State the divergence theorem. By applying this to $f(\mathbf{x}) \mathbf{k}$, where $f(\mathbf{x})$ is a scalar field in a closed region $V$ in $\mathbb{R}^{3}$ bounded by a piecewise smooth surface $S$, and $\mathbf{k}$ an arbitrary constant vector, show that

$$\int_{V} \nabla f d V=\int_{S} f d \mathbf{S}$$

A vector field $\mathbf{G}$ satisfies

$$\begin{gathered} \boldsymbol{\nabla} \cdot \mathbf{G}=\rho(\mathbf{x}) \\ \text { with } \quad \rho(\mathbf{x})= \begin{cases}\rho_{0} & |\mathbf{x}| \leqslant a \\ 0 & |\mathbf{x}|>a\end{cases} \end{gathered}$$

By applying the divergence theorem to $\int_{V} \nabla \cdot \mathbf{G} d V$, prove Gauss's law

$$\int_{S} \mathbf{G} \cdot d \mathbf{S}=\int_{V} \rho(\mathbf{x}) d V$$

where $S$ is the piecewise smooth surface bounding the volume $V$.

Consider the spherically symmetric solution

$$\mathbf{G}(\mathbf{x})=G(r) \frac{\mathbf{x}}{r}$$

where $r=|\mathbf{x}|$. By using Gauss's law with $S$ a sphere of radius $r$, centre $\mathbf{0}$, in the two cases $0<r \leqslant a$ and $r>a$, show that

$$\mathbf{G}(\mathbf{x})= \begin{cases}\frac{\rho_{0}}{3} \mathbf{x} & r \leqslant a \\ \frac{\rho_{0}}{3}\left(\frac{a}{r}\right)^{3} \mathbf{x} & r>a\end{cases}$$

The scalar field $f(\mathbf{x})$ satisfies $\mathbf{G}=\nabla f$. Assuming that $f \rightarrow 0$ as $r \rightarrow \infty$, and that $f$ is continuous at $r=a$, find $f$ everywhere.

By using a symmetry argument, explain why $(*)$ is clearly satisfied for this $f$ if $S$ is any sphere centred at the origin.