Compute Triple integral Compute $$\iiint\limits_{x^2+y^2+(z-2)^2\leq1}\frac{dx\,dy\,dz}{x^2+y^2+z^2}$$
I tried to convert it to spherical cordinates and I got $$\iiint \sin\theta \,dr\,d\theta \,d\varphi$$
However I am lost on how to find the boundaries of the new domain, can anyone help?
 A: Change coordinates by letting $z'=z-2$.  Then, we have
$$\begin{align}
\iiint_{x^2+y^2+(z-2)^2\le 1} \frac{1}{x^2+y^2+z^2}\,dx\,dy\,dz&=\iiint_{x^2+y^2+z^2\le 1} \frac{1}{x^2+y^2+(z+2)^2}\,dx\,dy\,dz\\\\
&=\int_0^{2\pi}\int_0^\pi\int_0^1 \frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,dr\,d\theta\,d\phi \\\\
&=2\pi\int_0^1 \int_0^\pi\frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,d\theta\,dr \tag 1\\\\
&=\frac{\pi}{2}\int_0^1 r \log\left(\frac{(r+2)^2}{(r-2)^2}\right)\,dr \tag 2
\end{align}$$
Can you finish now?
Note that in going from $(1)$ to $(2)$ we enforced the substitution $x=\cos(\theta)$.  Then, $dx=-\sin(\theta)\,d\theta$ and we have
$$\begin{align}
\int_0^\pi \frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,d\theta&=r^2\int_{-1}^1 \frac{1}{r^2+4rx+4}\,dx\\\\
&=\frac{r}{4}\left.\left(\log(r^2+4rx+4)\right)\right|_{-1}^1\\\\
&=\frac{r}{4}\left(\log(r^2+4r+4)-\log(r^2-4r+4)\right)\\\\
&=\frac{r}{4}\log\left(\frac{(r+2)^2}{(r-2)^2}\right)
\end{align}$$
A: Hint: 
Write
$$I=\int_1^3 \left ( \iint\limits_{x^2+y^2\leq 1-(z-2)^2}\frac{dx\,dy}{x^2+y^2+z^2} \right) \, dz$$
and use polar coordinates:
$$\iint\limits_{x^2+y^2\leq 1-(z-2)^2}\frac{dx\,dy}{x^2+y^2+z^2} = \iint\limits_{D_z}\frac{\rho \, d\rho \, d\theta}{\rho^2+z^2}$$
where $D_z=\{(\rho, \theta); 0\leq \rho\leq \sqrt{1-(z-2)^2}, 0\leq \theta\leq 2\pi\}$. 
