Surface integral over paraboloid and disk

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Let $S$ be the surface composed of the paraboloid $y=x^2+z^2$, with $y \in [0,1]$ and the disk $x^2+z^2 \leq 1$ and $y=1$, oriented with outward normal vectors. Find $\iint_S \mathbf{F} d\mathbf{S}$, where $\mathbf{F}=(0,y,-z)$. (The answer should be 0)

My attempt: Parameterized the paraboloid using parameters $r$ and $\theta$: $x=r\cos\theta$, $z=r\sin\theta$, $y=r^2$. Not sure if this is right and I'm not sure what to do with the disk. Any help is appreciated

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The evaluation is much easier by using the Divergence Theorem, $$\iint_S \mathbf{F} d\mathbf{S}=\iiint_V\text{div}(\mathbf{F})dV$$ where $V=\{(x,y,z): x^2+z^2\leq y\leq 1\}$. Can you take it from here?