Answer
$-\dfrac{32\pi}{3}$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_D [P \dfrac{\partial h}{\partial x}-Q+R\dfrac{\partial h}{\partial z}]dA $
where, $D$ is projection of $S$ onto xz-plane.
$\iint_S F \cdot n dS=-\iint_D 2\sqrt {4-x^2-y^2} dA$
$= \int_{0}^{2 \pi} \int_0^{2} -2r \sqrt {4-r^2} dr d\theta$
Plug $a=4-r^2 \implies da=-2r dr$
$= \int_{0}^{2 \pi} \int_4^{0} \sqrt a da d\theta$
$= \int_{0}^{2 \pi} [-d \theta] \times \int_4^{0} \sqrt a da$
$=[-2 \pi -0] \times [(2/3) (4^{3/2}-0]$
Hence, $Flux=-\dfrac{32\pi}{3}$
