1. Let S be the surface $\displaystyle z=x, x^2+y^2<=1$ Find

$\displaystyle \iint (x^2 +y^2)dS$

I know the answer is supposed to be $\displaystyle \sqrt{2}*\pi/ 2$ but can't seem to get there. I've tried evaluating it with the divergence theorem, but nothing I do seems to work correctly. I know that $\displaystyle x^2+y^2dS$ is equal to $\displaystyle F \cdot n$ and I have $\displaystyle n = <-1/\sqrt{2},0,1/\sqrt{2}> $, which would make $\displaystyle F=<-\sqrt{2} x^2,0,\sqrt{2} y^2>$. Then the integral becomes $\displaystyle \sqrt{2} \int_0^{2\pi}} \int_0^{1} \int_0^{rcos\theta} } -r^2cos(\theta)^2+r^2sin(\theta)^2 rdzdrd\theta$.

Am I doing this right? I can't figure out where I'm messing up.

These next two I have the same issues with:

2. Let S be the surface $\displaystyle z = x + y, 0\leqslant x \leqslant 1, 0\leqslant x \leqslant 1$. Find the upward flux of the vector field $\displaystyle F = <z,x,y>$ across S.

I think that the divergence is 0, and wouldn't that make the whole integral 0? But I'm told the correct answer is -1. Why is this?

3. Let S be the portion of the cylinder given by $\displaystyle 0\leqslant z \leqslant 3, r=1, 0\leqslant \theta \leqslant \pi/2 $.

Orient S by normal vectors pointing away from the z-axis and compute the flux of $\displaystyle F = <2x,y,-3z>$ across S.

In this case, wouldn't the divergence also be 0, since $\displaystyle \nabla \cdot F = 2 + 1 -3 = 0$?

But I know the answer to be $\displaystyle 9\pi/4$.

I'm obviously missing the same sort of thing in all these problems so any help would be appreciated very much!