Answer
$\dfrac{15625\pi \sqrt 2}{6}$
Work Step by Step
When the surface is orientable then the flux through a surface is defined as: $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ is the unit vector and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
Consider $\iint_S y^2 z^2 dS =\iint_{D} y^2 z^2 \times \sqrt 2 dx dz=\iint_{D}(x^2+z^2) \times z^2 dx dz$
$\iint_S y^2 z^2 dS= \int_{0}^{2 \pi} \int_0^5 (r^2) \cdot (r^{2}) \times \sin^2 \theta (r dr d\theta) \times (\sqrt 2)=\ \int_{0}^{2 \pi} \sin^2 \theta \times \int_0^5 r^5 dr \times \sqrt 2 $
Hence, we have $\iint_S y^2 z^2 dS= \int_{0}^{2 \pi} (\dfrac{1}{2})-\dfrac{ \cos (2 \theta) d \theta}{2} \times \sqrt 2 \times \dfrac{r^6}{6}=\dfrac{15625\pi \sqrt 2}{6}$
