= Solution
The exact gap between the plane and the circle is
$$
h(x)=h_0+a-\sqrt{a^2-x^2}.
$$
In the narrow region relevant to <lubrication theory>,
$$
\boxed{h(x)\simeq h_0+\frac{x^2}{2a}},
$$
so changes occur over the streamwise scale $x=O(\sqrt{ah_0})$; below use $x=\sqrt{2ah_0}X$.
Choose the positive $x$ direction to be the direction of the cylinder surface velocity $U=a\Omega$ in the gap, and write $\mu=\rho\nu$ for the <dynamic viscosity>. The lubrication velocity profile and <volume flux> per unit axial length are
$$
u(y)=\frac{p_x}{2\mu}y(y-h)+\frac{Uy}{h},
\qquad
Q=\int_0^hu\,dy=-\frac{h^3p_x}{12\mu}+\frac{Uh}{2}.
$$
Since $Q$ is constant and the pressure returns to the same ambient value as $x\to\pm\infty$, $\int_{-\infty}^{\infty}p_xdx=0$. Therefore
$$
Q=\frac U2
\frac{\int_{-\infty}^{\infty}h^{-2}dx}
{\int_{-\infty}^{\infty}h^{-3}dx}.
$$
Using $h=h_0(1+X^2)$ and the supplied integrals gives
$$
\boxed{Q=\frac23Uh_0=\frac23a\Omega h_0},
$$
in the direction of the cylinder surface motion.
The <shear stress> exerted by the fluid on the cylinder is opposite to
$$
\mu u_y(h)=\frac h2p_x+\frac{\mu U}{h}.
$$
Eliminating $p_x$ and inserting $Q=2Uh_0/3$ gives
$$
\boxed{\tau(x)=-4\mu U\left(\frac1h-\frac{h_0}{h^2}\right)},
$$
where the sign is relative to $U$. Since
$$
\int_{-\infty}^{\infty}
\left(\frac1h-\frac{h_0}{h^2}\right)dx
=\frac\pi2\sqrt{\frac{2a}{h_0}},
$$
the resisting <torque> per unit axial length is
$$
\boxed{T=a\int_{-\infty}^{\infty}\tau\,dx
=-2\sqrt2\,\pi\rho\nu\Omega
\frac{a^{5/2}}{h_0^{1/2}}}.
$$
Finally, the ratio of streamwise fluid inertia $U^2/\sqrt{ah_0}$ to transverse viscous acceleration $\nu U/h_0^2$ is of order
$$
\frac{a^{1/2}\Omega h_0^{3/2}}{\nu}.
$$
The stated restriction makes this lubrication-scale <Reynolds number> small, justifying neglect of inertia.
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