I already saw a similar question here. But the method discussed there is not familiar to me. Rather, I am used to the formula below for calculating surface integral:
$$\int\int\vec{F}.\vec{n} dS=\int\int\vec{F}.\vec{n}\frac{dxdy}{|\vec{n}.\hat{k}|}$$ $$or, \int\int\phi dS$$
Taking projection in $xy$ plane, I have $z=0$, $\vec{n}=\frac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$. So, $|\vec{n}.\hat{k}|=\frac{2}{3}$. Now, the surface integral becomes:
$$\frac{3}{2}\int_{x=0}^{4}\int_{y=0}^{\sqrt{16-x^{2}}}(x+2y-2)dydx$$ $$=\frac{3}{2}\int_{x=0}^{4}(x\sqrt{16-x^{2}}+16-x^2-12\sqrt{16-x^2})dx$$
But my result does not seem to match with the answer $6\pi$. I don't know if I did wrong somewhere. Any help is appreciated. Thank you...