Answer
(a) the region is the upper hemisphere between radii $4$ and $9$.
(b) the region is in the first quadrant between $\frac{\pi }{4}$ and $\frac{\pi }{3}$. It is a part of the cylinder of radius $2$, bounded below by $z=-2$ and bounded above by $z=1$.
(c) the region is the lower hemisphere of radius $3$, bounded below by $z = - \sqrt {9 - {r^2}} $ and bounded above by $z=0$.
Work Step by Step
(a) From the triple integral $\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 4}^9 {\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $, we obtain the domain of integration:
${\cal W} = \left\{ {\left( {\rho ,\phi ,\theta } \right)|4 \le \rho \le 9,0 \le \phi \le \pi /2,0 \le \theta \le 2\pi } \right\}$
Since $\rho $ varies from $4$ to $9$, and $\phi $ varies from $0$ to $\frac{\pi }{2}$, the region is the upper hemisphere between radii $4$ and $9$.
(b) From the triple integral $\mathop \smallint \limits_{ - 2}^1 \mathop \smallint \limits_{\pi /3}^{\pi /4} \mathop \smallint \limits_0^2 r{\rm{d}}r{\rm{d}}\theta {\rm{d}}z$, we obtain the domain of integration:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 2,\frac{\pi }{4} \le \theta \le \frac{\pi }{3}, - 2 \le z \le 1} \right\}$
Since $\theta $ varies from $\frac{\pi }{4}$ to $\frac{\pi }{3}$, the region is in the first quadrant between $\frac{\pi }{4}$ and $\frac{\pi }{3}$. Since $r$ varies from $0$ to $2$, and $ - 2 \le z \le 1$, it is a part of the cylinder of radius $2$, bounded below by $z=-2$ and bounded above by $z=1$.
(c) From the triple integral $\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^3 \mathop \smallint \limits_{ - \sqrt {9 - {r^2}} }^0 r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $, we obtain the domain of integration:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 3,0 \le \theta \le 2\pi , - \sqrt {9 - {r^2}} \le z \le 0} \right\}$
Since $z$ varies from $ - \sqrt {9 - {r^2}} $ to $0$, it is bounded below by the lower hemisphere $z = - \sqrt {9 - {r^2}} $ and bounded above by $z=0$. So, ${\cal W}$ is the lower hemisphere of radius $3$.
