Gauss's Theorem

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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:

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


We turn now to examine $\mathbf n$:

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


Hence:

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


We also have:

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

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:

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

and in general:

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


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.