How would I approach the question:
A volume $V$ is enclosed by a closed surface $S$. Show that
$$\iiint_V \frac{1}{r^2} dV = \iint_S \frac{\underline{r}.d\underline{S}}{r^2} $$
where $\underline{r}=x\underline{i}+y\underline{j}+z\underline{k}$.
I'm unsure of where to start.
We have $$ \nabla\cdot\left(\frac{\mathbf{r}}{r^2}\right)=\sum_{i=1}^3\frac{\partial}{\partial x_i}\left(\frac{x_i}{r^2}\right)\\ =\frac{1}{r^4}\sum_{i=1}^3\left(r^2\frac{\partial}{\partial x_i}x_i-x_i\frac{\partial}{\partial x_i}r^2\right)\\ =\frac{1}{r^4}\sum_{i=1}^3\left(3r^2-2x_ix_i\right)=\frac{1}{r^2} $$ so by the divergence theorem $$ \iint_S\frac{\mathbf r\cdot d\mathbf S}{r^2}=\iiint_V \nabla\cdot\left(\frac{\mathbf{r}}{r^2}\right)dV=\iiint_V \frac{1}{r^2}dV $$