= Solution
The initially irrotational inviscid flow remains irrotational, so $u=\nabla\phi$; incompressibility gives $\nabla^2\phi=0$. At the plates, normal <velocity> matches their motion:
$$
\frac1r\phi_\theta(r,\alpha)=-\Omega r,\qquad
\frac1r\phi_\theta(r,-\alpha)=\Omega r.
$$
With $\phi=r^2f(\theta)$, Laplace's equation gives $f''+4f=0$. The symmetric solution is
$$
\phi=\frac{\Omega r^2\cos2\theta}{2\sin2\alpha}.
$$
Hence
$$
u_r=\frac{\Omega r\cos2\theta}{\sin2\alpha},\qquad
u_\theta=-\frac{\Omega r\sin2\theta}{\sin2\alpha}.
$$
A streamfunction is
$$
\psi=\frac{\Omega r^2\sin2\theta}{2\sin2\alpha},
$$
so <streamlines> are $r^2\sin2\theta=\text{constant}$ and fluid is expelled radially as the plates close. The outward flux through $r=R$ is
$$
\boxed{\int_{-\alpha}^{\alpha}u_rR\,d\theta=\Omega R^2.}
$$
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