Answer
$16 \pi$
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.
Since, $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
$\iint_S F \cdot n dS= \int_{0}^{ \pi/2} \int_0^{2 \pi} [8 \sin^2 u] \times [\cos u (4 \sin u ) ]dv du$
or, $=32 \int_{0}^{ \pi/2} \int_0^{2 \pi} \sin^3 (u) \times \cos (u) dv du$
or, $=32 \int_{0}^{ \pi/2} \int_0^{2 \pi} \sin^3 (u) \times \cos (u) dv du$
or, $=32 [\int_{0}^{ 2\pi} dV] \times [ \int_0^{\pi/2} \sin^3 (u) \times \cos (u) du] $
or, $=32 (2 \pi) [\dfrac{\sin^4 u}{4}]_{0}^{(\pi/2)}$
Hence, we have $\iint_S F \cdot n dS=16 \pi$
