Evaluate the Integral $ \iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz$

Question

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?

Answer 1

$x\geq0, y\geq0, z\geq0$ 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 $$\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _1^3\frac{r^2 \sin (\varphi )}{r}drd\varphi d\theta=\color{blue}{\pi}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\bracks{\cdots}}$ 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}

Answer 3

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 $${1\over16}\int_1^3 {1\over r}\>4\pi r^2\>dr=\pi\ .$$