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So I'm trying to evaluate the triple integral $$\displaystyle \iiint \limits_{R} \displaystyle \frac{1}{((x-a)^2+y^2+z^2)^{1/2}} \mathrm dV$$ for $a>1$ over the solid sphere $0 \leq x^2 + y^2 + z^2 \leq 1$.

Apparently, there's an interpretation that I should be able to draw from this to. Not too sure what it is.

So the first thing that came to mind when I saw the integral was to apply spherical coordinates, but this doesn't make the denominator of the integrand any less messy.

Using spherical coordinates, the integrand becomes

$$\displaystyle \iiint \limits_{R} \displaystyle\frac{1}{(\rho^2-2a\rho\sin\phi\cos\theta+a^2)^{1/2} } \rho^2\sin\phi \space\mathrm d\rho\mathrm d\phi\mathrm d\theta$$ (I haven't bothered to add the bounds yet), which doesn't look that much more friendly.

Any support for this question would be appreciated.

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    $\begingroup$ i think using cylindrical coordinates $(x,r,\theta)$ brings this down into a 2d integral with an extra factor of $2\pi$ out front $\endgroup$ – gt6989b Mar 26 '18 at 21:00
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    $\begingroup$ I don't remember this stuff well at all, so beware, but you're looking at (a constant times) the mean electric potential of a charge at $(a,0,0)$ over the unit sphere. The potential will be a harmonic function over the unit sphere, because the charge is outside it ($a>0$), and the mean value property of harmonic functions would imply that the integral is proportional to the value of the potential at the center of the sphere. $\endgroup$ – stochasticboy321 Mar 26 '18 at 21:00
  • $\begingroup$ @Kaynex it's the denominator that doesn't look friendly $\endgroup$ – user98937 Mar 26 '18 at 21:05
  • $\begingroup$ My apologies, misunderstood $\endgroup$ – Kaynex Mar 26 '18 at 21:10
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    $\begingroup$ @user98937 you represent the $(y,z)$ as $(r,\theta)$ $\endgroup$ – gt6989b Mar 26 '18 at 21:19
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Orienting ourselves better will simplify things.

ie. make $\phi = 0$ align with the x-axis.

$\iiint \frac {\rho^2\sin\phi}{(\rho^2 -2a\rho\cos\phi + a^2 )^\frac 12}\ d\rho\ d\phi\ d\theta$

But I think that cylindrical will be easier.

$y = r\cos\theta\\ z = r\sin\theta\\ x = x$

$\iiint \frac {r}{(r^2 + (x-a)^2)^\frac 12}\ dr\ dx\ d\theta$

$\int_0^{2\pi}\int_{-1}^{1}\int_0^{\sqrt {1-x^2}} \frac {r}{(r^2 + (x-a)^2)^\frac 12}\ dr\ dx\ d\theta$

$\int_0^{2\pi}\int_{-1}^{1} ((r^2 + (x-a)^2)^\frac 12|_0^{\sqrt {1-x^2}}\ dx\ d\theta\\ \int_0^{2\pi}\int_{-1}^{1} (1 - 2ax + a^2)^\frac 12 - |x-a| \ dx\ d\theta$

etc.

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  • $\begingroup$ I didn't get the $1/4$ constant when I integrated and evaluated $\displaystyle \frac {r}{(r^2+(x-a)^2)^{1/2}}$ with respect to $r$. I got a constant of 1. $\endgroup$ – user98937 Mar 27 '18 at 21:03
  • $\begingroup$ My mistake, thanks... $\endgroup$ – Doug M Mar 27 '18 at 21:08
  • $\begingroup$ no worries...also, how would I go about integrating $\left|{x-a}\right|$? $\endgroup$ – user98937 Mar 27 '18 at 21:09
  • $\begingroup$ It is given that $a>1$ and $-1\le x\le 1$ so $|x-a| = a-x$ $\endgroup$ – Doug M Mar 27 '18 at 21:27

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