\(E\) a region in \(\reals^3\) (of appropriate shape as discussed below), with boundary \(S = \partial E\) being smooth and with outward unit normal \(\N\text{,}\) and
\(\vec{F}\) a vector field whose components all have continuous partial derivatives on \(E\) and an open region containing it,
The theorem can be proved in a way similar to what was done for Green’s theorem Section 6.4:
Break the equation to a sum of three parts, one for the partial derivative in each variable: with \(\vec{F} = P\veci + Q\vecj + R\veck\text{,}\) the equation can be expanded in components as
\begin{align}
\iiint\limits_E \frac{\partial P}{\partial x}\ dV \amp= \iint\limits_S P\ \veci \cdot \N dS\tag{6.8.3}\\
\iiint\limits_E \frac{\partial Q}{\partial y}\ dV \amp= \iint\limits_S Q\ \vecj \cdot \N dS\tag{6.8.4}\\
\iiint\limits_E \frac{\partial R}{\partial z}\ dV \amp= \iint\limits_S R \veck\ \cdot \N dS.\tag{6.8.5}
\end{align}
For one of these parts (we will use Equation (6.8.3)) prove the result on a region of a “nice” shape (we will use Type 1 as introduced in Subsection 5.4.5).
Argue that any region can be divided into a collection of these nice regions, and that summing the integrals over these pieces gives the result in general.
Note that the same argument works for the other two parts, so that adding them give the result.
Remark6.8.2.
A region that is of all three types is called a simple solid region. Another approach to proving the divergence theorem is to prove it first for simple solid regions, and then argue as in Item 3. In fact, in OpenStax Calculus Volume 3, Section 6.8 3
\begin{equation}
E = \{(x, y, z): (x,y) \in D, b(x,y) \le z \leq t(x,y)\}\tag{6.8.6}
\end{equation}
and thus has boundary consisting of
a “top” \(T = \{(x, y, z): (x,y) \in D, z = t(x,y)\}\text{,}\)
a “bottom” \(B = \{(x, y, z): (x,y) \in D, z = b(x,y)\}\text{,}\)
and a vertical part \(V = \{(x, y, z): (x,y) \in \partial D, b(x,y) \le z \leq t(x,y)\}.\)
Also note that on the vertical part \(V\text{,}\) the unit normal is horizontal (it is the same as for the base \(D\)), so \(\N \cdot \veck = 0\) there.
\begin{equation*}
\iiint\limits_E \frac{\partial R}{\partial z}\ dV = \iint\limits_D \left[ \int_{z=b(x,y)}^{z=t(x,y)} \frac{\partial R}{\partial z}\ dz \right] dA
\end{equation*}
and the integral in brackets can be evaluated with the Fundamental Theorem of Calculus (which is the one-dimensional little brother of the divergence theorem) so
Starting at the other end, the above division of the boundary \(S\) and the fact that \(\N \cdot \veck = 0\) on the vertical sides \(V\) gives
\begin{align}
\iint\limits_S R \veck \cdot \N dS \amp = \iint\limits_T R \veck \cdot \N dS + \iint\limits_B R \veck \cdot \N dS
+ \iint\limits_V R \veck \cdot \N dS\notag\\
\amp = \iint\limits_T R \veck \cdot \N dS + \iint\limits_B R \veck \cdot \N dS\tag{6.8.8}
\end{align}
For the top surface \(T\text{,}\) the outward unit normal is the upward unit normal for the graph \(z=t(x,y)\) so the integral can be evaluated using Equation (6.6.15) in Section 6.7:
Subsection6.8.4Verifying Equation (6.8.5) for general regions.
Proving Equation (6.8.5) rigorously on more general regions is quite difficult, but the basic idea is to divide a solid region \(E\) into a collection of Type 1 regions; much as in the proof of Green’s theorem Green’s Theorem (Circulation Form) where a region in the plane was divided into a collection of Type 1 regions.
One way to approach this is to “dice” the region by cutting with planes parallel to all three coordinate directions at tiny spacing \(\epsilon\text{:}\) most will be entirely internal, so cubes of edge size \(\epsilon\text{,}\) and so of type 1; the ones that reach the boundary will generically have \(\N\cdot\veck \ne 0\) over all of its external boundary, so that boundary has either \(\N\cdot\veck \gt 0\) (and so is at the top) or \(\N\cdot\veck \lt 0\) (and so at the bottom).
The total volume integral is the sum of the volume integrals over each simple piece, and so is the sum of the surface integrals over all the pieces.
This sum of surface integrals from the external boundary \(\partial E\) is the desired integral over the surface of the whole region plus, but there are extra integrals over the new internal surfaces produced by the cutting into simple pieces. Fortunately, all these extra internal bits of surface come in pairs on either side of each cut, with the outward normal being in the opposite direction on the two sides of the cut. Thus the values of those extra internal surface integrals are negatives of each other, so these pairs of extra surface integrals cancel out.
That leaves just the desired surface integrals over the boundary of the domain.
Subsection6.8.5The Flux of an Electric Field Through a Surface
The electric field around a point charge of \(Q\) at the origin of the coordinates is
for any surface \(\partial D\) that is the boundary of a region \(D\) containing the origin (and thus containing this point charge).
Exercise6.8.3.
Verify this, as follows:
First compute the integral directly for the case of \(\partial D\) a sphere of radius \(a\) centered at the origin.
Then use the Divergence Theorem to get the general version.
THen extend to a collection of point charges \(Q_1 \dots Q_n\) at locations \(\vec{x}_1 \dots \vec{x}_n\) inside region \(D\text{,}\) with the integral then giving the total charge in the region.
Subsection6.8.6Gauss’s Law
Equation (6.8.11) is a case of Gauss’s Law of Electrostatics. In general, Gauss’s Law says that the above flux integral always gives the total charge inside a region with boundary the surface \(S\text{:}\) this allows the charge within a region to be determined by measurements only at the boundary of the region.
Indeed, Gauss discovered the Divergence Theorem through his work on electrostatics.
