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I would like to evaluate this integral over the surface of a sphere in 3D: $$ \int_{\partial D} \int_{\partial D} \frac{1}{|x-y|}dxdy. $$ It seems there is a lot of symmetry in this integral so I imagine there is a good chance there is an explicit solution. However, I usually deal with 2D Helmholtz problems so I have very little, that is to say 'no', experience with evaluating integrals in 3D. Does anyone know how to evaluate this?

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  • $\begingroup$ Is the sphere centered at the origin? $\endgroup$ – David H Jul 30 '17 at 8:18
  • $\begingroup$ @DavidH Yes that is the case. $\endgroup$ – eurocoder Jul 30 '17 at 8:18
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First we compute the inner integral $$\int_{\partial D}{1\over |x-y|}\>dx$$ when $y:=(0,0,1)$ is the north pole. Here $dx$ denotes the surface element on the sphere. Introduce spherical coordinates with $\theta=0$ at $(0,0,1)$. Then $$\int_{\partial D}{1\over |x-y|}\>dx=\int_0^\pi{1\over2\sin{\theta\over2}}\>2\pi\sin\theta\>d\theta=2\pi\int_0^\pi\cos{\theta\over2}\>d\theta=4\pi\ .$$ Due to rotational symmetry the inner integral is in fact independent of $y$. We therefore obtain $$\int_{\partial D}\int_{\partial D}{1\over |x-y|}\>dx\>dy=4\pi\cdot 4\pi\ .$$

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  • $\begingroup$ Excellent thanks! I am also wondering about the case wehre $x$ and $y$ are on different spheres - math.stackexchange.com/questions/2376494/… - it seems this one could be much more tricky now that the symmetry is broken? $\endgroup$ – eurocoder Jul 30 '17 at 9:36
  • $\begingroup$ In the case of a non-unit sphere, what is the value of the integral? I calculated $4\pi r^2$, but from looking at some code I found, it seems the person who wrote it used $4\pi r$? $\endgroup$ – eurocoder Jul 30 '17 at 11:52

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