[Math] Show that $\nabla\cdot\left(\dfrac{\mathbf{e}_r}{r^2}\right)=4\pi\delta(\mathbf{r})$ using the divergence theorem.

Question

The book answer goes as follows:

By the divergence theorem, in spherical coordinates we find $$\color{red}{\iiint_\limits{\large\text{volume}\,\tau}\nabla\cdot\left(\dfrac{\mathbf{e}_r}{r^2}\right)\mathrm{d}\tau}=\color{blue}{\iint_\limits{\large\text{surface enclosing}\, \tau}\dfrac{\mathbf{e}_r}{r^2}\cdot\mathbf{e}_r\,\mathrm{d}\sigma}=\color{#180}{\int_{\phi=0}^{2\pi}\int_{\theta=0}^{\pi}\frac{1}{r^2}r^2\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\phi}=4\pi$$ Thus $\nabla\cdot\left(\dfrac{\mathbf{e}_r}{r^2}\right)$ has the properties that it is zero $\forall\,r\gt 0$ but its integral over any volume including the origin is $4\pi$; this suggests that it is equal to $4\pi\delta(\mathbf{r})$.

As mentioned in a comment below; $\mathbf{e}_r$ is a unit radial vector.

I know that the $\color{red}{\mathrm{red}}$ and $\color{blue}{\mathrm{blue}}$ integrals are a statement of the divergence theorem. The only thing I can't understand is how the $\color{#180}{\mathrm{green}}$ integral was obtained from the $\color{blue}{\mathrm{blue}}$ integral.

I know that $$\mathrm{d}\sigma=\left|\frac{\partial r}{\partial \theta}\times\frac{\partial r}{\partial \phi}\right|\,\mathrm{d}\theta\,\mathrm{d}\phi\tag{1}$$ I think that equation $(1)$ has been used but I'm not sure how to use it. Could someone please explain how the $\color{#180}{\mathrm{green}}$ integral was reached?

Answer 1

The surface integral can be evaluated in spherical coordinates:

1). Set

\begin{align} x&=r\sin\theta\cos\phi\;,\\ y&=r\sin\theta\sin\phi\;,\\ z&=r\cos\theta\;, \end{align}

2). Write

$\vec r(\theta,\phi)=r\sin\theta\cos\phi \vec i+r\sin\theta\sin\phi\vec j+r\cos\theta \vec k$

3). Then

$\vec r_{\theta}(\theta, \phi)=r\cos\theta\cos\phi \vec i+r\cos\theta\sin\phi\vec j-r\sin\theta \vec k$

and

$\vec r_{\phi}(\theta, \phi)=-r\sin\theta\sin\phi \vec i+r\sin\theta\cos\phi\vec j$

4). Find the magnitude of the cross product:

$\color{red}{\vert \vec r_{\phi}\times \vec r_{\theta}\vert =r^2\sin \theta.} $ $(\sin \theta \geq 0$ since $0\leq \theta <\pi)$,

so that

$d\sigma =r^2\sin \theta d\theta d\phi$.

5). Find the limits of integration:

$0\leq \phi <2\pi;\ 0\leq \theta <\pi$ because the surface is a sphere.

6). Substitute into the integral:

$\color{blue}{\iint_\limits{\large\text{surface enclosing}\, \tau}\dfrac{\mathbf{e}_r}{r^2}\cdot\mathbf{e}_r\,\mathrm{d}\sigma }=\int \int _{\sigma }\frac{1}{r^{2}}d\sigma =\int^{2\pi }_0 \int^{\pi }_0 \frac{1}{r^{2}}(r^{2}\sin \theta )d\theta d\phi=$