Question

Let A = f(r) rˆ. Verify Gauss’s theorem for the sphere r = R.

Answer 1

Given

$\vec A=f(r)\hat r$

Therefore

$\vec \nabla.\vec A=\frac{1}{r^2}\frac{d}{dr}(r^2f(r))$

Hence

$\begin{align} \int_V\vec \nabla.\vec A\,dV=\int \frac{1}{r^2}\frac{d}{dr}(r^2f(r))4\pi r^2 dr=4\pi\int \frac{d}{dr}(r^2 f(r))dr=4\pi\int d(r^2 f(r))=4\pi r^2 f(r) \nonumber \end{align}$

Since the radius of the sphere is R, the above integral over r is from 0 to R which gives the result

$\int_V\vec \nabla.\vec A\, dV=4\pi R^2 f(R)$ ...(1)

And

$\begin{align} \oint_S\vec A.\hat r\,dS=\oint_S f(r)dS\nonumber \end{align}$

where the integral over the surface element is given by

$\oint dS=r^2\oint d\Omega=4\pi r^2$

The surface is at r = R. Therefore

$\oint_S\vec A.\hat r\,dS=R^2f(R)\oint d\Omega=4\pi R^2f(R)$ ....(2)

Since equation (1) and (2) are the same, Gauss' divergence is proved.