= Solution
Separate the closed cylinder into its curved side and its two end disks, all with outward <normal vectors>. On the side, $\rho=R$ and $\mathbf n=\widehat{\boldsymbol\rho}$, so
$$
\Phi_{\rm side}=(-\alpha R)(2\pi Rh)=-2\pi\alpha R^2h.
$$
At the top, $z=h/2$ and $\mathbf n=\mathbf k$, giving $(\boldsymbol\nabla\times\mathbf v)\cdot\mathbf n=\alpha h$. At the bottom, both $z$ and the axial <normal vector> change sign, so the same positive value $\alpha h$ results. Each disk therefore contributes $\pi\alpha R^2h$. Adding all three <surface integrals>,
$$
\boxed{\int_S(\boldsymbol\nabla\times\mathbf v)\cdot d\mathbf S
=-2\pi\alpha R^2h+\pi\alpha R^2h+\pi\alpha R^2h=0.}
$$
This explicit cancellation is consistent with the fact that the <divergence of a curl is zero>. The two disk contributions add rather than cancel.
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