Gauss-Ostrogradsky Theorem

Theorem
Suppose $$U \ $$ is a subset of $$\R^3 \ $$ which is compact and has a piecewise smooth boundary. If $$F:\R^3 \to \R^3 \ $$ is a smooth vector function defined on a neighborhood of $$U \ $$, then we have


 * $$\iiint\limits_U\left(\nabla\cdot\mathbf{F}\right)dV=\iint\limits_{\part U} \mathbf{F} \cdot \mathbf{n}\ dS \ $$

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 $$R = \left\{{(x,y,z)|a_1 \leq x \leq a_2, b_1 \leq y \leq b_2, c_1 \leq z \leq c_2 }\right\} \ $$ and let $$S = \partial R \ $$, oriented outward. Let $$\mathbf{F} = M\mathbf{i}+N\mathbf{j}+P\mathbf{k} \ $$, where $$M,N,P:\R^3 \to \R \ $$. Then

$$\iiint_R \left({ \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z} }\right) dx dy dz = \ $$