Computing $\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz$ in Spherical Coordinates Compute: $\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz$. 
Hint given: show that


*

*$\cos \theta> {r^2+3\over 4r}$

*$1<r<3$


What I already did: I shift the unit sphere two coordinates upwards. With $z=r\cos \theta$, assisted by a general drawn line from the sphere (suppose from $a=(x,y,z)$) to the origin, we get that $x^2+y^2$ equals $a$'s projection on the $z$-axis, which by simple computations equals $(r\sin\theta)^2$.
Then: $$x^2+y^2+z^2-4z+4=r^2-4r\cos\theta+4\le 1\Rightarrow \cos \theta \ge {r^2+3\over 4r}$$ (How is the strong inequality achieved? I tried showing it but the RHS are variables.) Obviously $1<r<3$. The change of variables is injective and surjective using the inequality in the hint, and the remaining term is $r^2\sin \theta$, but I have no how $\theta$ acts in a single round. I find myself starting from $\theta=0$ (r=1), and again $\theta =0 $ for $r=3$. Do you have any idea how one can proceed from here? While I do understand how to bound the variables in the unit sphere(or, obviously any sphere centered at the origin), here it all seems to collapse. 
Final answer is told to be: $\pi(2-{3\over 2}\log 3)$, for those who are interested or need it.
 A: I think the aim of the hint is to guide you through finding the limits of integration. To me, there is no need for shifting the coordinate system. You can proceed as follows
$$\begin{align}{}
I 
&=\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} \int_{\theta=0}^{ \cos^{-1} (\frac{r^2+3}{4r})} \frac{1}{r^2} r^2 \sin \theta \, d\theta \, dr \, d \phi \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} \int_{\theta=0}^{\cos^{-1} (\frac{r^2+3}{4r})}  \sin \theta \, d\theta \, dr \, d \phi \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3}  -\cos \theta|_{0}^{\cos^{-1} (\frac{r^2+3}{4r})} \, dr \, d \phi \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3}  (1-\frac{r^2+3}{4r}) \, dr \, d \phi \\
&=(\int_{0}^{2 \pi} d\phi)(\int_{r=1}^{3}  (1-\frac{1}{4}r-\frac{3}{4}\frac{1}{r}) \, dr) \\
&= (2 \pi) (1-\frac{3}{4}\log3) \\
&=\pi (2-\frac{3}{2}\log3)
\end{align} \\ 
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Lets consider
\begin{align}
\color{#f00}{%
\iiint_{\verts{\vec{r} - 2\hat{z}} < 1}{\dd^{3}\vec{r} \over r^{2}}} & =
\iiint_{r < 1}{\dd^{3}\vec{r} \over \verts{\vec{r} - 2\,\hat{z}}^{2}} =
\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}
{r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi \over
r^{2} + 4 - 4r\cos\pars{\theta}}
\\[3mm] & = 
2\pi\int_{0}^{1}\left.{\ln\pars{\verts{r^{2} + 4 - 4r\xi}} \over -4r}
\right\vert_{\ \xi\ =\ -1}^{\ \xi\ =\ 1}\,r^{2}\,\dd r
\\[3mm] & =
{\pi \over 2}\int_{0}^{1}
\ln\pars{\verts{{r^{2} + 4r + 4 \over r^{2} - 4r + 4}}}r\,\dd r
=\pi\ \overbrace{\int_{0}^{1}
\ln\pars{{2 + r \over 2 - r }}r\,\dd r}^{\ds{2 - {3 \over 2}\,\ln\pars{3}}}
\\[3mm ] & =
\color{#f00}{\half\bracks{4 - 3\ln\pars{3}}\pi} \approx 1.1061
\end{align}
