Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{7}{4}\pi $
Work Step by Step
We have $f\left( {x,y,z} \right) = z$ and the region:
${\cal W}:{x^2} + {y^2} + {z^2} \le 4,z \ge 0,{x^2} + {y^2} \le 1$
From the figure attached, we see that ${\cal W}$ is bounded below by $z=0$ and bounded above by ${x^2} + {y^2} + {z^2} = 4$.
Since ${r^2} = {x^2} + {y^2}$ and $z \ge 0$, in cylindrical coordinates, $z$ ranges from $z=0$ to $z = \sqrt {4 - {r^2}} $.
The projection of ${\cal W}$ onto the $xy$-plane is ${\cal D}$, the unit disk ${x^2} + {y^2} \le 1$. In polar coordinates, ${\cal D}$ is described by
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
Thus, the description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1,0 \le \theta \le 2\pi ,0 \le z \le \sqrt {4 - {r^2}} } \right\}$
Integrate $f\left( {x,y,z} \right) = z$ over ${\cal W}$ using cylindrical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^{\sqrt {4 - {r^2}} } zr{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {{z^2}|_0^{\sqrt {4 - {r^2}} }} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {4 - {r^2}} \right)r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {4r - {r^3}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {2{r^2} - \frac{1}{4}{r^4}} \right)|_0^1} \right){\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {2 - \frac{1}{4}} \right){\rm{d}}\theta $
$ = \frac{7}{8}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \frac{7}{4}\pi $
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \frac{7}{4}\pi $.
