Surface integral of $f(x) = \frac{1}{ \Vert x -x_0 \Vert } $ over sphere

Question

Let $S \subseteq \mathbb{R}^3$ the sphere of radius $r$ centered at the origin. Let $x_0 \in \mathbb{R}^3$ be such that $x_0 \notin S $.

Let $f:S \to \mathbb{R}$ be such that $f(x) = \dfrac{1}{ \Vert x -x_0 \Vert }. $

Calculate the surface integral $$ \int_S f \Vert dσ \Vert$$.

The parametrization of the sphere is given by

$$ σ(x,y) = r( \cos(x) \sin(y) , \sin(x) \sin(y), \cos(y)).$$

Developing the definition of the integral I get that I have to find:

$$ \int_{[0,2π] \times [0,π]} \dfrac{r^2 sen(y)}{\Vert σ(x,y) - x_0 \Vert}.$$ Now, I don't know how to solve the last integral. I think I'm supposed to use Gauss theorem, but I can't find a way to express it as the integral of a differentiable vector function.

Any help would be appreciated! Thanks!

Answer 1

The case $\lVert\mathbf{x}_0\rVert=0$ is just integrating a constant, so let's assume $\lVert\mathbf{x}_0\rVert>0$.

Choosing spherical polar coordinates with the $\mathbf{x}_0$-direction being the zenith (equivalently, use a rotation to put $\mathbf{x}_0$ on the positive $z$-axis and use the usual spherical polars) the required integral becomes \begin{align*} \int_Sf\,\mathrm{d}S &=2\pi\int_0^\pi\frac{r^2\sin\theta}{\sqrt{r^2+\lVert\mathbf{x}_0\rVert^2-2r\,\lVert\mathbf{x}_0\rVert\cos\theta}}\,\mathrm{d}\theta\\ &=2\pi \frac{r}{\lVert\mathbf{x}_0\rVert}\left[\sqrt{r^2+\lVert\mathbf{x}_0\rVert^2-2r\,\lVert\mathbf{x}_0\rVert\cos\theta}\right]_0^\pi\\ &=\frac{2\pi r[(r+\lVert\mathbf{x}_0\rVert)-\lvert r-\lVert\mathbf{x}_0\rVert\rvert]}{\lVert\mathbf{x}_0\rVert}\\ &=\begin{cases} 4\pi r & \lVert\mathbf{x}_0\rVert<r\\ \dfrac{4\pi r^2}{\lVert\mathbf{x}_0\rVert} & \lVert\mathbf{x}_0\rVert>r \end{cases} \end{align*}