Misapplication of the divergence theorem when calculating a surface integral? 
Let $\mathbf{F} = (3y, -xz, yz^2)$, and let  $S=\{(x,y,z): z=\frac{1}{2}(x^2+y^2), z\leq 2\}$.  Find $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.

Firstly, I know I can compute this quite easily using Stokes' theorem -- my question concerns only the divergence theorem. Using Stokes' theorem, I obtain an answer of $-20\pi$.
Now below is my working relating to the divergence theorem.

If I let $\Sigma$ be the disc $x^2+y^2\leq 4$ on the plane $z=2$, then $S \cup \Sigma$ is the boundary of  $\Omega = \{(x,y,z) : \frac{1}{2}(x^2+y^2)\leq z\leq 2\}$. So by the divergence theorem,
$$ \iiint_\Omega \nabla \cdot \mathbf{F} dV = \iint_S \mathbf{F}\cdot d\mathbf{S} + \iint_\Sigma \mathbf{F}\cdot d\mathbf{S}. $$ For the region $\Sigma$, the unit normal is $(0,0,1)$, so we can calculate the surface integral as
\begin{align*}
 \iint_\Sigma \mathbf{F}\cdot d\mathbf{S} &= \iint_{\Sigma} \mathbf{F}\cdot (0,0,1) \,dA \\ 
&= \iint_{x^2+y^2\leq 4 \\ z=2} yz^2 dA \\ 
&= 4 \iint_{x^2+y^2\leq 4} y dA \\
&= 4 \int_0 ^{2\pi} \int_0^2 (r\sin\theta) drd\theta \\ 
&= 4 \int_0 ^{2\pi} 2\sin \theta = 0.
\end{align*}
Also, $\nabla \cdot \mathbf{F} = (0,0,2zy)$, so
\begin{align*}
\iiint_\Omega 2zy dV &= 2\int_0^2 \iint _{x^2+y^2\leq 2z} yz\,dV \\ 
&=2\int_0^2\int_0^{2\pi} \int_0^\sqrt{2z} (r\sin\theta z) dr d\theta dz
\end{align*}
but this will still be $0$ because of the presence of $\sin \theta$. This would imply that $\iint_S \mathbf{F}\cdot d\mathbf{S}=0$.

My question is: what went wrong with the second computation? I think it might be the way I handled the volume integral $\iiint_\Omega$, but I'm not quite sure. Have I misunderstood the divergence theorem? Any guidance would be very much appreciated.
 A: I've figured out my mistake, thanks to @Event Horizon. My impression from the Help Center page is that I shouldn't delete my question, so I'll outline what went wrong:
I should've applied the divergence theorem to $\nabla \times \mathbf{F}$ instead, so that the statement of the theorem becomes
$$ \iiint_\Omega \nabla \cdot (\nabla \times \mathbf{F}) = \iint_S \nabla \times \mathbf{F} \cdot d\mathbf{S} + \iint_\Sigma \nabla \times \mathbf{F}\cdot d\mathbf{S}. $$
The left is $0$ by since the divergence of a curl is $0$. From $\nabla \times \mathbf{F} = (z^2+x,0,-z-3)$, the correct computation is
$$ \iint_\Sigma \nabla \times \mathbf{F}\cdot d\mathbf{S} = \iint_{x^2+y^2\leq 4 \\ z=2} (-z-3 )dA = \iint-5dA = -20\pi$$ which matches what I got using Stokes' theorem (up to sign - but this just depends on the orientation of $S$).
A: Just for reinforcement, I thought it would be good to show that your work is correct for $\iint_S \vec F \cdot d\vec S$, and we can show this by working directly:
$$\vec r (x, y) = \langle x, y, \frac12 (x^2 + y^2)\rangle$$
$$\vec n = \vec r_x \times \vec r_y = \langle 1, 0, x\rangle \times \langle 0, 1, y\rangle = \langle -x, -y, 1\rangle$$
$$\iint_S \vec F \cdot d\vec S = \iint_\Sigma \langle 3y, -\frac12 x (x^2 + y^2), \frac14 y(x^2 + y^2)^2\rangle \cdot \frac1{x^2 + y^2+1}\langle-x, -y, 1\rangle dA = \iint_\Sigma \frac1{x^2 + y^2 + 1} \left(-3xy + \frac12xy(x^2+y^2)+\frac14 y(x^2+y^2)^2 \right) dA$$
where $\Sigma$ is defined as the same disk you used.
At first this looks like a bit of a mess, but note that the integrand is odd in $y$ and that the region of integration is symmetric across the $x$-axis. Letting our messy integrand be $f,$ this means that $f(x, -y) = -f(x,y)$ and that if a point $(x,y)$ is in $\Sigma$ then so is $(x, -y).$ (both of which are trivial here)
We can use this by defining $R_1 = \{(x, y) \in \Sigma : y \geq 0\}$ and $R_2 = \{(x, y) \in \Sigma : y \leq 0\},$ noting that also $R_2 = \{(x, -y) : (x, y) \in R_1\}.$ So,
$$\iint_\Sigma f(x,y) dA = \iint_{R_1} f(x,y) dA + \iint_{R_2} f(x,y) dA = \iint_{R_1} f(x,y) dA + \iint_{R_1} f(x,-y) dA = \iint_{R_1} f(x,y) + f(x, -y) dA = \iint_{R_1} 0 dA = \boxed{0}$$
as you calculated.
