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