## A remark on the divergence theorem

The divergence theorem states that for a compact domain ${D}$ in ${\mathbb R^3}$ with piecewise smooth boundary ${\partial D}$, then for a smooth vector field ${\textbf{F}}$ on ${D}$, we have

$\displaystyle \begin{array}{rl} \displaystyle \iint_{\partial D}\textbf{F}\cdot \textbf{n}\,dS=\iiint_D \nabla \cdot\textbf{F}\,dV,\end{array}$

where ${\textbf{n}}$ is the unit outward normal and ${\nabla \cdot\textbf{F}}$ is the divergence of ${\textbf{F}}$.

In most textbooks, the divergence theorem is proved by the following strategy:

1. Prove that

$\displaystyle \begin{array}{rl} \displaystyle \iint _{\partial D}(0, 0, R)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial R}{\partial z}\;dV \ \ \ \ \ (1)\end{array}$

where ${D}$ is a so called type ${1}$ domain, i.e. ${D=\{(x, y, z): (x, y)\in \Omega, \underline f(x, y)\le z\le \bar f(x, y)\}}$ for some region ${\Omega\subset \mathbb R^2}$ and ${\bar f, \underline f\in C^1(R)}$. i.e. ${D}$ is bounded by the graphs of two functions with variables ${x, y}$. This can be quite easily done using the fundamental theorem of calculus.

2. Similarly prove

$\displaystyle \begin{array}{rl} \displaystyle \iint _{\partial D}(P, 0, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial P}{\partial x}\;dV \ \ \ \ \ (2)\end{array}$

and

$\displaystyle \begin{array}{rl} \displaystyle \iint _{\partial D}(0, Q, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial Q}{\partial y}\;dV \ \ \ \ \ (3)\end{array}$

for the so called type ${2}$ domain and type ${3}$ domain respectively.

3. Argue that for sufficiently nice ${D}$, it can be cut into finitely many pieces ${D_i}$, each of which are of both type ${1}$, ${2}$, and ${3}$ (the so called type ${4}$ domain). Thus we can compute the divergence integral ${\displaystyle \iiint_D \nabla \cdot \textbf{F}\,dV}$ by adding the integrals of the pieces and apply the previous result (using linearity). Argue that the orientation of the common face of these ${D_i}$ (if any) is opposite to each other and thus a cancellation argument will give the surface integral ${\displaystyle \iint _{\partial D}\textbf{F}\cdot \textbf{n}\;dS}$.

I am slightly disappointed by this approach because even for the so called type ${1}$ domain, we still cannot prove the divergence theorem very directly (say, by reducing to a double integral on some planar domain ${\Omega}$ and apply Green’s theorem), but have to further cut it into even smaller type ${4}$ domains.

In this note we modify the strategy above by proving (1), (2), (3) all hold on a type ${1}$ domain. In particular, I will show that (2) and (3) can be proved for a type ${1}$ domain by using Green’s theorem.

Proof:

1. We first prove (1). Let ${\textbf{F}=(0,0, R)}$, ${D=\{(x, y, z): (x, y)\in \Omega, \underline f(x, y)\le z\le \bar f(x, y)\}}$.

Let ${S_{\textrm{top}}=\{z=\bar f(x,y)\}}$ and ${S_{\textrm{bottom}}=\{z=\underline f(x,y)\}}$ be the top and bottom face respectively.

As an outward normal of ${S_{\textrm{top}}}$ and ${S_{\textrm{bottom}}}$ is given by ${(-\bar f_x, -\bar f_y, 1)}$ and ${({\underline f\,}_x, {\underline f\,}_y, -1)}$ respectively, it is easy to see that on the lateral face ${S_{\textrm{lateral}}}$, ${\textbf{F}\cdot n=0}$ and so

