Evaluate the Integral $ \iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz$ I am attempting to evaluate the following triple integral using spherical coordinated:
$$
\iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz \\
W = \{(x,y,z) \in \mathbb{R}^3 : 1 \le x^2+y^2+z^2 \le 9; 0 \le x \le y, z \ge 0\}
$$
This tells me that I have the following regions:
$$
x^2 + y^2 + z^2 = 1 \rightarrow \text{Sphere of radius 1}\\
x^2 + y^2 + z^2 = 9 \rightarrow \text{Sphere of radius 3} \\
x=y \rightarrow \text{line in the xy plane}
$$
I have also determined that:
$$
1\le r \le 3
$$
But I am not sure how to determine which part of the intersection of the regions I am to integrate and to solve for $\theta$ and $\phi$. How do I proceed?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\bracks{\cdots}}$ is an Iverson Bracket. With Spherical Coordinates:

\begin{align}
&\bbox[15px,#ffe]{\ds{\iiint_{\large\mathbb{R}^{3}}{\bracks{1 \leq x^{2} + y^{2} + z^{2} \leq 9}
\bracks{0 \leq x \leq y}\bracks{z \geq 0} \over \root{x^{2} + y^{2} + z^{2}}}\,
\dd x\,\dd y\,\dd z}}
\\[1cm] = &\
\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}{\bracks{1 \leq r^{2} \leq 9}
\bracks{0 \leq r\sin\pars{\theta}\cos\pars{\phi} \leq r\sin\pars{\theta}\sin\pars{\phi}}\bracks{r\cos\pars{\theta} \geq 0} \over \root{r^{2}}}
\,\,\times
\\ & \phantom{\int_{0}^{2\pi}\int_{0}^{\pi}\int_{1}^{3}\,\,\,\,\,} r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi
\\[1cm] = &\
\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}
\bracks{1 \leq r \leq 3}\bracks{0 \leq \cos\pars{\phi} \leq \sin\pars{\phi}}\bracks{\cos\pars{\theta} \geq 0}
r\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd\phi
\\[1cm] = &\,\,\,
\overbrace{\pars{\int_{1}^{3}r\,\dd r}}^{\ds{=\ 4}}\
\overbrace{\bracks{\int_{0}^{\pi/2}\sin\pars{\theta}\,\dd\theta}}^{\ds{=\ 1}}
\,\times
\\ &\
\braces{\int_{0}^{\pi/2}\bracks{\cos\pars{\phi} \leq \sin\pars{\phi}}\,\dd\phi\ +\
\overbrace{\int_{3\pi/2}^{2\pi}\bracks{\cos\pars{\phi} \leq \sin\pars{\phi}}\,\dd\phi}^{\ds{=\ 0}}}
\\[1cm] = &\
4\int_{0}^{\pi/2}\bracks{\tan\pars{\phi} \geq 1}\,\dd\phi =
4\int_{\pi/4}^{\pi/2}\,\dd\phi = \bbx{\pi} \approx 3.1416
\end{align}
A: $x\geq0, y\geq0, z\geq0$ show $0\leq\theta\leq\dfrac{\pi}{2}$ and $0\leq\phi\leq\dfrac{\pi}{2}$. Also $x\leq y$ states $r\cos\theta\sin\phi\leq r\sin\theta\sin\phi$ or $\tan\theta\geq1$ so $\dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{2}$ then
$$\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _1^3\frac{r^2 \sin (\varphi )}{r}drd\varphi d\theta=\color{blue}{\pi}$$
A: The integrand ${1\over\sqrt{x^2+y^2+z^2}}={1\over r}$ is fully rotationally symmetric, and the integration is over ${1\over16}$ of the spherical shell with inner radius $1$ and outer radius $3$. It follows that the value of the integral is given by
$${1\over16}\int_1^3 {1\over r}\>4\pi r^2\>dr=\pi\ .$$
