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Section 2.8 Divergence theorem

Definition 2.8.1.

A piecewise smooth surface \(\mathcal{S}\subseteq\R^3\) is closed if it has no boundary curve.

Example 2.8.3. Flux out of cube.

Let \(\mathcal{S}\) be the cube in the first octant bounded by the planes \(x=1,y=1,z=1\text{.}\) Compute the outward flux of \(\boldF(x,y,z)=x^2\boldi+4xyz\boldj+ze^x\boldk\text{.}\)

Solution.

Let \(D\) be the solid cube \([0,1]\times [0,1]\times [0,1]\text{.}\) Since \(\boldF\) clearly satisfies the conditions for the divergence theorem, we have

\begin{align*} \iint\limits_{\mathcal{S}}\boldF\cdot \boldn\, d\sigma \amp = \iiint\limits_D\diver\boldF\, dV\\ \amp = \int_0^1\int_0^1\int_0^12 2x+4xz+e^x\, dy\, dz\, dx\\ \amp =e+1\text{.} \end{align*}

An easy consequence of the divergence theorem:

\begin{align*} \iint\limits_\mathcal{S}\boldF\cdot \boldn\, d\sigma \amp = \iiint\limits_D\diver\boldF\, dV\\ \amp =\iiint\limits_D 0\, dV\\ \amp =0\text{.} \end{align*}

Let \(D_1, D_2\) be the solid regions enclosed by \(\mathcal{S}_1\) and \(\mathcal{S}_2\text{,}\) respectively (so that \(\mathcal{D}=D_2-D_1\)). Notice that if \(\boldn_2\) is the outward orientation on \(\mathcal{S}_2\text{,}\) considered as the boundary of \(D_2\text{,}\) then \(-\boldn_2\) is the outward orientation on \(\mathcal{S}_2\) considered as a boundary of \(D\text{.}\) By the divergence theorem we have

\begin{align*} 0\amp=\iiint\limits_D \diver \boldF\, dV \\ \amp=\iint\limits_{\mathcal{S}_1}\boldF\cdot \boldn_1\, d\sigma + \iint\limits_{\mathcal{S}_2}\boldF\cdot (-\boldn_2)\, d\sigma\\ \amp = \iint\limits_{\mathcal{S}_1}\boldF\cdot \boldn_1\, d\sigma-\iint\limits_{\mathcal{S}_2}\boldF\cdot \boldn_2\, d\sigma\text{.} \end{align*}

We conclude that

\begin{equation*} \iint\limits_{\mathcal{S}_1}\boldF\cdot \boldn_1\, d\sigma=\iint\limits_{\mathcal{S}_2}\boldF\cdot \boldn_2\, d\sigma\text{,} \end{equation*}

as desired.

We have

\begin{align*} \diver\curl \boldF \amp =\nabla\cdot \Bigl\langle \frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} , \frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \Bigr\rangle \\ \amp=\frac{\partial}{\partial x}\left( \frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} \right)+ \frac{\partial}{\partial y}\left( \frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x} \right)+\frac{\partial}{\partial z}\left( \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \right)\\ \amp = \frac{\partial F_3}{\partial x \partial y}- \frac{\partial F_2}{\partial x \partial z}+\frac{\partial F_3}{\partial y \partial x}- \frac{\partial F_1}{\partial y \partial z}+\frac{\partial F_2}{\partial z \partial x}- \frac{\partial F_1}{\partial z \partial y}\\ \amp = 0\text{,} \end{align*}

where in the last step we have used Clairaut's theorem to identify the various “mixed partials”: i.e., \(\frac{\partial F_3}{\partial x \partial y}=\frac{\partial F_3}{\partial y \partial x}\text{,}\) \(\frac{\partial F_2}{\partial x \partial z}=\frac{\partial F_3}{\partial z \partial x}\text{,}\) \(\frac{\partial F_1}{\partial y \partial z}=\frac{\partial F_1}{\partial z \partial y}\text{.}\)

Remark 2.8.7. Gradient, curl, divergence sequence.

Let \(D\subseteq \R^n\) be an open set. Define \(C^\infty(D, \R^m)\) as the set of all functions \(\boldF\colon D\rightarrow \R^m\) whose component functions have partial derivatives of all orders. We can think of \(\grad\text{,}\) \(\curl\text{,}\) \(\diver\) as defining a sequence of operators on such collections of functions.

We use this sequence to conveniently summarize some of our theoretical results about these operators as follows:

  1. Applying two operators in sequence results in a zero function (\(\curl\grad f=\boldzero\) and \(\diver\curl\boldF=0\)). In other words the image of any one of these operators lies within the set of a functions mapping to zero when applying the next operator.

  2. If \(D\) is simply connected, then any vector field that is mapped to \(\boldzero\) by \(\curl\) is in the image of \(\grad\text{.}\)

  3. If \(D\) is contractible (too technical to define here), then any vector field that is mapped to \(0\) by \(\diver\) is in the image of \(\curl\text{.}\)

Example 2.8.9. Gauss's law.

