Gauss-Ostrogradsky 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:

We turn now to examine $\mathbf n$:

Hence:

We also have:

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:

and in general:

Hence the result:


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

Also known as
This result is also known as the divergence theorem.