Answer
$\dfrac{\sqrt {2}}{10}$
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, $\iint_S F \cdot dS =\int_{0}^1 \int_{0}^{\pi/2} u^3 \sin v \cos v \times \sqrt 2 u dA=(\sqrt {2}) \int_{0}^1 u^4 du \int_{0}^{\pi/2} \sin v \cos v dv$
or, $=\sqrt {2} \times [\dfrac{u^5}{5}] \times (\sin^2 v/2]_0^{\pi/2}$
or, $=\sqrt {2}(\dfrac{1}{5}) (\dfrac{1}{2})$
Thus, we have $\iint_S F \cdot dS =\dfrac{\sqrt {2}}{10}$
