Gauss-Ostrogradsky Theorem/Informal Proof

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.


 * Gauss-Ostrogradsky-theorem-proof-1.png

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

In the above diagram, a small cube 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 cube 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 cube cancels out the flux through the adjacent face of the next cube.

This argument can be applied by imagining a sequence of such cubes 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:

Also see

 * Green's Theorem