Answer
$\dfrac{15625\pi \sqrt 2}{6}$
Work Step by Step
Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ denotes the unit vector.
and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
The flux through a surface can be defined only when the surface is orientable.
Since, $\iint_S y^2 z^2 dS =\iint_{D} y^2 z^2 \sqrt 2 dx dz$
$=\iint_{D}(x^2+z^2) \times z^2 dx dz$
$=(\sqrt 2) \times \int_{0}^{2 \pi} \int_0^5 (r^2) \cdot (r^{2}) \sin^2 \theta (r dr d\theta)$
$=\sqrt 2 \times \int_{0}^{2 \pi} \sin^2 \theta \times \int_0^5 r^5 dr$
$=\sqrt 2 \times \int_{0}^{2 \pi} (1/2)-\dfrac{ \cos (2 \theta) d \theta}{2} \times ( r^6/6)$
$=\dfrac{15625\pi \sqrt 2}{6}$
