= Solution
This is a squeezing motion, not the rotational motion of the previous part. The gap changes at rate $h_t=V\sin\theta$. The <Reynolds lubrication equation> therefore gives
$$
\frac1a q_\theta=-h_t,\qquad q=aV\cos\theta+C.
$$
The outer wall also has tangential speed $V\cos\theta$, but its local <Couette flow> flux is $O(V\Delta)$ compared with the squeezing flux $O(Va)$ and does not affect the leading <force>. Thus $p_\theta=-12\mu a q/h^3$ at leading order. Periodic <pressure> gives $C=0$, because the <integral> of $\cos\theta/h^3$ vanishes over a full circle. Balancing <pressure> traction on the fixed inner cylinder and using <integration by parts>,
$$
F_z=a\int_0^{2\pi}p\sin\theta\,d\theta
=a\int_0^{2\pi}p_\theta\cos\theta\,d\theta
=-\frac{12\mu a^3V}{\Delta^3}\int_0^{2\pi}\frac{\cos^2\theta}{(1+\alpha\sin\theta)^3}\,d\theta.
$$
The supplied <integral> gives the <squeeze resistance of an eccentric journal bearing>:
$$
\boxed{F_z=-\frac{12\pi\mu a^3V}{\Delta^3(1-\alpha^2)^{3/2}}.}
$$
The fluid tends to pull the inner cylinder upwards with the moving outer wall, so a downward holding <force> is required when $V>0$. At concentricity the translation resistance is $R_0=12\pi\mu a^3/\Delta^3$. The omitted wall-tangential flux and shear traction give smaller thin-gap contributions.
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