Evaluate the integral 5 Evaluate the integral $ \iiint_{R} \frac{dxdydz}{\sqrt{(x-a)^{2}+y^{2}+z^{2}}} $ over the solid sphere $ 0 \leq x^{2}+y^{2}+z^{2} \leq 1 \ $   for  a>1 . $$ $$ I know that the required  integral is  $ \int_{-1}^{1} \int_{- \sqrt{1^{2}-x^{2}}}^{\sqrt{1^{2}-x^{2}}} \int_{-\sqrt{1^{2}-x^{2}-y^{2}}}^{\sqrt{1^{2}-x^{2}-y^{2}}}f(x,y,z) dxdydz $ . But I can't evaluate it , say using polar coerdinate. Please someone help me. 
 A: I hope you don't mind if I change your integral to
$$ \iiint_R \frac{dx dy dz}{\sqrt{x^2 + y^2 + (z - a)^2}}. $$
So the denominator is the distance from $(x,y,z)$ to $(0,0,a)$.
Now use spherical polar coordinates $(r, \theta, \phi)$. If you draw a diagram, then hopefully you can see that the distance from $(x,y,z)$ to $(0,0,a)$ is
$$ \sqrt{r^2 + a^2 - 2ar \cos \theta},$$
by the cosine rule in trigonometry!
So the integral reduces to
$$ \int_{r = 0}^{r = 1} \int_{\theta = 0}^{\theta = \pi} \int_{\phi = 0}^{\phi = 2\pi} \frac{r^2 \sin \theta dr d\theta d \phi}{\sqrt{r^2 + a^2 - 2ar \cos \theta}}.$$
Making the substitution $u = \cos \theta$, this simplifies to
$$ \int_{r = 0}^{r = 1} \int_{u = -1}^{u = 1} \int_{\phi = 0}^{\phi = 2\pi} \frac{r^2  dr du d \phi}{\sqrt{r^2 + a^2 - 2ar u}}.$$
I hope this is now manageable enough for you to finish off.

Here is a nicer method: $1/\sqrt{x^2 + y^2 + (z-a)^2}$ is a harmonic function. All harmonic functions obey the mean value property, which states that the average value of a harmonic function on a spherical shell equals its value at the centre of the shell. Hence your triple integral equals the volume of the unit sphere times the value of $1/\sqrt{x^2 + y^2 + (z-a)^2}$ at the origin, i.e. the answer is $4\pi/3a$.
