Integral over a sphere How to calculate $$I=\int\int_S\frac{d^2S}{r^n}$$ where S is a surface of the sphere $S=\{x^2+y^2+z^2=R^2\}$ and $r=dist((x,y,z),(0,0,c))$, $c>R,c\in\mathbb{R}$ ?
I believe $r=\sqrt{x^2+y^2+(\sqrt{R^2-x^2-y^2}-c)^2}$ so $$I=\int\int_{x^2+y^2<R^2}\frac{1}{\Big(x^2+y^2+(\sqrt{R^2-x^2-y^2}-c)^2\Big)^\frac{n}{2}}*\sqrt{G}$$ where $G = \left| \begin{array}{cc}
1 & 0 \\
0 & 1 \\
\frac{-x}{\sqrt{R^2-x^2-y^2}} & \frac{-y}{\sqrt{R^2-x^2-y^2}} \end{array} \right|$
I made few calculations using $x=tcos\alpha,y=tsin\alpha$ and then $t=Rsin\beta$ substitutions but I'm stuck with some ugly looking integral, so I'm not sure if I got it all right.

Work in spherical coordinates $(R, \theta, \phi)$.  Note that
$$r = \sqrt{x^2 + y^2 + (z-c)^2} = \sqrt{x^2 + y^2 + z^2 - 2cz + c^2} = \sqrt{R^2 - 2Rc \cos \phi + c^2}$$ and $$d^2 S = R^2 \sin \phi\, d\theta\, d\phi$$
so
$$\iint_S \frac{d^2S}{r^n} = \int_0^\pi \int_0^{2\pi} \frac{R^2 \sin \phi}{(R^2 - 2Rc\cos \phi + c^2)^{n/2}}\, d\theta\, d\phi$$
You should be able to calculate this.
