Answer
$\approx 4.4506$
Work Step by Step
We have $A(S)= \iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|$
and $\iint_{D} dA$ is the area of the region $D$
Now, $\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=\dfrac{9}{4} \cos^3 v \sin 2v \sin 2u (sin u \sin vi+\cos u \sin v j+\cos v \cos u \sin u k) $
and $|\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|=\dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}}$
Therefore, $A(S)=\iint_{D} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dA=\int_0^{\pi} \int_{0}^{2 \pi} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dv du$
By using a calculator, we get
$A(S) = \int_0^{\pi} \int_{0}^{2 \pi} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dv du \approx 4.4506$
