Answer
$\dfrac{1}{6} \pi a^3$
Work Step by Step
We have: $Normal \ Vector ; n=\vec{r_u} \times \vec{r_v}=a^2 \cos u \sin^2 v i+a^2 \sin u \sin^2 v j+a^2 \sin v \cos v$
Now, $F \cdot n \ d \theta =a^3 \sin v \cos^2 v$
and $\iint_{S} F (u,v) \ d S=\int_{0}^{\pi/2} \int_{0}^{\pi/2} a^3 \sin v \cos^2 v \ dv \ du$
or, $=(\dfrac{-1}{3} ) a^3 \int_{0}^{\pi/2}[\cos^3 v]_{0}^{\pi/2} \ du$
or, $=(\dfrac{-1}{3} ) a^3 \int_{0}^{\pi/2}[\cos^3 (\pi/2) -\cos^3 (0) ] \ du$
or, $=\dfrac{1}{3} a^3 [u]_{0}^{\pi/2}$
We evaluate the integral as:
$\int_{0}^{\pi/2} \int_{0}^{\pi/2} a^3 \sin v \cos^2 v \ dv \ du=\dfrac{1}{6} \pi a^3$
