Please help in calculating Integral on the unit ball.

Question

For an arbitrary vector R with length $$R= |\overrightarrow{R}| > 1$$, we define the integral I(R), which is taken over a ball of unit radius: $$I(R)=\int_{|r|\leq 1} \frac{dxdydz}{|R-r|^{2}}$$ where r=(x,y,z). Your task in this exercise is to calculate the integral I(R) exactly, and see how accurate the approximation $$I(R)\simeq \frac{4*pi}{3*R^{2}}$$ works for $$(R\sim 1)$$

Find I(R) and enter the ratio in the answer $$ \frac{I(R)}{\frac{4*pi}{3*R^{2}}} $$ at R=2

I tried solving it by using spherical coordinates and this is what i got

Answer 1

Your answer is an approximation, not the correct solution since you are not taking a limit. The trick is to do the integral in spherical coordinates, but centered at the point $(0,0,R)$. In other words

$$\begin{cases}x = r\sin\theta\cos\varphi \\ y = r\sin\theta\sin\varphi \\ z = R + r\cos\theta\end{cases}$$

which orients the sphere in the $z$ direction, giving the equation

$$x^2+y^2+(z-R)^2 = 1$$

Can you take it from here?

($\textbf{HINT}$: convert the above equation into spherical coordinates and solve for $\theta$, $\textbf{not}$ $r$. Then integrate with respect to $\theta$ first.)

Answer 2

Wlog we can assume that the vector $\overrightarrow{R}$ is $$ (0,0,R)\,. $$ The integral to calculate is \begin{align} I(\overrightarrow{R})&=\int_{\|\mathbf{x}\|\le 1}\frac{dx\,dy\,dz}{|\overrightarrow{R}-\mathbf{x}|^2}=\int_{\|\mathbf{x}\|\le 1}\frac{dx\,dy\,dz} {x^2+y^2+(R-z)^2}\tag{1}\\ &=\int_0^1\int_{x^2+y^2\le 1-z^2} \frac{dx\,dy\,dz} {x^2+y^2+(R-z)^2}\,. \end{align} For fixed $z\,,$ the $dx\,dy$-integral becomes in two dimensional polar coordinates \begin{align} &\int_0^{\sqrt{1-z^2}}\int_0^{2\pi}\frac{r\,d\varphi\,dr}{r^2+(R-z)^2} =2\pi\int_0^{\sqrt{1-z^2}}\frac{r\,dr}{r^2+(R-z)^2}\\[2mm] &=\pi\int_{(R-z)^2}^{1-z^2+(R-z)^2}\frac{du}{u}=\pi\log\frac{1-z^2+(R-z)^2}{(R-z)^2}\,.\tag{2} \end{align} Therefore,

\begin{align} I(\overrightarrow{R})&=\pi\int_0^1\log\Big(1-z^2+(R-z)^2\Big)\,dz-\pi\int_0^1\log\Big((R-z)^2\Big)\,dz\,.\tag{3} \end{align} From \begin{align} 1-z^2+(R-z)^2=1+R^2-2Rz \end{align} we get the first integral in (3) as $$ \frac{1}{2R}\int_{(R-1)^2}^{R^2+1}\log u\,du=\Bigg[\frac{u\log u-u}{2R}\Bigg]_{(R-1)^2}^{R^2+1}=\frac{(R^2+1)\log(R^2+1)-(R-1)^2\log((R-1))^2+2R}{2R}\,. $$ The second integral in (3) we get as $$ \int_{R-1}^R\log(u^2)\,du=\Bigg[u\log(u^2)-2u\Bigg]_{R-1}^R=R\log(R^2)-(R-1)\log((R-1)^2)-2\,. $$

Someone thinks that $I(R)\approx \frac{4\pi}{3R^3}$ at least when $R$ is large. However I believe that it should be $$\boxed{\quad I(R)\approx \frac{4\pi}{3R^\color{red}{2}}\,.\quad} $$ Proof. Pulling out $R^2$ from the denominator of (1) we get for large $R$ $$ I(\overrightarrow{R})=\frac{1}{R^2} \int_{\|\mathbf{x}\|\le 1}\frac{dx\,dy\,dz} {\frac{x^2+y^2}{R^2}+(1-z/R)^2} \approx\frac{1}{R^2}\int_{\|\mathbf{x}\|\le 1}\,dx\,dy\,dz =\frac{4\pi}{3R^2}\,. $$

Here is a log-log graph: