Integral of Reciprocal of r Squared Over a Region Excluding the Origin

The Divergence Theorem states  
\[\int \int \int \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int \frac{\mathbf{r} \cdot \mathbf{r}} {r^2}dS\]

Let  
\[\mathbf{F} = \frac{\mathbf{r}}{r^2}\]

\[ \begin{equation} \begin{aligned} \mathbf{\nabla} \cdot ( \frac{\mathbf{r}}{r^2}) &= (\frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k}) \cdot (\frac{x}{x^2 + y^2+z^2} \mathbf{i} + \frac{y}{x^2 + y^2+z^2} \mathbf{j} + \frac{z}{x^2 + y^2+z^2} \mathbf{k}) \\ &= \frac{1}{x^2 +y^2 +z^2} - \frac{2x^2}{(x^2 +y^2 +z^2)^2} + \frac{1}{x^2 +y^2 +z^2} - \frac{2y^2}{(x^2 +y^2 +z^2)^2} + \frac{1}{x^2 +y^2 +z^2} - \frac{2z^2}{(x^2 +y^2 +z^2)^2} \\ &= \frac{3(x^2 +y^2+z^2) - 2(x^2 +y^2 +z^2)}{(x^2+y^2+z^2)^2} \\ &= \frac{1}{r^2} \end{aligned} \end{equation}\]
|
Hence  
\[\int \int \int \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int \frac{\mathbf{r} \cdot \mathbf{r}} {r^2}dS\]