The Divergence Theorem and Stokes’ Theorem

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We introduced Green’s theorem and observed that it is a two-dimensional analogue of the fundamental theorem of calculus. The culmination of calculus is to extend this to higher dimensions; the unifying principle is again that the integral over the interior of a region is linked to the integral over its boundary.

Divergence Theorem

We first present the divergence theorem. This concerns integrals over three-dimensional space bounded by a two-dimensional surface.

Theorem 1 (Divergence theorem) If \(E\) is a solid region in space bounded by a piecewise smooth closed surface \(\partial E\), and \(\mathbf{F}\) is a \(C^1\) vector field on an open set containing \(E\), then with \(\partial E\) oriented outward,

\[\iint_{\partial E} \mathbf{F} \cdot d\mathbf{S} = \iiint_E \divergence \mathbf{F}\dd{V}\]

holds.

Proof

Write \(\mathbf{F} = (P, Q, R)\). If we show

\[\iint_{\partial E} (0,0,R)\cdot d\mathbf{S} = \iiint_E \partial R/\partial z\dd{V}\]

for the \(z\)-component, then \(P\) and \(Q\) are handled symmetrically, and adding the three contributions yields the theorem.

Suppose \(E\) is a solid simple in all three coordinate directions; in particular, in the \(z\)-direction it is \(E = \{(x,y,z) \mid (x,y) \in D,\ u_1(x,y) \leq z \leq u_2(x,y)\}\). The right-hand triple integral, integrating \(z\) first by §Multiple Integrals, ⁋Theorem 2 (Fubini), becomes

\[\iiint_E \frac{\partial R}{\partial z}\dd{V} = \iint_D \bigl(R(x,y,u_2) - R(x,y,u_1)\bigr)\dd{A}.\]

On the other hand, \(\partial E\) consists of the top face \(z = u_2\), the bottom face \(z = u_1\), and the side faces. On the side faces the outward normal is horizontal, so \((0,0,R)\cdot \mathbf{n} = 0\) and there is no contribution. The top face has outward normal pointing upward, giving flux \(+\iint_D R(x,y,u_2)\dd{A}\), and the bottom face points downward, giving \(-\iint_D R(x,y,u_1)\dd{A}\); their sum equals the double integral above. For a general solid, cutting it into such pieces and adding them causes the flux across interior boundary faces to cancel, since they appear twice with opposite orientations, so the theorem holds.

The divergence theorem states that the amount flowing out through a closed surface equals the total amount produced inside by \(\divergence \mathbf{F}\). Thus the intuition that divergence is “outflow per unit volume” is established as a theorem. It is also practical in that flux over a closed surface can be computed by a volume integral instead of integrating over the surface directly.

Example 2 (Reducing flux to a volume integral) In §Surface Integrals and Flux, ⁋Example 6, we directly computed the flux of \(\mathbf{F} = (x,y,z)\) through a sphere of radius \(R\) by a surface integral and obtained \(4\pi R^3\). By the divergence theorem, since \(\divergence \mathbf{F} = 3\),

\[\iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E 3\dd{V} = 3\cdot\frac{4}{3}\pi R^3 = 4\pi R^3,\]

giving the same value.

Stokes’ Theorem

Stokes’ theorem is essentially the same as Green’s theorem. In fact, since it is fundamentally a theorem about integrals over two-dimensional regions bounded by one-dimensional boundaries, it is merely a modification of Green’s theorem that remains valid even when the region of integration is curved inside three-dimensional space.

Theorem 3 (Stokes) If \(S\) is an oriented piecewise smooth surface whose boundary \(\partial S\) is a piecewise smooth simple closed curve, and \(\mathbf{F}\) is \(C^1\) on an open set containing \(S\), then with \(\partial S\) oriented compatibly with \(S\) (so that the surface lies to the left),

\[\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint_S \curl \mathbf{F} \cdot d\mathbf{S}\]

holds.

Proof

It suffices to prove the case where \(S\) is the graph \(z = g(x,y)\) of a \(C^2\) function \(g\), oriented upward; a general surface is then handled by cutting it into such pieces and verifying that interior boundaries cancel. So let us treat only this special case. First, on the boundary we have \(z = g(x,y)\), so \(\dd{z} = g_x\dd{x} + g_y\dd{y}\), and therefore

\[\oint_{\partial S} \mathbf{F}\cdot d\mathbf{r} = \oint_{\partial D} P\dd{x} + Q\dd{y} + R\dd{z} = \oint_{\partial D} (P + R g_x)\dd{x} + (Q + R g_y)\dd{y},\]

and applying §Green’s Theorem, ⁋Theorem 1 (Green) to the planar region \(D\) gives

\[\iint_D \bigl[\partial_x(Q + R g_y) - \partial_y(P + R g_x)\bigr]\dd{A}.\]

Noting that \(P, Q, R\) are evaluated at \((x, y, g(x,y))\) and differentiating by the chain rule, the terms \(R g_{xy}\) and \(R g_{yx}\) cancel by §Functions of Several Variables and Partial Derivatives, ⁋Theorem 7 (Clairaut), and simplifying yields the integrand

\[(Q_x - P_y) + (Q_z - R_y)g_x + (R_x - P_z)g_y.\]

On the other hand, the upward normal of the graph is \(\mathbf{N} = (-g_x, -g_y, 1)\) and \(\curl \mathbf{F} = (R_y - Q_z,\ P_z - R_x,\ Q_x - P_y)\), so \(\curl \mathbf{F} \cdot \mathbf{N}\) is exactly this expression. Hence the above double integral equals

\[\iint_D \curl \mathbf{F}\cdot \mathbf{N}\dd{A} = \iint_S \curl \mathbf{F}\cdot d\mathbf{S}.\]

As in the plane, the following also holds.

Corollary 4 In a simply connected open region \(D \subseteq \mathbb{R}^3\), a \(C^1\) vector field \(\mathbf{F}\) that is irrotational (\(\curl \mathbf{F} = 0\)) is conservative.

Proof

Since \(D\) is simply connected, any closed curve \(C\) in \(D\) can be filled by a surface \(S\) in \(D\) whose boundary is \(C\). By Stokes’ theorem,

\[\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \curl \mathbf{F}\cdot d\mathbf{S} = 0,\]

and since the integral vanishes over every closed curve, \(\mathbf{F}\) is conservative by §Line Integrals, ⁋Theorem 4.

Example 5 Let us compute the circulation of the vector field \(\mathbf{F} = (-y, x, z)\) around the unit circle

\[C\colon \mathbf{r}(t) = (\cos t, \sin t, 0).\]

We have \(\curl \mathbf{F} = (0, 0, 2)\). Choosing the unit disk \(S\) in the \(xy\)-plane (with upward normal \(\mathbf{k}\)) as the surface with boundary \(C\), Stokes’ theorem gives

\[\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \curl \mathbf{F}\cdot d\mathbf{S} = \iint_S 2\dd{A} = 2\pi.\]

Even without using Stokes’ theorem, direct computation yields

\[\mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t) = (-\sin t, \cos t, 0)\cdot(-\sin t, \cos t, 0) = 1,\]

so the integral is \(\oint_C = \int_0^{2\pi} \dd{t} = 2\pi\), agreeing with the above. One can also verify that choosing another surface with the same boundary, such as a hemisphere, does not change the value of the integral.

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.