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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.

Jules
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    Calculating dS is the issue for you. Check this similar post :

    http://math.stackexchange.com/questions/131735/surface-element-in-spherical-coordinates

     – DiffeoR Apr 01 '14 at 13:00
  • Your approach with trig functions looks good, and the resulting integral should be easy (even if it looks messy). – Andrew D. Hwang Apr 01 '14 at 13:01
  • Doing this in rectangular, with $r = \sqrt{x^2 + y^2 + (\sqrt{R^2 - x^2 - y^2}-c)^2}$ only handles a hemisphere (namely, the portion where $z > 0$). The other half can easily be handled using $r = \sqrt{x^2 + y^2 + (-\sqrt{R^2 - x^2 - y^2}-c)^2}$. But as the answer below shows, spherical coordinates are more convenient (and only requires computing a single integral). – Nicholas Stull Apr 01 '14 at 13:04

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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.

Paul Siegel
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