I want to evaluate the surface intgeral of $$\vec{F}=\langle 2x^3,5xz^2,x^2+6y^2z\rangle$$ in the $xy$-plane and $x^2+y^2\le4$.
I know I should use $$\iint_S\vec F\cdot d\vec{S}=\iint_S\vec F\cdot \vec nd\vec S.$$
My approach:
(1) Given this is in the $xy$-plane, then $z=0$ so $\vec r=\langle x,y,0\rangle$ and our boundary is $x^2+y^2\le4$.
(2) Next, $\vec F\cdot \vec n=\langle 2x^3,5xz^2,x^2+6y^2z\rangle\cdot\big(\langle 1,0,0\rangle\times \langle 0,1,0\rangle\big)=x^2+6y^2z$.
(3) And then, I was told that since $z=0$, I can just take the double integral: $\int\int x^2dA$ over $x^2+y^2\le4$ which is
$$\int_0^{2\pi}\int_0^2(r\cos{\theta})^2rdrd\theta=4\pi.$$ My confusion comes from step (3). Why are we able to say $6y^2z=0$ just because our boundary says that $z=0$? I mean, following the formula above, dotting $\vec F$ with $\vec n$ gives $x^2+6y^2z$, why do we integrate something else?