Answer
$\dfrac{\pi(1+391\sqrt{17})}{60}$
Work Step by Step
$dS=\iint_D\sqrt {1+(2x)^2 +(2y)^2} dy dx=\iint_D\sqrt {1+4x^2 +4y^2} dy dx$
Now, $\iint_S z dS=\iint_D (x^2+y^2) \sqrt {1+4x^2 +4y^2} dy dx$
or, $=\int_0^{2\pi}\int_0^{2}r^2\sqrt{1+4r^2}r dr d\theta$
or, $=[\theta]_0^{2\pi}\int_0^2 r^2\sqrt{1+4r^2}r dr$
Plug $1+4r^2 =u $ then we have $du=8rdr$
Then $\iint_S z dS=(\dfrac{\pi}{4})\int_1^{17}(\dfrac{u-1}{4}) [u^{1/2}] du$
or, $=(\dfrac{\pi}{16})\int_1^{17}p^{3/2}-p^{1/2} dp$
or, $=(\dfrac{\pi}{16})[\dfrac{2}{5}p^{5/2}-\dfrac{2}{3}p^{3/2}]_1^{17}$
Hence, $\iint_S z dS=\dfrac{\pi(1+391\sqrt{17})}{60}$
