Answer
$\dfrac{\pi (37^{3/2}-1)}{6}$
Work Step by Step
We have $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA= \iint_{D} \sqrt {1+4y^2+4z^2} dA $
and $\iint_{D} dA$ is the area of the region $D$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA=\iint_{D} \sqrt {1+4y^2+4z^2} dA =\int_0^{2 \pi} \int_{0}^{3} \sqrt {1+(2r)^2} r dr d \theta $
Substitute $a=1+4r^2 $ and $da= 8 r dr$
Now, $A(S)= \dfrac{1}{8} \int_0^{2 \pi} \int_{1}^{ 37} \sqrt {a} da d \theta=\dfrac{1}{8} \int_0^{2 \pi} [\dfrac{2a^{3/2}}{3}]_{1}^{37} d \theta$
This implies that,
$A(S)=\dfrac{\pi (37^{3/2}-1)}{6}$
