<s>[INST] Triple Integrals in Spherical Coordinates, problem with boundaries Calculate 
$$\iiint \frac{1}{{x^2+y^2+z^2}}dA$$
Where
$x^2+y^2+(z-2)^2\le1$
I've used spherical coordinates, like this:
$x=\rho\sin\phi\cos\theta$; $y=\rho\sin\phi\sin\theta$; $z=\rho\cos\phi$ and $J=\rho^2\sin\phi$
but then I am having a rough time with the boundaries.
I am stuck at-
$$\rho^2\sin^2\phi\cos^2\theta + \rho^2\sin^2\phi\sin^2\theta +(\rho \cos\phi-2)^2 \le 1$$
$$ \rho^2-4\rho\cos\phi \ +4\le1$$
and I don't know where to go from here.
Thank you
 [/INST] By letting $z=2+w$ the problem boils down to computing
$$\iiint_{x^2+y^2+w^2\leq 1}\frac{d\mu}{x^2+y^2+w^2+4w+4} $$
or, by setting $w=\rho\cos\theta,y=\rho\sin\theta\sin\varphi,x=\rho\sin\theta\cos\varphi$,
$$ 2\pi\int_{0}^{1}\int_{0}^{\pi}\frac{\rho^2\sin\theta}{\rho^2+4\rho\cos\theta+4}\,d\theta\,d\rho.$$
Let us focus on the inner integral:
$$ \int_{0}^{\pi}\frac{\rho^2\sin\theta}{\rho^2+4\rho\cos\theta+4}\,d\theta = \frac{\rho}{2}\,\log\left(\frac{2+\rho}{2-\rho}\right) $$
by the tangent half-angle substitution. By integration by parts, the final outcome is $2\pi-\frac{3\pi}{2}\log(3)$.
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