Answer
$\pi$
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.
Here, $\iint_S F \cdot n dS=\iint_D (vi+u \sin v j+u \cos v k) \cdot (\sin v i-\cos v j+uk) dA$
$= \int_{0}^{1} \int_0^{\pi} v \sin v -u \sin v \cos v +u^2 \cos v dv du$
$= \int_{0}^{1} [(- v \cos v + \sin v) +\dfrac{u \cos 2 v}{4} +u^2 \sin v]_0^{\pi} du$
or, $= \int_{0}^{1} \pi du$
Hence, $Flux=\pi$
