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