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


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

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.

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, $A_3, A_4$ 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:

Thus


 * $\displaystyle\iiint_R \nabla \cdot \mathbf{F} dV =\iint_{A_2} M\ dydz - \iint_{A_1} M\ dydz + \iint_{A_4} N\ dxdz - \iint_{A_3} N\ dxdz + \iint_{A_6} P\ dxdy - \iint_{A_5} P\ dxdy$

We turn now to examine $\mathbf{n}$:


 * On $A_1, \mathbf{n} = (-1,0,0)$
 * On $A_2, \mathbf{n} = (1,0,0)$
 * On $A_3, \mathbf{n} = (0,-1,0)$
 * On $A_4, \mathbf{n} = (0,1,0)$
 * On $A_5, \mathbf{n} = (0,0,-1)$
 * On $A_6, \mathbf{n} = (0,0,1)$.

Hence:


 * On $A_1, \mathbf{F} \cdot \mathbf{n} = -M$
 * On $A_2, \mathbf{F} \cdot \mathbf{n} = M$
 * On $A_3, \mathbf{F} \cdot \mathbf{n} = -N$
 * On $A_4, \mathbf{F} \cdot \mathbf{n} = N$
 * On $A_5, \mathbf{F} \cdot \mathbf{n} = -P$
 * On $A_6, \mathbf{F} \cdot \mathbf{n} = P$.

We also have:


 * On $A_1$ and $A_2$, the area element is $dS=dydz$
 * On $A_3$ and $A_4$, the area element is $dS = dxdz$
 * On $A_5$ and $A_6, dS= dxdy$

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


 * $\displaystyle \iint_{A_2} M\ dydz= \iint_{A_2} \mathbf{F} \cdot \mathbf{n}\ dS$

and in general

and so

$\displaystyle \iiint_R \nabla \cdot \mathbf{F}\ dV = \iint_{\partial R} \mathbf{F}\cdot \mathbf{n}\ dS$