Let
\[V\]
be a region enclosed by a surface \[S\]
.If
\[\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\]
, then\[\int \int_S \frac{\mathbf{r} \cdot \mathbf{n}}{r^3} dS = \left\{ \begin{array}{cc} 0 & (0,0,0) \notin S \\ 4 \pi & (0,0,0) \in S \end{array} \right. \]
This result is known as Gauss's Theorem.
Proof
To prove the first part, use the Divergence Theorem,
\[\int \int_S \mathbf{F} \cdot \mathbf{n} dS = \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV\]
where \[S\]
the the surface of the volume \[V\]
and \[\mathbf{F}\]
is twice differentiable.Let
\[\mathbf{F} = \frac{\mathbf{r}}{r^3}\]
then\[\int \int_S \frac{\mathbf{r}}{r^3} \cdot \mathbf{n} dS = \in \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^3}) dV\]
\[\begin{equation} \begin{aligned} \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^3}) &=
( \frac{\partial }{\partial x} \mathbf{i} + \frac{\partial }{\partial y} \mathbf{j} + \frac{\partial }{\partial z} \mathbf{k} ) \cdot (\frac{x \mathbf{i}}{(x^2 + y^2 +z^2)^{3/2}} + \frac{y \mathbf{j}}{(x^2 + y^2 +z^2)^{3/2}} + \frac{z \mathbf{k}}{(x^2 + y^2 +z^2)^{3/2}}) \\ &=
\frac{\partial}{ \partial x}(\frac{x}{(x^2 + y^2 +z^2)^{3/2}}) + \frac{\partial}{ \partial y}(\frac{y}{(x^2 + y^2 +z^2)^{3/2}})+ \frac{\partial}{ \partial z}(\frac{z}{(x^2 + y^2 +z^2)^{3/2}} ) \\ &=
\frac{1}{(x^2 + y^2 +z^2)^{3/2}} - \frac{3x^2}{(x^2 + y^2 +z^2)^{5/2}} + \frac{1}{(x^2 + y^2 +z^2)^{3/2}} - \frac{3y^2}{(x^2 + y^2 +z^2)^{5/2}} + \frac{1}{(x^2 + y^2 +z^2)^{3/2}} - \frac{3z^2}{(x^2 + y^2 +z^2)^{5/2}} \\ &=
\frac{3}{r^3} - \frac{3 r^2}{r^5} \\ &=
0 \end{aligned} \end{equation}\]
To prove the second part let
\[V=V_1 +B(O, r_0 ), \: B(O, r_0 ) \in V\]
thenand let
\[S_0\]
be the surface of \[B(O, r_0 )\]
.\[\int \int_{S-S_0} \frac{\mathbf{r}}{r^3} \cdot \mathbf{n} dS = 0\]
Hence
\[\int \int_{S} \frac{\mathbf{r}}{r^3} \cdot \mathbf{n} dS = \int \int_{S_0} \frac{\mathbf{r}}{r^3} \cdot \mathbf{n} dS\]
For the surface
\[S_0 , \: \mathbf{n} = \frac{\mathbf{r}}{r_0}\]
Hence
\[\int \int_{S} \frac{\mathbf{r}}{r^3} \cdot \mathbf{n} dS = \int \int_{S_0} \frac{\mathbf{r}}{r_0^3} \cdot \frac{\mathbf{r}}{r_0} dS = \int \int_{S_0} \frac{r_0^2}{r^3_0} dS = \frac{1}{r_0} \times 4 \pi \r_0 = 4 \pi\]
For