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

$\displaystyle\iiint_R \nabla \cdot \mathbf{F} dV = \iiint_R \left({ \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z} }\right) dx dy dz = \iiint_R \frac{\partial M}{\partial x} dxdydz + \iiint_M \frac{\partial N}{\partial y} dxdydz + \iiint_M \frac{\partial P}{\partial z} dxdydz $

$\displaystyle= \int_{c_1}^{c_2} \int_{b_1}^{b_2} \left({ M(a_2,y,z)-M(a_1,y,z) }\right)     dydz + \int_{c_1}^{c_2} \int_{a_1}^{a_2}     \left({ N(x,b_2,z)-N(x,b_1,z) }\right)      dxdz + \int_{b_1}^{b_2} \int_{a_1}^{a_2} \left({ P(x,y,c_2)-P(x,y,c_1) }\right) dxdy $

But note that this is precisely

$\displaystyle =\iint_{A_2} M dydz - \iint_{A_1} M dydz + \iint_{A_4} N dxdz - \iint_{A_3} N dxdz + \iint_{A_6} P dxdy - \iint_{A_5} P dxdy$

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 $dS = dxdz$, and 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

$\displaystyle \iint_{A_2} Mdydz= \iint_{A_2} \mathbf{F} \cdot \mathbf{n} dS$

and in general

$\displaystyle =\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 = \sum_{i=1}^6 \iint_{A_i} \mathbf{F}\cdot \mathbf{n} dS $

and so

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