Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{5\pi }}{8}$
Work Step by Step
We have $f\left( {x,y,z} \right) = z$ and the region defined by
${\cal W} = \left\{ {\left( {\rho ,\varphi ,\theta } \right)|1 \le \rho \le 2,0 \le \varphi \le \frac{\pi }{2},0 \le \theta \le \frac{\pi }{3}} \right\}$
Since $x = \rho \sin \varphi \cos \theta $, $y = \rho \sin \varphi \sin \theta $ and $z = \rho \cos \varphi $, in spherical coordinates:
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = \rho \cos \varphi $
Using Eq. (8), the triple integral of $f\left( {x,y,z} \right)$ over ${\cal W}$ in spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{\pi /3} \mathop \smallint \limits_{\varphi = 0}^{\pi /2} \mathop \smallint \limits_{\rho = 1}^2 {\rho ^3}\cos \varphi \sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{1}{4}\mathop \smallint \limits_{\theta = 0}^{\pi /3} \mathop \smallint \limits_{\varphi = 0}^{\pi /2} \left( {{\rho ^4}|_1^2} \right)\cos \varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{15}}{4}\mathop \smallint \limits_{\theta = 0}^{\pi /3} \mathop \smallint \limits_{\varphi = 0}^{\pi /2} \cos \varphi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{{15}}{8}\mathop \smallint \limits_{\theta = 0}^{\pi /3} \mathop \smallint \limits_{\varphi = 0}^{\pi /2} \sin 2\varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - \frac{{15}}{{16}}\mathop \smallint \limits_{\theta = 0}^{\pi /3} \left( {\cos 2\varphi |_0^{\pi /2}} \right){\rm{d}}\theta $
$ = \frac{{15}}{8}\mathop \smallint \limits_{\theta = 0}^{\pi /3} {\rm{d}}\theta = \frac{{5\pi }}{8}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{{5\pi }}{8}$.
