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, $D$ shows the projection of $S$ onto xz-plane.
$\iint_S F \cdot n dS=-\iint_D 2\sqrt {4-x^2-y^2} dA= \int_{0}^{2 \pi} \int_0^{2} -2r \sqrt {4-r^2} dr d\theta$
Suppose $4-r^2=p; dp=-2r dr$
$\iint_S F \cdot n dS= \int_{0}^{2 \pi} \int_4^{0} \sqrt p d d\theta= \int_{0}^{2 \pi} [-d \theta] \times \int_4^{0} \sqrt p dp$
Hence, we have $\iint_S F \cdot n dS=[-2 \pi -0] \times [(\dfrac{2}{3}) (4^{3/2}-0)]=(-2 \pi) (\dfrac{2}{3}) (4^{3/2})=-\dfrac{32\pi}{3}$
