Answer
$\dfrac{364 \pi \sqrt 2}{3}$
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, $\iint_S x^2z^2 dS =\iiint_{D} (v \sin u)^2 v^2 v\sqrt 2 dA$
$=\int_{1}^{3} \int_{0}^{2 \pi} (\sin^2 u) v^5 \sqrt 2 du dv$
$=( \int_{0}^{2 \pi} \sin^2 u du) \times (\int_{1}^{3} v^5 \sqrt 2 dv)$
$=[(1/2) \int_{0}^{2 \pi} \sin^2 u du] \times (\int_1^3 v^5 \sqrt 2 dv)$
$=(1/2) [u-\dfrac{\sin 2u}{2}]_0^{2 \pi} \times (\dfrac{729\sqrt 2}{6}-\dfrac{\sqrt 2}{6})$
$=\dfrac{364 \pi \sqrt 2}{3}$
