I was given the following task calculate:
$\iiint_V{(xz^2+z)dxdydz} \space$ where $\space V = \begin{cases} x^2+y^2+z^2\geq1 & \text{inner sphere} \\ x^2+y^2+z^2\leq4 & \text{sphere} \\ z\geq \sqrt{x^2+y^2} & \text{cone} \\ \end{cases}$
I calculated the spherical coordinates :
$$\begin{cases} x=r\cos\phi{}\cos\psi{} \\ y=r\sin\phi{}\cos\psi \\ z=r\sin\psi \\ \end{cases}$$
$$[J]=r^2\cos\psi$$
$$\iiint_V{(xz^2+z)dxdydz}=\iiint_\Omega [r\cos\phi \cos\psi(r\sin\psi)^2+ r\sin\psi] r^2\cos\psi] drdydz$$
$$\Omega = \begin{cases} 0\leq \phi \leq 2\pi \\ 0 \leq\psi \leq \frac{\pi}{2} \\ 1 \leq r \leq 2 \\ \end{cases}$$
I did the further computations and solved the problem but I am looking for an explanation why is :
$$0 \leq\psi \leq \frac{\pi}{2}$$
Are my assumptions correct?
I have found a similar example in a book and I am looking for an explanation of $\Omega$ and checking if it is correct.
Here is my whole solution but what I need help with is a range of $\psi$ in $\Omega$ and why is it to $\frac{\pi}{2}$
The constraint $z>\sqrt {x^2 +y^2}$ becomes $r\sin\psi > r\cos\psi$
$\tan \psi > 1$
$\frac \pi 4 < \psi < \frac\pi 2$