Gauss-Ostrogradsky Theorem

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.

Informal Proof

Let $S$ be the surface of $U$.

By definition, the surface integral of $\mathbf V$ over $S$ is therefore defined as:

$\ds \iint_S \mathbf V \cdot \mathbf n \rd S = \iint_S \mathbf V \cdot \rd \mathbf S$

where $\rd \mathbf S$ is the differential of vector area.

Let $\d v$ be an element of volume within $U$.

In the above diagram, a small cuboid has been used for convenience.

The total flux emerging from $\d v$ is:

$\operatorname {div} \mathbf V \rd v$

where in this context $\mathbf V$ is the vector quantity at the center of $\d v$.

For the face $abcd$, the positive direction of the $x$ component of $\mathbf V$ and the outward normal are in the same sense, and so the flux is positive.

Consider another small cuboid contiguous to $\d v$, shown dotted in the above diagram.

This has the same component of $\mathbf V$, but this is acting inward, and so is negative.

Hence the flux through the face of one cuboid cancels out the flux through the adjacent face of the next cuboid.

This argument can be applied by imagining a sequence of such cuboids until you reach the surface of $U$.

This is the only part of the construction we have just built which contributes to the flux through $S$.

By applying the same treatment to the surfaces of all such small elements of volume, we arrive at a total flux:

$\ds \iint_S \mathbf V \cdot \rd \mathbf S$

But at the same time we have integrated $\operatorname {div} \mathbf V \rd v$ throughout the whole of the volume of $U$.

This also measures the total flux through $S$.

Hence:

$\ds \iint_S \mathbf V \cdot \rd \mathbf S$

$=$

$\ds \iint_S \mathbf V \cdot \mathbf n \rd S$

$\ds $

$=$

$\ds \iiint \operatorname {div} \mathbf V \rd v$

$\ds $

$=$

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

Formal 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: