# 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?

• Is it $z$? or $z^2$? May 28 '17 at 18:34
• it is $z^2$. thank you for informing. @caverac May 28 '17 at 18:49

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