Answer
$0$
Work Step by Step
Here, we have $x=\sqrt{4-y^2}$
Re-arrange as $x^2=4-y^2$
or, $x^2+y^2=4$
or, $r=2$
For the cylindrical coordinate system, we get
$\int_0^{2\pi} \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) (\cos \theta dz dr d\theta) =\int_0^{2\pi} (\cos \theta d\theta) \cdot [ \int_{0}^{2}\int_{r}^{2} (z)(r^2) dz dr ] $
$=[\sin \theta]_0^{2 \pi}\times [ \int_{0}^{2}\int_{r}^{2} (z)(r^2) dz dr ] $
$=[\sin 2 \pi-\sin 0] [ \int_{0}^{2}\int_{r}^{2} (z)(r^2) dz dr ] $
Thus, we have $(0)[ \int_{0}^{2}\int_{r}^{2} (z)(r^2) dz dr ]=0$
