Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xz{\rm{d}}V = \frac{4}{3}$
Work Step by Step
We have $f\left( {x,y,z} \right) = xz$ and the region:
${\cal W} = \left\{ {\left( {x,y,z} \right)|{x^2} + {y^2} \le 1,x \ge 0,0 \le z \le 2} \right\}$
In polar coordinates, we get $f\left( {r\cos \theta ,r\sin \theta ,z} \right) = rz\cos \theta $.
Since $x \ge 0$, the projection of ${\cal W}$ onto the $xy$-plane is half of an unit disk, located at the right-hand side of the $y$-axis (namely, the first and the fourth quadrant). Thus, the cylindrical description of ${\cal W}$:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1, - \frac{\pi }{2} \le \theta \le \frac{\pi }{2},0 \le z \le 2} \right\}$
Using Eq. (5) of Theorem 2, we evaluate the integral of the function $f\left( {x,y,z} \right) = xz$ over ${\cal W}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xz{\rm{d}}V = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^2 {r^2}z\cos \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{2}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 {r^2}\left( {{z^2}|_0^2} \right)\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = 2\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \left( {{r^3}|_0^1} \right)\cos \theta {\rm{d}}\theta $
$ = \frac{2}{3}\mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} \cos \theta {\rm{d}}\theta $
$ = \frac{2}{3}\left( {\sin \theta |_{ - \pi /2}^{\pi /2}} \right) = \frac{4}{3}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} xz{\rm{d}}V = \frac{4}{3}$.
