Let $D$ be the region in $\mathbb{R}^3$ below $z=-\sqrt{x^2 + y^2}$ and above $z=-\sqrt{4-x^2 -y^2}$.
Rewrite \begin{align*}\iiint \limits_D z^2 dV\end{align*}
using Spherical Coordinates.
I rewrote the integral in spherical coordinates, and now know that it is as follows:
\begin{align*}\iiint \limits_D z^2 dV &= \int\limits_{\theta_0}^{\theta_1}\int \limits_{\phi_1}^{\phi_2}\int\limits_{\rho_1}^{\rho_2}(\rho^4\cos^2{\phi}sin{\phi})d\rho d\phi d\theta \end{align*}
My problem, however, is finding the limits for $\rho, \phi$ and $\theta$ in the triple integral.
I am having some trouble finding these limits. Can somebody please assist me, as I honestly have no idea how to go about doing so for this question.
EDIT:
$\textbf{I have to draw a sketch of the region $D$ in the given question (but I am struggling to do so)}$ and use that to find the limits.
Best Answer
The lower cone begins at $\phi=\dfrac{3\pi}{4}$ and runs all the way down to $\phi=\pi$. If the starting point for $\phi$ isn't obvious, then you can find it as follows:
Let $x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi, z=r\cos\theta$. Substituting this into the cone equation, squaring, and rearranging, you obtain:
$$r^2\cos^2\theta = r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi\iff \tan^2\phi=1.$$
Since we're in the lower octants, this means $\tan\phi=-1$, i.e. $\phi=\dfrac{3\pi}{4}$.
Obviously, $0\le \theta\le 2\pi$.
$\rho$ will fun from the origin to the hemisphere. $0 \le \rho\le 2$.