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.

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

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

$$= \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

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

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

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

and in general

$$=\iint_{A_2} M dydz - \iint_{A_1} M dydz + \iint_{A_4} N dxdz - \iint_{A_3} N dxdz + \iint_{A_6} P dxdy - \iint_{A_5} P dxdy = \sum_{i=1}^6 \iint_{A_i} \mathbf{F}\cdot \mathbf{n} dS $$

and so

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