Answer
$2 \pi(\dfrac{16}{3}-3 \sqrt 3)$
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.
$\iint_S F \cdot n dS=\iint_S y^2 dS=\iint_D y^2 \dfrac{2}{\sqrt {4-x^2-y^2}} dA$
$= \int_{0}^{2 \pi} \int_0^{1} (r \sin \theta)^2 \dfrac{2}{ \sqrt {4-r^2}} dr d\theta$
$=(2) \int_{0}^{1} \dfrac{r^3}{ \sqrt {4-r^2}} dr \int_0^{2 \pi}
\sin^2 \theta d\theta$
$=(2) [(-1/3)(r^2+8)(\sqrt {4-r^2})]_0^1 \times [(1/2) (\theta-\dfrac{\sin 2 \theta }{2}]_0^{2 \pi}$
Hence, $Flux=2 \pi(\dfrac{16}{3}-3 \sqrt 3)$
