Answer
$\dfrac{\sqrt {2}}{10}$
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.
Since, we have $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f[\overline{x}, \overline{y}, \overline{z}] AS_i$
Consider $\iint_S F \cdot dS =(\sqrt 2) \int_{0}^1 \int_{0}^{\pi/2} u^3 \sin (v) \cos (v) u dA= \int_{0}^1 u^4 du \cdot \int_{0}^{\pi/2} \sin v \cos v dv \times (\sqrt 2) $
$\iint_S F \cdot dS= [\dfrac{u^5}{5}] \times (\sin^2 v/2]_0^{\pi/2} \times \iint_S F \cdot dS=\sqrt {2} \times (\dfrac{1}{5}) \times (\dfrac{1}{2})$
Hence, $\iint_S F \cdot dS =\dfrac{\sqrt {2}}{10}$
