Triple integral question with spherical coordinates 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?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: 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
$$\int_E(x^2+y^2+z^2)^{3/2} dxdydz = \int_0^{\pi / 2} \int_0^{\pi / 4} \int_3^4 \rho^5 sin\varphi  d\rho d\varphi d\theta $$
$$=1/6\int_0^{\pi / 2} sin \varphi \int_0^{\pi / 4} [p^6]_3^4 \space d\rho d\varphi d\theta $$
$$1/6\int_0^{\pi / 4} 3367 d\theta = 3367\pi/24 $$
