(Redirected from 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:

$\displaystyle \iiint \limits_U \paren {\nabla \cdot \mathbf F} \rd V = \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \rd S$

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:

$R = \set {\tuple {x, y, z}: a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le c_2}$

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:

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

We turn now to examine $\mathbf n$:

 $\displaystyle \mathbf n$ $=$ $\displaystyle \tuple {-1, 0, 0}$ on $A_1$ $\displaystyle \mathbf n$ $=$ $\displaystyle \tuple {1, 0, 0}$ on $A_2$ $\displaystyle \mathbf n$ $=$ $\displaystyle \tuple {0, -1, 0}$ on $A_3$ $\displaystyle \mathbf n$ $=$ $\displaystyle \tuple {0, 1, 0}$ on $A_4$ $\displaystyle \mathbf n$ $=$ $\displaystyle \tuple {0, 0, -1}$ on $A_5$ $\displaystyle \mathbf n$ $=$ $\displaystyle \tuple {0, 0, 1}$ 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:

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

where $\d S$ is the area element.

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

Hence:

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

and in general:

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

Hence the result:

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

$\blacksquare$

## Also known as

The Gauss-Ostrogradsky Theorem is also known as:

the Divergence Theorem
Gauss's Theorem

## Source of Name

This entry was named for Carl Friedrich Gauss and Mikhail Vasilyevich Ostrogradsky.

## Historical Note

The Gauss-Ostrogradsky Theorem was first discovered by Joseph Louis Lagrange in $1762$.

It was the later independently rediscovered by Carl Friedrich Gauss in $1813$ during the course of his investigations into electromagnetism.

Mikhail Vasilyevich Ostrogradsky, who also gave the first proof of the general theorem, rediscovered it in $1826$.

It was also rediscovered by George Green in $1828$, Siméon-Denis Poisson in $1824$ and Pierre Frédéric Sarrus in $1828$.