How to evaluate $S=\int_{\partial D}{\frac{1}{|x-y|^2}}ds(y)$?

Question

Let $D=B(0,r_0), x,y \in \mathbb{R}^3$ and $x\in D$. \begin{align*} & S=\int_{\partial D}{\frac{1}{|x-y|^2}}ds(y)=\int_{\partial D}{\frac{1}{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}}ds(y) \end{align*} I want to prove that $S\leq k$ ($k>0$ and $k$ only depends on $r_0$).

In case $x \ne 0$, my idea is \begin{split} S^+ &=\int_{y_1^2+y_2^2\leq r_0^2}{\frac{1}{(x_1-y_1)^2+(x_2-y_2)^2+\Big(x_3-\sqrt{r_0^2-y_1^2-y_2^2}\Big)^2}}\frac{r_0}{\sqrt{r_0^2-y_1^2-y_2^2}}dy_1dy_2\\ &\leq \int_{y_1^2+y_2^2\leq r_0^2}{\frac{1}{\Big(x_3-\sqrt{r_0^2-y_1^2-y_2^2}\Big)^2}}\frac{r_0}{\sqrt{r_0^2-y_1^2-y_2^2}}dy_1dy_2\: \text{ (assume } x_3 \ne 0) \end{split} Let $y_1=r\cos\phi, y_2=r\sin\phi, 0\leq \phi \leq 2\pi$ \begin{split} S^+&=\int_0^{2\pi}d\phi\int_0^{r_0}\frac{1}{\Big(x_3-\sqrt{r_0^2-r^2}\Big)^2}\frac{r_0}{\sqrt{r_0^2-r^2}}rdr\\ &=2r_0\pi\left(\frac{1}{-x_3}-\frac{1}{r_0-x_3}\right)\\ \end{split} with $-r_0<x_3<r_0, x_3 \ne 0$, I can't show that $\left(\frac{1}{-x_3}-\frac{1}{r_0-x_3}\right)$ is bounded. Please help me, thank you so much!

Answer 1

You may use spherical coordinates and the cosine-theorem: $$ \frac{1}{|x-y|^2} = \frac{1}{|x|^2+|y|^2-2|x||y|\cos(\theta_{x,y})} $$ where $\theta_{x,y}$ is the angle between $x$ and $y$. Putting it into the integral you have: $$ \int\limits_{|y|= r_0}\frac{ds}{|x-y|^2} = \int\limits_{|y|= r_0}\frac{ r_0^2\sin(\theta) d\phi d\theta}{|x|^2+r_0^2-2|x|r_0\cos(\theta)}=2\pi r_0^2\int\limits_{-1}^1\frac{ d\cos(\theta)}{|x|^2+r_0^2-2|x|r_0\cos(\theta)} $$ where we chose to allign $x$ along the z-axis, which we are free to do due to symmetry. Then $\theta_{x,y}$ simply becomes the coordinate $\theta$.

Answer 2

Clearly, $$ F(y)=\int_{|y|=r_0}\frac{ds}{|s-y|^2} $$ is a function of $|y|$. For simplicity, assume that $y=(|y|,0,0)$.

Since the sphere is a solid by revolution, and in particular of revolution of the graph of the function $f(x)=\sqrt{r^2-x^2}$ around the $x-$axis, then $$ F(y)=2\pi\int_{-r_0}^{r_0} f(x)\sqrt{1+\big(f'(x)\big)^2}g(x)\,dx =\cdots=2\pi r_0\int_{-r_0}^{r_0} g(x)\,dx $$ where $$ g(x)=\frac{1}{\big|\big(x,\sqrt{r_0^2-x^2},0\big)-(|y|,0,0)\big|^2}=\frac{1}{(x-|y|)^2+\big(\sqrt{r_0^2-x^2}\big)^2}=\frac{1}{r_0^2-2x|y|+|y|^2} $$ and hence $$ F(y)=2\pi r_0\int_{-r_0}^{r_0}\frac{dx}{r_0^2+|y|^2-2|y|x}= \left.-\frac{\log(r_0^2+|y|^2-2|y|x)}{2|y|}\right|_{-r_0}^{r_0} =\frac{\log\left(\frac{r^2_0+2r_0|y|+|y|^2}{r^2_0-2r_0|y|+|y|^2}\right)}{2|y|}= \frac{ \log\left( \frac{r_0+|y|}{r_0-|y|} \right) }{|y|} $$ Clearly, $$ \lim_{|y|\to r_0^-}F(y)=\infty $$