Gauss-Ostrogradsky Theorem/Formal Proof

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Theorem

Let $U$ be a subset of $\R^3$ which is compact and has a piecewise smooth boundary $\partial U$.

Let $\mathbf V: \R^3 \to \R^3$ be a smooth vector field defined on a neighborhood of $U$.

Then:

$\ds \iiint \limits_U \paren {\nabla \cdot \mathbf V} \rd v = \iint \limits_{\partial U} \mathbf V \cdot \mathbf n \rd S$

where $\mathbf n$ is the unit normal to $\partial U$, directed outwards.

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.

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Let:

$R = \set {\tuple {x, y, z}: a_1 \le x \le a_2, b_1 \le y \le b_2, c_1 \le z \le c_2}$

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.

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Let:

$\mathbf V = M \mathbf i + N \mathbf j + P \mathbf k$

where $M, N, P: \R^3 \to \R$.

Then:

$\ds \iiint_R \nabla \cdot \mathbf V \rd v$

$=$

$\ds \iiint_R \paren {\frac {\partial M} {\partial x} + \frac {\partial N} {\partial y} + \frac {\partial P} {\partial z} } \rd x \rd y \rd z$

$\ds $

$=$

$\ds \iiint_R \frac {\partial M} {\partial x} \rd x \rd y \rd z + \iiint_M \frac {\partial N} {\partial y} \rd x \rd y \rd z + \iiint_M \frac {\partial P} {\partial z} \rd x \rd y \rd z$

$\ds $

$=$

$\ds \int_{c_1}^{c_2} \int_{b_1}^{b_2} \paren {\map M {a_2, y, z} - \map M {a_1, y, z} } \rd y \rd z$

$\ds $

$\, \ds + \, $

$\ds \int_{c_1}^{c_2} \int_{a_1}^{a_2} \paren {\map N {x, b_2, z} - \map N {x, b_1, z} } \rd x \rd z$

$\ds $

$\, \ds + \, $

$\ds \int_{b_1}^{b_2} \int_{a_1}^{a_2} \paren {\map P {x, y, c_2} - \map P {x, y, c_1} } \rd x \rd y$

$\ds $

$=$

$\ds \iint_{A_2} M \rd y \rd z - \iint_{A_1} M \rd y \rd z$

$\ds $

$\, \ds + \, $

$\ds \iint_{A_4} N \rd x \rd z - \iint_{A_3} N \rd x \rd z$

$\ds $

$\, \ds + \, $

$\ds \iint_{A_6} P \rd x \rd y - \iint_{A_5} P \rd x \rd y$

We turn now to examine $\mathbf n$:

$\ds \mathbf n$

$=$

$\ds \tuple {-1, 0, 0}$

on $A_1$

$\ds \mathbf n$

$=$

$\ds \tuple {1, 0, 0}$

on $A_2$

$\ds \mathbf n$

$=$

$\ds \tuple {0, -1, 0}$

on $A_3$

$\ds \mathbf n$

$=$

$\ds \tuple {0, 1, 0}$

on $A_4$

$\ds \mathbf n$

$=$

$\ds \tuple {0, 0, -1}$

on $A_5$

$\ds \mathbf n$

$=$

$\ds \tuple {0, 0, 1}$

on $A_6$

Hence: