Answer
$$\dfrac{4 \pi}{3}$$
Work Step by Step
$\vec{r} (u,v) =\cos u \sin v i+\sin u \sin v j+\cos v k$
$\implies \vec{r_u} \times \vec{r_v}=-\cos u \sin^2 v i-\sin u \sin^2 v j -\sin v \cos v k$
$\implies |\vec{r_x} \times \vec{r_y}| =\sqrt {(-\cos u \sin^2 v)^2 +(-\sin u \sin^2 v)^2 +( -\sin v \cos v )^2}=\sqrt {\sin^2 v(\sin^2 v +\cos^2 v) } \\ =\sqrt {\sin^2 v} \\=\sin v$
$\int_{0}^{\pi} \int_{0}^{2 \pi} \cos^2 u \sin^3 v \ du \ dv =(1/2)\int_{0}^{\pi} \int_{0}^{2 \pi} \cos^2 u \sin^3 v \ du \ dv \\=\dfrac{1}{2} \int_{0}^{\pi} [u+\dfrac{\sin 2u}{2} ]_{0}^{2 \pi} \implies \sin^3 v \ dv \\=\pi \times \int_{0}^{\pi} \sin^3 v \ dv \\= \pi \implies \int_{0}^{\pi} \sin v \times \sin^2 v \ dv=\pi \int_{0}^{\pi} \sin v (1-\cos^2 v) \ dv\\= \pi \implies [-\cos v +\dfrac{\cos^3 v}{3}]_0^{\pi} \\=\dfrac{4 \pi}{3}$$
