Triple integral question with spherical coordinates

Question

I have this triple integral question. I'm pretty sure that it is solvable with spherical coordinates. I found the upper and lower limits of $\theta$ and $\rho$ but I couldn't find the limits of $\varphi$.

Here is the question:

$$\int_E (x^2+y^2+z^2)^{3/2}dxdydz$$ where E is in the first octant bounded by the plane $z=0$ and the hepisphere $x^2+y^2+z^2=9$ bounded above by the hepisphere $x^2+y^2+z^2=16$ and the planes $y=0$ and $y=x$.

I have a sketch so far. And these are the limits I have found. $3 \le \rho \le 4$ and $0 \le \theta \le \pi /4$ and also the inside part is $\rho^5sin(\varphi)d\rho d\varphi d\theta$.

Edit: typing mistake.

2nd Edit: I think $o \le \varphi \le \pi /2$ am I right?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} V & \,\,\,\stackrel{\mbox{def.}}{=}\, \iiint_{\large \mathbb{R}^{3}}\pars{x^{2} + y^{2} + z^{2}}^{3/2} \bracks{9 < x^{2} + y^{2} + z^{2} < 16}\bracks{0 < y < x}\dd x\,\dd y\,\dd z \\[1cm] & \stackrel{\mbox{Sph. Cord.}}{=} \iiint_{\atop {\!\!\Large\mathbb{R}^{3}}}r^{3}\bracks{9 < r^{2} < 16} \bracks{0 < r\sin\pars{\theta}\sin\pars{\phi} < r\sin\pars{\theta}\cos\pars{\phi}} \times \\[3mm] & \phantom{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\stackrel{\mbox{Sph. Cord.}}{=}} r^{2}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd \phi \\[1cm] & = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{3}^{4} \bracks{0 < \sin\pars{\phi} < \cos\pars{\phi}} r^{5}\sin\pars{\theta}\,\dd r\,\dd\theta\,\dd \phi \\[5mm] & = \bracks{\int_{0}^{\pi}\sin\pars{\theta}\,\dd\theta} \pars{\int_{3}^{4}r^{5}\,\dd r} \int_{-\pi}^{\pi}\bracks{0 < \sin\pars{\phi} < \cos\pars{\phi}}\,\dd\phi \\[5mm] & = {3367 \over 3}\braces{% \int_{0}^{\pi}\bracks{0 < \sin\pars{\phi} < \cos\pars{\phi}}\,\dd\phi + \int_{0}^{\pi}\bracks{0 < -\sin\pars{\phi} < \cos\pars{\phi}}\,\dd\phi} \\[5mm] & = {3367 \over 3}\braces{% \int_{-\pi/2}^{\pi/2}\!\!\!\!\!\bracks{0 < \cos\pars{\phi} < -\sin\pars{\phi}}\,\dd\phi + \int_{-\pi/2}^{\pi/2}\!\!\!\!\!\bracks{0 < -\cos\pars{\phi} < -\sin\pars{\phi}}\,\dd\phi} \\[1cm] & = {3367 \over 3}\braces{% \int_{0}^{\pi/2}\!\!\!\!\!\bracks{0 < \cos\pars{\phi} < -\sin\pars{\phi}}\,\dd\phi + \int_{0}^{\pi/2}\!\!\!\!\!\bracks{0 < -\cos\pars{\phi} < -\sin\pars{\phi}}\,\dd\phi} \label{1}\tag{1} \\[3mm] & + {3367 \over 3}\braces{% \int_{0}^{\pi/2}\bracks{0 < \cos\pars{\phi} < \sin\pars{\phi}}\,\dd\phi + \int_{0}^{\pi/2}\bracks{0 < -\cos\pars{\phi} < \sin\pars{\phi}}\,\dd\phi} \end{align}

Integrals in line \eqref{1} vanishes out.

Then, \begin{align} V & = {3367 \over 3}\pars{\int_{\pi/4}^{\pi/2}\dd\phi + \int_{0}^{\pi/2}\dd\phi} = \bbx{{3367 \over 4}\,\pi} \approx 2644.4356 \end{align}

Answer 2

Using spherical coordinates, $x=\rho sin\varphi cos\theta, y= \rho sin\varphi sin\theta, z=\rho cos\varphi$ and $p=||(x,y,z)||=\sqrt{x^2+y^2+z^2} $ where $dxdydz=\rho^2 sin\varphi d\rho d\varphi d\theta $, we obrain

$$\int_E(x^2+y^2+z^2)^{3/2} dxdydz = \int_0^{\pi / 2} \int_0^{\pi / 4} \int_3^4 \rho^5 sin\varphi d\rho d\varphi d\theta $$ $$=1/6\int_0^{\pi / 2} sin \varphi \int_0^{\pi / 4} [p^6]_3^4 \space d\rho d\varphi d\theta $$ $$1/6\int_0^{\pi / 4} 3367 d\theta = 3367\pi/24 $$