every general piecewise smooth surface.

Green's circulation theorem is the planar case with upward normal. The surface need not be flat: if two oriented surfaces share the same compatible boundary and the hypotheses hold for each, either surface may be used to compute the common circulation.

Worked example

Choose the simplest spanning surface

Let $C$ be $x^2+y^2=4$ in the $xy$-plane, counterclockwise viewed from above, and $\mathbf F=(-y,x,z^2)$. Here $\nabla\times\mathbf F=(0,0,2)$. Choose the disk with upward normal, giving

$$\oint_C\mathbf F\cdot d\mathbf r=\iint_{x^2+y^2\le4}2\,dA=8\pi.$$

The orientation must match: the downward normal would require clockwise boundary orientation.

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.

Flux and divergence

Theorem

Divergence theorem

Let $E$ be a bounded solid with compact closure $\overline E$ and piecewise smooth closed boundary $\partial E$ oriented outward, and let $\mathbf F$ be $C^1$ on an open neighbourhood of $\overline E$. Then

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

Every boundary component must be included. At a cavity the outward normal of the solid points into the cavity.

Proof

How interior contributions cancel

On a rectangular box, integrate $\partial P/\partial x$ with respect to $x$: the fundamental theorem gives the flux through the two faces normal to the $x$-axis. Do the same for the $y$ and $z$ components, producing the total outward flux. For a suitable subdivided solid, the flux through shared internal faces cancels between adjacent pieces. The box calculation alone does not justify curved boundaries; the localization proof linked below supplies that step for its explicitly stated regular classes.

The integral-theorem supplement now proves the precisely stated regular classes there, including solids whose boundaries consist of finitely many regular faces, edges and vertices, as well as the stated planar regions and oriented surfaces. Local charts, a finite partition of unity and corner/edge estimates justify the curved-boundary step for those classes; no unproved finite coordinate-simple decomposition is assumed. Arbitrary nonregular or weak boundaries remain outside the claim.

Worked example

Closed surface flux

Let $E$ be the ball of radius $R>0$ and $\mathbf F=(x,y,z)$. Its divergence is $3$, so

$$\iint_{\partial E}\mathbf F\cdot\mathbf n\,dS=3\cdot\frac{4\pi R^3}{3}=4\pi R^3.$$

If instead only the upper hemisphere is requested, this theorem applies only after adding a disk to close the surface; compute and subtract the disk's outward flux with the correct normal.

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.”

Choosing the theorem and checking hypotheses

Use Stokes for an oriented boundary curve and a spanning surface; use divergence for a closed oriented surface and the enclosed solid. The latter cannot be applied directly to an open surface, and neither theorem permits ignoring singularities inside the relevant region. To use a closing surface, account explicitly for its flux and its outward direction. Verify a purported boundary orientation before changing surfaces.

Practice

1. Evaluate the circulation of $(-y,x,0)$ around the counterclockwise unit circle in the $xy$-plane.
2. Find the outward flux of $(x,y,z)$ through the unit cube.
3. A solid has a cavity. Which way does its outward normal point on the cavity boundary?
4. Why cannot the divergence theorem be used directly on an open hemisphere alone?

Full solutions

1. Curl is $(0,0,2)$. The upward disk is compatible, so Stokes gives $2\pi$.
2. Divergence is $3$ and the unit cube has volume $1$, so outward flux is $3$.
3. Into the cavity: outward means away from the solid, not necessarily away from a chosen origin.
4. Its boundary is a circle, so the hemisphere is not a closed surface bounding a solid by itself. Add the base disk, apply the theorem to the resulting closed boundary, then subtract the disk flux.

Synthesis

Stokes relates boundary circulation to curl through a spanning surface; divergence relates outward flux over a closed surface to divergence throughout its volume. Their geometry and orientation conditions determine when the conversion is legitimate.