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