How to show that $\iint_{S}\vec{F}\cdot d\vec{S}=0$ with the vector field $\vec{F}=\big\langle0,0,z\big\rangle$? Problem: If $S$ is the cylindrical surface parametrized by $\phi(\theta,u)=(\cos{(\theta)},\sin{(\theta)},u)$, $\theta\in[0,2\pi]$ and $u\in[0,1]$, then $\iint_{S}\vec{F}\cdot d\vec{S}=0$ for the vector field $\vec{F}=\big\langle0,0,z\big\rangle$.
Solution:
$\iint_{S}\vec{F}\cdot d\vec{S}=\iint_{D}\vec{F}(\phi(\theta,u))\cdot \vec{n}\ dA$
I know that $\vec{F}(\phi(\theta,u))=F(\cos{(\theta)},\sin{(\theta)},u)=\big\langle0,0,u\big\rangle$
But I don't know how to get $\vec{n}$. What it is?

$d\vec{S} = \frac{\partial \phi}{\partial u} \times \frac{\partial \phi}{\partial v} \  du \ dv$ this will be in the direction perpendicular to the $\vec{F}$ which you are given.
