Answer
$16 \pi$
Work Step by Step
When the surface is orientable then the flux through a surface is defined as: $\iint_S F \cdot dS=\iint_S F \cdot n dS$
where, $n$ is the unit vector and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$
$\iint_S F \cdot n dS= \int_{0}^{ \pi/2} \int_0^{2 \pi} 8 \times \sin^2 u \times \cos u \times (4 \sin u ) dv du$
or, $\iint_S F \cdot n dS=\int_{0}^{ \pi/2} \int_0^{2 \pi} 32 \times \sin^3 u \times \cos u dv du$
or, $\iint_S F \cdot n dS= \int_{0}^{ \pi/2} \int_0^{2 \pi} (\sin^3 u) (\cos u) dv du \times 32$
or, $\iint_S F \cdot n dS=(32) [\int_{0}^{ 2\pi} dV] \times [ \int_0^{\pi/2} \sin^3 u \cos u du] = (2 \pi) [\dfrac{\sin^4 u}{4}]_{0}^{\pi/2} \times 32=16 \pi$
