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.
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.