= Solution
In the lubrication regime the thin region has longitudinal scale $\ell\sim\sqrt{ab}$, since the height increase $x^2/(2a)$ becomes comparable with $b$ there. The characteristic shear is $\mu V/b$, so force per axial length scales as $\mu V\ell/b\sim\mu V\sqrt{a/b}$. Negligible inertia, no slip, no cylinder rotation, and full wetting are understood.
Use a frame translating with the cylinder. The wall moves with <velocity> $-V$, the cylinder is stationary, and the gap is $h(x)=b+x^2/(2a)$. Lubrication momentum gives $\mu u_{yy}=p_x$, $p_y=0$, so
$$
u_{\rm frame}=-V(1-y/h)+\frac{p_x}{2\mu}y(y-h),\qquad q=-\frac{Vh}{2}-\frac{h^3p_x}{12\mu}.
$$
Continuity makes $q$ constant. Equal ambient <pressures> at both infinities give $\int p_xdx=0$, hence $q=-(V/2)(\int h^{-2}dx)/(\int h^{-3}dx)=-2Vb/3$. Therefore
$$
\boxed{p_x=\frac{2\mu V(4b-3h)}{h^3},\qquad p=\frac{2\mu Vx}{h^2},\qquad u_{\rm lab}=\frac{Vy}{h}+\frac{p_x}{2\mu}y(y-h).}
$$
The <vertical velocity>, with the wall condition $w(0)=0$, follows explicitly from incompressibility:
$$
w=\frac{Vh'y^2}{2h^2}-\frac{p_{xx}}{2\mu}\left(\frac{y^3}{3}-\frac{hy^2}{2}\right)+\frac{p_xh'y^2}{4\mu}.
$$
It gives $w(h)=0$, consistently with a horizontally translating cylinder in the lab frame.
The wall shear is $\mu u_y(0)=4\mu V(1/h-b/h^2)$. With $x=\sqrt{2ab}\,s$, its integral is
$$
F=4\mu V\sqrt{2ab}\left(\frac\pi b-\frac\pi{2b}\right),\qquad\boxed{F\sim2\pi\mu V\sqrt{\frac{2a}{b}}\quad(b/a\to0).}
$$
<Force balance> equates this wall force to the required cylinder force at leading order; the <pressure> and <viscous stresses> on remote boundaries make no leading contribution. This is the <cylinder translating parallel to a wall in a thin gap> result. The asymptotic sign denotes the approximation implicit in the parabolic gap, rather than an exact finite-gap cylinder formula.
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