2 Gauss’ theorem

This is sometimes also known as the divergence theorem and is similar to Stokes’ theorem but equates a surface integral to a volume integral. Gauss’ theorem states that for a volume $V$ , bounded by a closed surface $S$ , any ‘well-behaved’ vector field $\underset{̲}{F}$ satisfies

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$

Notes

1. $\underset{̲}{dS}$ is a unit normal pointing outwards.
2. Both sides of the equation are scalars.
3. The theorem is often a useful way of calculating a surface integral over a surface composed of several distinct parts (e.g. a cube).
4. $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ is a scalar field representing the divergence of the vector field $\underset{̲}{F}$ and may, alternatively, be written as div $\underset{̲}{F}$ .
5. Gauss’ theorem can be justified in a manner similar to that used for Stokes’ theorem (i.e. by proving it for a small volume element, then summing up the volume elements and allowing the internal surface contributions to cancel.)

Example 34

Verify Gauss’ theorem for the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ and the function $\underset{̲}{F}=x\underset{̲}{i}+z\underset{̲}{j}$

Solution

To find $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}$ , the integral must be evaluated for all six faces of the cube and the results summed.

On the left face, $x=0$ , $\underset{̲}{F}=z\underset{̲}{j}$ and $\underset{̲}{dS}=-\underset{̲}{i}dydz$ so $\underset{̲}{F}\cdot \underset{̲}{dS}=0$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}0dydz=0$

On the right face, $x=1$ , $\underset{̲}{F}=\underset{̲}{i}+z\underset{̲}{j}$ and $\underset{̲}{dS}=\underset{̲}{i}dydz$ so $\underset{̲}{F}\cdot \underset{̲}{dS}=1dydz$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}1dydz=1$

On the front face, $y=0$ , $\underset{̲}{F}=x\underset{̲}{i}+z\underset{̲}{j}$ and $\underset{̲}{dS}=-\underset{̲}{j}dxdz$ so $\underset{̲}{F}\cdot \underset{̲}{dS}=-zdxdz$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=-{\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}zdxdz=-\frac{1}{2}$

On the back face, $y=1$ , $\underset{̲}{F}=x\underset{̲}{i}+z\underset{̲}{j}$ and $\underset{̲}{dS}=\underset{̲}{j}dxdz$ so $\underset{̲}{F}\cdot \underset{̲}{dS}=zdxdz$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}zdxdz=\frac{1}{2}$

On the bottom face, $z=0$ , $\underset{̲}{F}=x\underset{̲}{i}$ and $\underset{̲}{dS}=-\underset{̲}{k}dydz$ so $\underset{̲}{F}\cdot \underset{̲}{dS}=0dxdy$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}0dxdy=0$

On the top face, $z=1$ , $\underset{̲}{F}=x\underset{̲}{i}+\underset{̲}{j}$ and $\underset{̲}{dS}=\underset{̲}{k}dydz$ so $\underset{̲}{F}\cdot \underset{̲}{dS}=0dxdy$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}0dxdy=0$

Thus, summing over all six faces, $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=0+1-\frac{1}{2}+\frac{1}{2}+0+0=1.$

To find $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$ note that $\underset{̲}{\nabla }\cdot \underset{̲}{F}=\frac{\partial }{\partial x}x+\frac{\partial }{\partial y}z=1+0=1$ .

So $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{1}1dxdydz=1$ .

So $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV=1.$

Note that the volume integral required just one triple integral while the surface integral required six double integrals. Reducing the number of integrals is often the motivation for using Gauss’ theorem.

Example 35

Use Gauss’ theorem to evaluate the surface integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}$ where $\underset{̲}{F}$ is the vector field $x^{2}y\underset{̲}{i}+2xy\underset{̲}{j}+z^3\underset{̲}{k}$ and $S$ is the surface of the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ .

Solution

Note that to carry out the surface integral directly will involve the evaluation of six double integrals (one for each face of the cube). However, by Gauss’ theorem, the same result comes from the surface integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_V\underset{̲}{\nabla }\cdot \underset{̲}{F}dV$ . As $\underset{̲}{\nabla }\cdot \underset{̲}{F}=2xy+2x+3z^2$ , the surface integral becomes the triple integral

The six double integrals also sum to $\frac{5}{2}$ but this approach requires a greater amount of work.