Answer
$32 \pi \sqrt 3$
Work Step by Step
We are given that $z=4+x+y$
Now, $\iint_S x dS=\iint_D x \sqrt {3} dy dx=\iint_D (x^2+y^2) (\sqrt {4+x+y}) (\sqrt 3) dx dy$
$\iint_S x dS=\int_0^{2\pi}\int_0^{2}(r^2) (4+r \cos \theta+r \sin \theta)(\sqrt 3r) dr d\theta$
or, $=\sqrt 3 \int_0^{2\pi}\int_0^2 4r^3+r^4\cos \theta+r^4 \sin \theta dr d \theta$
or, $=\sqrt 3 \int_0^{2 \pi} [r^4+\dfrac{r^5(\cos \theta+\sin \theta)}{5}]_0^2 d\theta$
Thus, $\iint_S x^2 z+y^2 z dS=\sqrt 3 [16 (\theta)+\dfrac{32(\sin \theta-\cos \theta)}{5}]_0^{2 \pi}=32 \pi \sqrt 3$