$\displaystyle \begin{array}{rl} \displaystyle \iint_{\partial D} \textbf{F}\cdot \textbf{n} \;dS =& \displaystyle \left(\iint_{S_{ \textrm{top} }}+\iint _{S_{\textrm{bottom}}}\right)\textbf{F}\cdot \textbf{n} \;dS\\ =& \displaystyle \iint_\Omega \left[R(x, y, \bar f(x, y))-R(x, y, \underline f(x, y))\right] dA\\ =& \displaystyle \iint_\Omega \left(\int_{\underline f(x,y)}^{\bar f(x,y)}\frac{\partial R}{\partial z}dz\right) dxdy\\ =& \displaystyle \iiint_D \frac{\partial R}{\partial z}dV. \end{array}$

2. Now we prove the remaining cases. For simplicity let us prove only (2), as (3) is similar. With the same setting as in the previous part except that ${\textbf{F}=(P, 0, 0)}$, we compute

$\displaystyle \begin{array}{rl} \displaystyle \left(\iint_{S_{ \textrm{top} }}+\iint _{S_{\textrm{bottom}}}\right)\textbf{F}\cdot \textbf{n} \;dS =\iint_\Omega \left[-P(x, y, \bar f (x, y))\bar f_x +P(x, y, \underline f (x, y)) {\underline f\,} _x\right] dA \ \ \ \ \ (4)\end{array}$

On the other hand, let ${\partial \Omega}$ be parametrized by ${\textbf{x}(t)=(x(t), y(t))}$ for ${t\in I}$ (with positive orientation), then the lateral face ${S_{\textrm{lateral}}}$ is given by

$\displaystyle \begin{array}{rl} \displaystyle S_{\textrm{lateral}}=\{(\textbf{x}(t), z): t\in I, \underline f(\textbf{x}(t))\le z\le \bar f(\textbf{x}(t))\}\end{array}$

and an outward normal is then given by ${(y', -x', 0)}$. From this we compute

$\displaystyle \begin{array}{rl} \displaystyle \iint _{S_{\textrm{lateral}}} \textbf{F}\cdot \textbf{n} \;dS =\int_{\partial \Omega}\left(\int_{\underline f(\textbf{x}(t))}^{\bar f(\textbf{x}(t))}P(x, y, z)dz\right)dy. \end{array}$

Define ${\phi(x, y): \Omega\rightarrow \mathbb R}$ by

$\displaystyle \begin{array}{rl} \displaystyle \phi(x, y)= \int_{\underline f(x,y)}^{\bar f(x, y)}P(x, y, z)dz. \end{array}$

Then by Green’s theorem,

$\displaystyle \begin{array}{rl} \displaystyle \iint _{S_{\textrm{lateral}}} \textbf{F}\cdot \textbf{n} \;dS =\int_{\partial \Omega}\phi \;dy =\iint_{\Omega}\frac{\partial \phi}{\partial x} \;dA. \ \ \ \ \ (5)\end{array}$

We compute

$\displaystyle \begin{array}{rl} \displaystyle \frac{\partial \phi}{\partial x}= P(x, y, \bar f(x, y))\bar f_x -P(x, y, \underline f(x, y)){\underline f\,}_x +\int_{\underline f(x,y)}^{\bar f (x, y)}\frac{\partial P}{\partial x}dz. \ \ \ \ \ (6)\end{array}$

 Exercise 1 Show that $\displaystyle \begin{array}{rl} \displaystyle \frac{d}{dx}\left(\int_{a(x)}^{b(x)}c(x, t)dt\right)=c(x, b(x))b'(x)-c(x, a(x))a'(x)+ \int_{a(x)}^{b(x)}\frac{\partial c}{\partial x}(x, t)dt. \end{array}$ Hint: let ${\psi(u, v, x)=\int_{u}^{v}c(x, t)dt}$ and use the chain rule and the fundamental theorem of calculus.

Adding (4) and (5) together and comparing with (6), we obtain

$\displaystyle \begin{array}{rl} \displaystyle \iint_{\partial D}\textbf{F}\cdot \textbf{n}\;dS =\iint_\Omega \left(\int_{\underline f(x,y)}^{\bar f(x, y)}\frac{\partial P}{\partial x}dz\right)dx\;dy =\iiint _D\frac{\partial P}{\partial x}dV. \end{array}$

$\Box$