 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 $\mathbf{n}=\hat{\mathbf{\rho }}$, soAt the top, $z=h/2$ and $\mathbf{n}=\mathbf{k}$, giving $(\nabla \times \mathbf{v})\cdot \mathbf{n}=\alpha h$. At the bottom, both $z$ and the axial normal vector change sign, so the same positive value $\alpha h$ results. Each disk therefore contributes $\pi \alpha R^{2}h$. Adding all three surface integrals,This explicit cancellation is consistent with the fact that the divergence of a curl is zero. The two disk contributions add rather than cancel.