Answer
$0$
Work Step by Step
$x=\sqrt{4-y^2}$
$x^2=4-y^2$
$x^2+y^2=4 $
That is, $ r=2$
In cylindrical coordinate system, we have $I=\int_0^{2\pi} \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) \cos \theta \times dz dr d\theta$
$=\int_0^{2\pi} \cos \theta d\theta \times [ \int_{0}^{2} (1) \times \int_{r}^{2} (z)(r^2) dz dr ] $
$=[\sin (\theta) ]_0^{2 \pi} \times [ \int_{0}^{2}(1) \times \int_{r}^{2} \times (z) \times (r^2) dz dr ] $
$=[\sin (2 \pi)-\sin (0) ] \times [ \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) dz dr ] $
$=0 \times [ \int_{0}^{2}\int_{r}^{2} (z) \times (r^2) dz dr ] $
$=0$
