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Any idea for solving this?

Compute the integral of $((x-a)^2+(y-b)^2+(z-c)^2)^{-1/2}$ over the sphere of radius $R$ centered at the origin. Hint: Spherical coordinates.

I don't know how to make the transformation in order to obtain an "easier" problem.

Thanks everyone

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  • $\begingroup$ There's no good way to do this. By symmetry, you can reduce to the case where $a=b=0$. Do you know whether $a^2+b^2+c^2$ is less than or greater than $R^2$? $\endgroup$ – Ted Shifrin Oct 11 '16 at 3:33
  • $\begingroup$ Yes, the point $(a,b,c)$ is exterior to the sphere. $\endgroup$ – user106153 Oct 11 '16 at 3:35
  • $\begingroup$ OK, put the point at $(0,0,C)$, use spherical coordinates, and use the law of cosines to rewrite that denominator. Have fun! $\endgroup$ – Ted Shifrin Oct 11 '16 at 3:38
  • $\begingroup$ How can I take this point? I don`t understand the symmetry you're talking about. $\endgroup$ – user106153 Oct 11 '16 at 3:55
  • $\begingroup$ why not shift the origin to $(a, b, c )$ and then use spherical coodinates? $\endgroup$ – vidyarthi Oct 11 '16 at 4:07
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It's not clear whether you mean integration over the two-dimensional sphere of radius $R$ (with respect to area measure) or over the full ball of radius $R$. In any case put $\sqrt{a^2+b^2+c^2}=:\rho$. Then you may assume that $(a,b,c)=(0,0,\rho)$, since everything is rotationally symmetric with respect to the axis through $O$ and $(a,b,c)$. Now use spherical coordinates $(r,\phi,\theta)$, and distinguish the cases $\rho<R$ and $\rho>R$.

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