# Gauss's Theorem

## Theorem

Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.

Let $\mathbf F: \R^3 \to \R^3$ be a smooth vector function defined on a neighborhood of $U$.

Then:

$\displaystyle \iiint \limits_U \left({\nabla \cdot \mathbf F} \right) \mathrm d V = \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \ \mathrm d S$

where $\mathbf n$ is the normal to $\partial U$.

## Proof

It suffices to prove the theorem for rectangular prisms; the Riemann-sum nature of the triple integral then guarantees the theorem for arbitrary regions.

Let:

$R = \left\{ {\left({x, y, z}\right): a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le c_2}\right\}$

and let $S = \partial R$, oriented outward.

Then :

$S = A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6$

where:

$A_1, A_2$ are those sides perpendicular to the $x$-axis
$A_3, A_4$ are those sides perpendicular to the $y$ axis

and

$A_5, A_6$ are those sides perpendicular to the $z$-axis

and in all cases the lower subscript indicates a side closer to the origin.

Let:

$\mathbf F = M \mathbf i + N \mathbf j + P \mathbf k$

where $M, N, P: \R^3 \to \R$.

Then:

 $\displaystyle \iiint_R \nabla \cdot \mathbf F \ \mathrm d V$ $=$ $\displaystyle \iiint_R \left({ \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z} }\right) \ \mathrm d x \ \mathrm d y \ \mathrm d z$ $\displaystyle$ $=$ $\displaystyle \iiint_R \frac{\partial M}{\partial x} \ \mathrm d x \ \mathrm d y \ \mathrm d z + \iiint_M \frac{\partial N}{\partial y} \ \mathrm d x \ \mathrm d y \ \mathrm d z + \iiint_M \frac{\partial P}{\partial z} \ \mathrm d x \ \mathrm d y \ \mathrm d z$ $\displaystyle$ $=$ $\displaystyle \int_{c_1}^{c_2} \int_{b_1}^{b_2} \left({M \left({a_2, y, z}\right) - M \left({a_1, y, z}\right)}\right) \ \mathrm d y \ \mathrm d z$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \int_{c_1}^{c_2} \int_{a_1}^{a_2} \left({N \left({x, b_2, z}\right) - N \left({x, b_1, z}\right)}\right) \ \mathrm d x \ \mathrm d z$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \int_{b_1}^{b_2} \int_{a_1}^{a_2} \left({P \left({x, y, c_2}\right) - P \left({x, y, c_1}\right)}\right) \ \mathrm d x \ \mathrm d y$ $\displaystyle$ $=$ $\displaystyle \iint_{A_2} M \ \mathrm d y \ \mathrm d z - \iint_{A_1} M \ \mathrm d y \ \mathrm d z$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \iint_{A_4} N \ \mathrm d x \ \mathrm d z - \iint_{A_3} N \ \mathrm d x \ \mathrm d z$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \iint_{A_6} P \ \mathrm d x \ \mathrm d y - \iint_{A_5} P \ \mathrm d x \ \mathrm d y$

We turn now to examine $\mathbf n$:

 $\displaystyle \mathbf n$ $=$ $\displaystyle \left({-1, 0, 0}\right)$ on $A_1$ $\displaystyle \mathbf n$ $=$ $\displaystyle \left({1, 0, 0}\right)$ on $A_2$ $\displaystyle \mathbf n$ $=$ $\displaystyle \left({0, -1, 0}\right)$ on $A_3$ $\displaystyle \mathbf n$ $=$ $\displaystyle \left({0, 1, 0}\right)$ on $A_4$ $\displaystyle \mathbf n$ $=$ $\displaystyle \left({0, 0, -1}\right)$ on $A_5$ $\displaystyle \mathbf n$ $=$ $\displaystyle \left({0, 0, 1}\right)$ on $A_6$

Hence:

 $\displaystyle \mathbf F \cdot \mathbf n$ $=$ $\displaystyle -M$ on $A_1$ $\displaystyle \mathbf F \cdot \mathbf n$ $=$ $\displaystyle M$ on $A_2$ $\displaystyle \mathbf F \cdot \mathbf n$ $=$ $\displaystyle -N$ on $A_3$ $\displaystyle \mathbf F \cdot \mathbf n$ $=$ $\displaystyle N$ on $A_4$ $\displaystyle \mathbf F \cdot \mathbf n$ $=$ $\displaystyle -P$ on $A_5$ $\displaystyle \mathbf F \cdot \mathbf n$ $=$ $\displaystyle P$ on $A_6$

We also have:

 $\displaystyle \mathrm d S$ $=$ $\displaystyle \mathrm d y \ \mathrm d z$ on $A_1$ and $A_2$ $\displaystyle \mathrm d S$ $=$ $\displaystyle \mathrm d x \ \mathrm d z$ on $A_3$ and $A_4$ $\displaystyle \mathrm d S$ $=$ $\displaystyle \mathrm d x \ \mathrm d y$ on $A_5$ and $A_6$

where $\mathrm d S$ is the area element.

This is true because each side is perfectly flat, and constant with respect to one coordinate.

Hence:

 $\displaystyle \iint_{A_2} \mathbf F \cdot \mathbf n \ \mathrm d S$ $=$ $\displaystyle \iint_{A_2} M \ \mathrm d y \ \mathrm d z$

and in general:

 $\displaystyle \sum_{i \mathop = 1}^6 \iint_{A_i} \mathbf F \cdot \mathbf n \ \mathrm d S$ $=$ $\displaystyle \iint_{A_2} M \ \mathrm d y \ \mathrm d z - \iint_{A_1} M \ \mathrm d y \ \mathrm d z + \iint_{A_4} N \ \mathrm d x \ \mathrm d z - \iint_{A_3} N \ \mathrm d x \ \mathrm d z + \iint_{A_6} P \ \mathrm d x \ \mathrm d y - \iint_{A_5} P \ \mathrm d x \ \mathrm d y$

Hence the result:

$\displaystyle \iiint_R \nabla \cdot \mathbf F \ \mathrm d V = \iint_{\partial R} \mathbf F \cdot \mathbf n \ \mathrm d S$

$\blacksquare$

## Also known as

This result is also known as the divergence theorem.

## Source of Name

This entry was named for Carl Friedrich Gauss.

## Historical Note

Gauss's Theorem was proved by Carl Friedrich Gauss during the course of his investigations into electromagnetism.