Answer
$$\dfrac{\pi}{6} (65 \sqrt {65} -1)$$
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+(2x)^2+(2z)^2} dA = \iint_{D} \sqrt {1+4x^2+4z^2} \ dA $
and $\iint_{D} dA$ is the area of the region $D$
Use the polar coordinates as follows: $x=r \cos \theta; y= r \sin \theta$
Therefore, $A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA \\=\int_0^{2\pi} d \theta \int_{0}^{4} r \sqrt {1+4r^2} \ dr$
Substitute $a=1+4r^2 $ and $ r \ dr =\dfrac{da}{ 8}$
Now, $A(S)=2 \pi \int_{1}^{65} \sqrt a \dfrac{da}{ 8}$
This implies that, $A(S)=\dfrac{\pi}{4} [\dfrac{2a^{3/2}}{3}]_1^{65} =\dfrac{\pi}{6} [a\sqrt a]_1^{65}=\dfrac{\pi}{6} (65 \sqrt {65} -1)$
