So I have to solve the following integral:
$$\iiint_G\,dV,$$
where G is defined as:
$$x^2+y^2-z^2 \geq 6R^2,\; x^2+y^2+z^2\leq 12R^2,\;z \geq 0.$$
So since I have a sphere I am thinking of using spherical coordinates:
\begin{align*} x&=r\cos(\theta)\cos(\phi)\\ y&=r\cos(\theta)\sin(\phi)\\ z&=r\sin(\theta). \end{align*}
So since the graph is halved, I presume $\theta\in[\frac{\pi}{4},\frac{\pi}{3}]$ at first look, not involving structure yet. $\phi\in[0,2\pi]$
And for $r$ I got:
$$\left[\frac{\sqrt6R}{\cos(2\theta)},2\sqrt3R\right] \;.$$
Is this the right way I took, or did I make any mistakes? Any help would be appreciated.
$\begin{align*} x&=r\cos(\theta)\cos(\phi)\\ y&=r\cos(\theta)\sin(\phi)\\ z&=r\sin(\theta). \end{align*}$
$x^2+ y^2 - z^2 =r^2 (\cos^2\theta -\sin^2\theta) \ge 6R^2\\ r \ge \frac {R\sqrt 6}{\sqrt {\cos 2\theta}}$
and
$ 12R^2\ge r^2 \ge \frac {6R^2}{2\cos\theta}\\ \cos 2\theta \ge \frac 12\\ \theta \le \frac {\pi}{6}$
$z = 0$ coresponds to $\theta = 0$
$\displaystyle\int_0^{2\pi}\int_0^{\frac \pi6}\int_\frac {R\sqrt6}{\sqrt{\cos 2\theta}}^{R\sqrt 12} r\ dr\ d\theta\ d\phi$