Answer
$\dfrac{\pi}{6} (65 \sqrt {65} -1)$
Work Step by Step
$Surface \ Area ; 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 $D$
Apply polar coordinates: $x=r \cos \theta; y= r \sin \theta$
Therefore, $Surface \ Area ; A(S)=\iint_{D} \sqrt {1+(\dfrac{\partial z}{\partial x} )^2+(\dfrac{\partial z}{\partial y} )^2 } dA \\=\int_0^{2\pi} d \theta \times \int_{0}^{4} r \sqrt {1+4r^2} \ dr$
Set $1+4r^2 = u $ and $ r \ dr =\dfrac{1}{ 8} du$
Now, $Surface \ Area ; A(S)=2 \pi \times \dfrac{1}{ 8} \times \int_{1}^{65} u^{1/2} du =\dfrac{\pi}{4}\times \dfrac{2}{3}[u^{3/2}]_1^{65} =\dfrac{\pi}{6} [u\sqrt u]_1^{65}=\dfrac{\pi}{6} (65 \sqrt {65} -1)$
