TITLE: explanation of a triple integral $\iiint _{x^2+y^2+z^2<R^2}\left|xz\right|y^2dxdydz\:$
QUESTION [6 upvotes]: I'm trying to figure out how the my book solve this exercise:
$$\iiint _{x^2+y^2+z^2<R^2}\left|xz\right|y^2dxdydz\:$$
BOOK SOLUTION:
$$\iiint _{x^2+y^2+z^2<R^2}\left|xz\right|y^2dxdydz\: = 8\iiint _Sxzy^2dxdydz\:$$
with:
$$S=\{(x,y,z):x^2+y^2+z^2<R^2, x\ge0, y\ge0, z\ge0\}$$
We transform in spherical coordinates:
$$S=\{(\rho,\phi,\theta): \rho \in [0,R], \phi\in[0,\frac{\pi}{2}], \theta \in [0,\frac{\pi}{2}]\}, dxdydz=\rho^2\sin\phi d\phi d\theta d\rho$$
then, the exercise continues with the triple integral solution after the transformation (which is not difficult).
MY DOUBT
i will report below, what I am not clear (outlined in red)
$$\iiint _{x^2+y^2+z^2<R^2}\left|xz\right|y^2dxdydz\: = \color{red}{8\iiint_Cxzy^2dxdydz\:}$$
I just can not understand how these two integrals can be equal.
Then:
$$S=\{(\rho,\phi,\theta): \rho \in [0,R], \color{red}{\phi\in[0,\frac{\pi}{2}]}, \color{red}{\theta \in [0,\frac{\pi}{2}]}\}, dxdydz=\rho^2\sin\phi d\phi d\theta d\rho$$
why in that range?
Thanks

REPLY [2 votes]: Let
$$
\begin{aligned}
D_1&=\{(x,y,z)\in\mathbb R^3~|~x^2+y^2+z^2\leq R^2\},\\
D_2&=\{(x,y,z)\in\mathbb R^3~|~x^2+y^2+z^2\leq R^2,\ x\geq 0\}\\
D_3&=\{(x,y,z)\in\mathbb R^3~|~x^2+y^2+z^2\leq R^2,\ x\geq 0,\ y\geq 0\}\\
D_4&=\{(x,y,z)\in\mathbb R^3~|~x^2+y^2+z^2\leq R^2,\ x\geq 0,\ y\geq 0,\ z\geq 0\}.
\end{aligned}
$$
Then, since your function $f(x,y,z)=|xz|y^2$ is even in all three variables,
$$
\begin{aligned}
\iiint_{D_1}f(x,y,z)\,dx\,dy\,dz
&=
2\iiint_{D_2}f(x,y,z)\,dx\,dy\,dz\\
&=
4\iiint_{D_3}f(x,y,z)\,dx\,dy\,dz\\
&=
8\iiint_{D_4}f(x,y,z)\,dx\,dy\,dz
\end{aligned}
$$
Moreover, in $D_4$ you have $f(x,y,z)=xy^2z$.
For the second question, in spherical coordinates the first octant satisfies the bound for the angles, i.e. $0\leq\phi\leq\pi/2$ and $0\leq\theta\leq\pi/2$. Draw a picture and make sure you understand that.