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

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!

• – HK Lee May 14 '19 at 5:10

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\rVertr \end{cases} \end{align*}
• Isn't this the potential of a charged spherical shell with uniform charge density? Shouldn't the answer for $\mathbf{x}_0$ in the exterior of $S$ be $4\pi r^2/\|\mathbf{x}_0\|$ which would agree with Gauss' law (and the duplicate answer)? – RRL May 14 '19 at 5:20