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
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 $2\pi-\frac{3\pi}{2}\log(3)$.