Answer
$$\dfrac{1}{6} \pi a^3$$
Work Step by Step
$\vec{r_u} \times \vec{r_v}=a^2 \cos u \sin^2 v i+a^2 \sin u \sin^2 v j+a^2 \sin v \cos v k$
$$I=\iint_{S} F (u,v) \ d S=\int_{0}^{\pi/2} \int_{0}^{\pi/2} a^3 \sin v \cos^2 v \ dv \ du \\=(\dfrac{-1}{3} ) a^3 \int_{0}^{\pi/2}[\cos^3 v]_{0}^{\pi/2} \ du \\=(\dfrac{-1}{3} ) \times a^3 \int_{0}^{\pi/2}[\cos^3 (\pi/2) -\cos^3 (0) ] \ du \\=\dfrac{1}{3} \times a^3 [u]_{0}^{\pi/2}$$
Now, we evaluate the integral
$I=\int_{0}^{\pi/2} \int_{0}^{\pi/2} a^3 \sin v \cos^2 v \ dv \ du \\=\dfrac{1}{6} \pi a^3$
