Answer
$\dfrac{\pi(1+391\sqrt{17})}{60}$
Work Step by Step
As we are given that $z=x^2+y^2$
Now, $\iint_S z dS=\iint_D z\sqrt {1+4x^2 +4y^2} dy dx=\iint_D (x^2+y^2) [\sqrt {1+4x^2 +4y^2}] dy dx$
or, $\iint_S z dS=\int_0^{2\pi}\int_0^{2}r^2\sqrt{1+4r^2}r dr d\theta=[\theta]_0^{2\pi}\int_0^2 (r^2) (\sqrt{1+4r^2})(r) dr$
Suppose $1+4r^2 =k$
and $dk=8rdr$
Then $\iint_S z dS=(\dfrac{\pi}{4})\int_1^{17}(\dfrac{k-1}{4}) [k^{1/2}] dk$
or, $\iint_S z dS=(\dfrac{\pi}{8})[\dfrac{(17)^2\sqrt{17}}{5}-\dfrac{(17)\sqrt {17}}{3}]-(\dfrac{\pi}{8})[\dfrac{1}{5}-\dfrac{1}{3}]=\dfrac{\pi(1+391\sqrt{17})}{60}$
