Answer
(a) the limits of integration:
$0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le \phi \le \pi $, ${\ \ \ }$ $0 \le \rho \le 4$
Thus, the triple integral:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 0}^4 f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
(b) the limits of integration:
$0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le \phi \le \pi $, ${\ \ \ }$ $4 \le \rho \le 5$
Thus, the triple integral:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 4}^5 f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
(c) the limits of integration:
$0 \le \theta \le 2\pi $, ${\ \ \ }$ $\frac{\pi }{2} \le \phi \le \pi $, ${\ \ \ }$ $0 \le \rho \le 2$
Thus, the triple integral:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = \pi /2}^\pi \mathop \smallint \limits_{\rho = 0}^2 f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
Work Step by Step
We have a triple integral in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
(a) In spherical coordinates, we have $0 \le \phi \le \pi $ and $0 \le \theta \le 2\pi $.
Since the sphere has radius $4$, so $\rho $ varies from $0$ to $4$, that is, $0 \le \rho \le 4$.
So, the limits of integration:
$0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le \phi \le \pi $, ${\ \ \ }$ $0 \le \rho \le 4$
Thus, the triple integral:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 0}^4 f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
(b) Since the region is between the spheres of radii $4$ and $5$, $\rho$ varies from $4$ to $5$, that is, $4 \le \rho \le 5$.
So, the limits of integration:
$0 \le \theta \le 2\pi $, ${\ \ \ }$ $0 \le \phi \le \pi $, ${\ \ \ }$ $4 \le \rho \le 5$
Thus, the triple integral:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = 0}^\pi \mathop \smallint \limits_{\rho = 4}^5 f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
(c) Since the region is a lower hemisphere of the sphere of radius $2$, we have $\rho$ varies from $0$ to $2$, $\phi$ varies from $\frac{\pi }{2}$ to $\pi$.
So, the limits of integration:
$0 \le \theta \le 2\pi $, ${\ \ \ }$ $\frac{\pi }{2} \le \phi \le \pi $, ${\ \ \ }$ $0 \le \rho \le 2$
Thus, the triple integral:
$\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\phi = \pi /2}^\pi \mathop \smallint \limits_{\rho = 0}^2 f\left( {\rho ,\phi ,\theta } \right){\rho ^2}\sin \phi {\rm{d}}\rho {\rm{d}}\phi {\rm{d}}\theta $
