## Calculus: Early Transcendentals 8th Edition

$\dfrac{\pi(1+391\sqrt{17})}{60}$
The surface area is calculated as: $S=\iint_S dS=\iint_D\sqrt {1+z_x^2 +z_y^2} dA$ Given: $z=x^+y^2$ Thus, $dS=\iint_D\sqrt {1+(2x)^2 +(2y)^2} dy dx=\iint_D\sqrt {1+4x^2 +4y^2} dy dx$ Now, $\iint_S z dS=\iint_D z\sqrt {1+4x^2 +4y^2} dy dx=\iint_D (x^+y^2) \sqrt {1+4x^2 +4y^2} dy dx$ Now, in polar co-ordinates we have $\iint_S z dS=\int_0^{2\pi}\int_0^{2}r^2\sqrt{1+4r^2}r dr d\theta=[\theta]_0^{2\pi}\int_0^2 r^2\sqrt{1+4r^2}r dr$ Plug $1+4r^2 =p \implies dp=8rdr$ Then $\iint_S z dS=(\dfrac{\pi}{4})\int_1^{17}(\dfrac{p-1}{4}) p^{1/2} dp=(\dfrac{\pi}{16})\int_1^{17}p^{3/2}-p^{1/2} dp$ or, $(\dfrac{\pi}{16})[\dfrac{2}{5}p^{5/2}-\dfrac{2}{3}p^{3/2}]_1^{17}=(\dfrac{\pi}{8})[\dfrac{(17)^2\sqrt{17}}{5}-\dfrac{(17)\sqrt {17}}{3}]-(\dfrac{\pi}{8})[\dfrac{1}{5}-\dfrac{1}{3}]$ Hence, $\iint_S z dS=\dfrac{\pi(1+391\sqrt{17})}{60}$