Answer
$\approx 4.4506$
Work Step by Step
$Surafce \ area ; A(S)= \iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}| \ dA$
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) $
Therefore, $Surface \ area ; A(S)=\iint_{D} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} \ dA
\\
=\iint_{D} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dA \\ =\int_0^{\pi} \int_{0}^{2 \pi} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} dv du$
Now, we will use calculator to get as follows:
$Surface \ Area; A(S) = \int_0^{\pi} \int_{0}^{2 \pi} \dfrac{9}{4} |\cos^3 v \sin 2v \sin 2u| \sqrt {\sin^2 v+\dfrac{\cos^2 v \sin^2 2u}{4}} \ dv \ du \approx 4.4506$
