Answer
$-\dfrac{4 \pi}{3}$
Work Step by Step
The flux through a surface can be defined only when the surface is orientable.
We know that $\iint_S F \cdot dS=\iint_S F \cdot n dS$
Here, $n$ denotes the unit vector.
Since, $\iint_S F \cdot n dS=-\iint_D \dfrac{x^2}{\sqrt {4-x^2-y^2}}dA$
or, $=- \int_{0}^{\pi/2} \int_0^{2} \dfrac{r^3 \cos^2 \theta}{\sqrt {4-r^2}} dr d\theta$
or, $=[- \int_{0}^{\pi/2} \cos^2 \theta] [\int_0^{2} \dfrac{r^3}{\sqrt {4-r^2}} dr]$
or, $=0.5 \times \int_{0}^{\pi/2} 1+\cos( 2 \theta) d \theta \times (-0.5) \times \int_0^{2} (-2r) \dfrac{r^2}{\sqrt {4-r^2}} dr$
or, $=(0.5) \int_{0}^{\pi/2}[\theta \times (0.5) \times \sin ( 2 \theta)]_0^{\pi/2}+\dfrac{16}{3}$
Hence, the flux is $\iint_S F \cdot n dS=-\dfrac{\pi}{4} \times \dfrac{16}{3}=-\dfrac{4 \pi}{3}$
