Answer
$$\approx 4.4506$$
Work Step by Step
$\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=\dfrac{9 (\cos^3 v \sin 2v \sin 2u)}{4} (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, $A(S)= \iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}| \\=\iint_{D} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dA \\=\int_0^{\pi} \int_{0}^{2 \pi} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} \ dv \ du$
Now, by using a calculator, we get:
$$A(S) =\dfrac{9}{4} \times \int_0^{\pi} \int_{0}^{2 \pi} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dv du
\\ \approx 4.4506$$