Given a a point charge \(q\) located at the origin in \(\R^3\text{,}\) the electrostatic field it produces is given by

\begin{equation*} \boldE(x,y,z)=\frac{q}{4\pi \epsilon_0}\frac{\angvec{x,y,z}}{(x^2+y^2+z^2)^{3/2}}=\frac{q}{4\pi \epsilon_0}\frac{\boldr}{\abs{\boldr}^3}, \end{equation*}

where \(\boldr=\angvec{x,y,z}\) and \(\epsilon_0\) is a physical constant called the permittivity of free space. Prove Gauss's law: if \(\mathcal{S}\subseteq\R^3\) is a smooth closed surface with outward orientation then

\begin{equation*} \iint\limits_\mathcal{S}\boldE\cdot \boldn\, d\sigma=\begin{cases} 0 \amp \text{if } \mathcal{S} \text{ does not enclose the point charge;}\\ \frac{q}{\epsilon_0}\amp \text{if } \mathcal{S} \text{ encloses the point charge.} \end{cases} \end{equation*}

Gauss's law can be generalized to a situation where the surface \(\mathcal{S}\) encloses a region \(D\) with continuous charge density \(\delta(x,y,z)\text{.}\) If \(\boldE\) is the resulting electrostatic field, then

\begin{equation*} \iint\limits_\mathcal{S}\boldE\cdot \boldn\, d\sigma=\frac{1}{\epsilon_0}\iiint\limits_D\delta(x,y,z)\, dV=\frac{Q}{\epsilon_0}\text{,} \end{equation*}

where \(Q=\iiint_D\delta(x,y,z)\, dV\) is the total charge enclosed by \(\mathcal{S}\text{.}\)

Solution.

First observe that \(\boldE\) satisfies the conditions of the divergence theorem on any solid region that does not contain the origin. Furthermore, a (somewhat onerous) computation shows that

\begin{equation*} \diver\boldE (x,y,z)=0 \end{equation*}

for all \((x,y,z)\ne (0,0,0)\text{.}\) It follows that if \(\mathcal{S}\) does not enclose the origin, then the divergence theorem applies to the solid region \(D\) it encloses, and we have

\begin{align*} \iint\limits_\mathcal{S}\boldE\cdot\boldn\, d\sigma \amp= \iiint\limits_D\diver\boldE\, dV \amp \text{(divergence thm.)}\\ \amp=\iiint\limits_D 0\, dV=0 \text{.} \end{align*}

Next, assume \(\mathcal{S}\) does enclose the origin. Pick \(R> 0\) small enough so that the solid ball \(B_R\) centered at the origin lies inside \(\mathcal{S}\text{,}\) and let \(D\) be the region within \(\mathcal{S}\) and outside of \(B_R\text{.}\) Since \(D\) does not contain the origin, \(\boldE\) satisfies the conditions of the divergence theorem. By Corollary 2.8.5, it follows that

\begin{equation*} \iint\limits_\mathcal{S}\boldE\cdot\boldn\, d\sigma= \iint\limits_{S_R}\boldE\cdot\boldn\, d\sigma\text{,} \end{equation*}

where \(S_R\) is the sphere of radius \(R\) centered at the origin. It remains only to show that

\begin{equation*} \iint\limits_{S_R}\boldE\cdot \boldn\, d\sigma=\frac{q}{\epsilon_0}\text{,} \end{equation*}

which one can show directly using the definition of the surface integral and the standard parametrization of \(S_R\text{.}\) Indeed, the outward unit normal vector on \(S_R\) at a point \((x,y,z)\) is given by \(\boldn=\frac{1}{R}\angvec{x,y,z}\text{,}\) and thus for all \((x,y,z)\in S_R\) we have

\begin{align*} \boldE\cdot \boldn(x,y,z)\amp=\frac{q}{4\pi R\epsilon_0}\frac{1}{(x^2+y^2+z^2)^{3/2}}(x^2+y^2+z^2)\\ \amp = \frac{q}{4\pi R\epsilon_0}\frac{1}{(x^2+y^2+z^2)^{1/2}}\\ \amp =\frac{q}{4\pi R\epsilon_0}\frac{1}{R} \\ \amp =\frac{q}{4\pi R^2\epsilon_0}\text{.} \end{align*}

Finally, using the usual parametrization

\begin{equation*} \boldr(\phi, \theta) =R(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)\text{,} \end{equation*}

which satisfies

\begin{equation*} \abs{\boldr_\phi\times \boldr_\theta}=R^2\sin\phi\text{,} \end{equation*}

we have

\begin{align*} \iint\limits_{S_R}\boldE\cdot\boldn\, d\sigma \amp = \int_0^{2\pi}\int_0^\pi \frac{q}{4\pi R^2\epsilon_0}\abs{\boldr_\phi\times r_\theta}\, d\phi\, d\theta \\ \amp = \int_0^{2\pi}\int_0^\pi \frac{q}{4\pi \cancel{R^2}\epsilon_0}\cancel{R^2}\sin\phi\, d\phi\, d\theta \\ \amp=\frac{q}{4\pi\epsilon_0}4\pi \\ \amp =\frac{q}{\epsilon_0}\text{,} \end{align*}

as desired!

For all \(R > 0\) let \(B_R\) be the solid ball of radius \(R\) centered at \(P\text{.}\) We have \(S_R=\partial B_R\) and so by the divergence theorem

\begin{equation*} \iiint\limits_{B_R}\diver\boldF\, dV=\iint\limits_{S_R}\boldF\cdot \boldn\, d\sigma\text{.} \end{equation*}

Next a generalization of the integral average value theorem to triple integrals imlies

\begin{equation*} \iiint\limits_{B_R}\diver\boldF\, dV=\diver\boldF(P_R)\cdot \operatorname{vol} B_R, \end{equation*}

for some point \(P_R\) in the ball \(B_R\) centered around \(P\text{.}\) Since \(\diver\boldF\) is continuous at \(P\) we have

\begin{align*} \diver\boldF(P) \amp =\lim_{R\to 0} P_R\\ \amp=\lim_{R\to 0}\frac{1}{\operatorname{vol} B_R}\iiint\limits_{B_R}\diver \boldF\, dV \amp \text{(avg. value thm.)}\\ \amp =\lim_{R\to 0}\frac{1}{(4/3)\pi R^3}\iint\limits_{\mathcal{S_R}}\boldF\cdot \boldn\, d\sigma \amp \text{(divergence thm.)} \text{.} \end{align*}