= Solution
In this continuation $\lambda$ measures eccentricity, not the viscosity ratio used in Question 2. Take $\epsilon\to0$ with $0\leq\lambda<1$ fixed. For the <nearly occluding sphere in a cylindrical tube>, the narrow gap has thickness $H\sim\epsilon a$ and axial length $\ell\sim a\sqrt\epsilon$, obtained by balancing $h_0$ and $x^2/(2a)$. In the sphere frame the tangential velocity scale is $U$. The <lubrication approximation> then gives the \b[gap scales]
$$
\boxed{|p_x|\sim\frac{\mu U}{a^2\epsilon^2},\qquad
|p-p_{\mathrm{far}}|\sim\frac{\mu U}{a\epsilon^{3/2}},\qquad
|\sigma_{xy}|\sim\frac{\mu U}{a\epsilon}.}
$$
The large local pressure excursion is distinct from the eventual net pressure jump. Away from the sphere the velocity varies across a radius $a$, so the far <Hagen-Poiseuille flow> has
$$
\boxed{|p_x|\sim\frac{\mu U}{a^2},\qquad|\sigma_{rx}|\sim\frac{\mu U}{a}.}
$$
These scales also show why the gap pressure can be much larger than its shear stress while the leading net jump can be smaller than that pressure.
Use $y=a-r$, so $y=0$ is the moving tube wall and $y=h$ is the locally stationary sphere. The <no-slip boundary conditions> are $u(0)=-U$, $u(h)=0$. Integrating $\mu u_{yy}=p_x$, with pressure independent of $y$ to leading order, gives the <Couette-Poiseuille flow in a thin gap>
$$
u(y)=\frac{p_x}{2\mu}y(y-h)+U\left(\frac yh-1\right),\qquad
q=\int_0^hu\,dy=-\frac{h^3}{12\mu}p_x-\frac{Uh}{2}.
$$
Differentiate at both surfaces and eliminate $p_x$ using this <moving-boundary lubrication flux>:
$$
\boxed{\frac{\sigma_{xy}(0)}\mu=\frac{4U}h+\frac{6q}{h^2},\qquad
\frac{\sigma_{xy}(h)}\mu=-\frac{2U}h-\frac{6q}{h^2}.}
$$
The coordinate reversal means $\sigma_{rx}=-\sigma_{xy}$ at the tube wall.
At leading order $q$ is independent of $x$ in each angular strip, though it may depend on $\theta$. Both boundaries are stationary as shapes in the sphere frame and impermeable, so the integrated <continuity equation> is $q_x+a^{-1}(q_\theta^{\mathrm{circ}})_\theta=0$. The pressure changes along $x$ over $a\sqrt\epsilon$ but along the circumference over $a$. Thus circumferential velocity is only $O(U\sqrt\epsilon)$, and its flux is $O(Ua\epsilon^{3/2})$. Circumferential flux divergence is smaller than the possible leading axial divergence by a factor $\epsilon$. Consequently $q_x=0$ to the order needed. This is not an assumption of equal flux at different angular positions.
For fixed $\theta$ define the <lubrication resistance integrals for a nearly occluding sphere>
$$
J_n(\theta)=\int_{-\infty}^{\infty}\frac{dx}{h(\theta,x)^n}
=\sqrt{2a}\,h_0^{1/2-n}I_n.
$$
In particular, $J_1=\pi\sqrt{2a/h_0}$, $J_2=J_1/(2h_0)$ and $J_3=3J_1/(8h_0^2)$. The local pressure relation gives
$$
\int p_xdx=-12\mu qJ_3-6\mu UJ_2.
$$
A putative generic value $q=O(Uh_0)$ makes this jump $O(\mu U/(a\epsilon^{3/2}))$. However, the integrated wall shear in part (ii) is only $O(\mu Ua/\sqrt\epsilon)$, so the <force-free> global balance permits a common net pressure jump only of order $\mu U/(a\sqrt\epsilon)$. The outer flow has circumferentially uniform pressures on both sides. Therefore the $O(\epsilon^{-3/2})$ pressure jump must cancel in every angular strip, not merely after angular averaging. The <pressure recovery condition in lubrication flow> determines
$$
-12\mu qJ_3-6\mu UJ_2=0,\qquad
\boxed{q=-\frac U2\frac{J_2}{J_3}=-\frac23Uh_0(\theta).}
$$
This explains how the leading-order part of the global force balance selects the local flux.
The total sphere-frame <volume flux> is now
$$
Q_{\mathrm{frame}}=a\int_0^{2\pi}q(\theta)d\theta
=-\frac43\pi\epsilon a^2U.
$$
Transforming back to the laboratory adds $\pi a^2U$, so $Q_{\mathrm{lab}}=\pi a^2U(1-4\epsilon/3)$ to this order. For physical pressure, <Hagen-Poiseuille flow> gives $Q_{\mathrm{lab}}=-\pi a^4p_x/(8\mu)$. Thus
$$
\boxed{-p_x=\frac{8\mu U}{a^2}\left(1-\frac43\epsilon\right)+o\left(\frac{\epsilon\mu U}{a^2}\right).}
$$
The printed positive expression is the pressure-drop gradient in the direction of the laboratory flow; the Poiseuille hint omits the minus sign when $p_x$ is read as the derivative of physical pressure. With the specified wall velocity $-U$, the laboratory flow is positive and its physical pressure gradient is negative.
Next insert the leading flux into the wall <shear stress>:
$$
\sigma_{xy}(0)=4\mu U\left(\frac1h-\frac{h_0}{h^2}\right),\qquad
\int\sigma_{xy}(0)dx=4\mu U(J_1-h_0J_2)=2\mu UJ_1.
$$
Let $\Delta p>0$ be the localized extra pressure drop, after subtracting the regular far Poiseuille gradient. Contributions outside the neck are only $O(\mu U/a)$ and are smaller than the $O(\epsilon^{-1/2})$ jump. Since $\sigma_{rx}=-\sigma_{xy}$, part (ii) gives $\pi a^2\Delta p=a\int\!\int\sigma_{xy}(0)dxd\theta$. Hence the \b[leading pressure drop] is
$$
\boxed{\Delta p\sim\sqrt{\frac2\epsilon}\frac{2\mu U}a
\int_0^{2\pi}\frac{d\theta}{\sqrt{1+\lambda\cos\theta}}.}
$$
For example, the concentric limit is $\Delta p\sim4\pi\mu U\sqrt{2/\epsilon}/a$. The integral grows as the minimum gap shrinks; the fixed-$\lambda<1$ qualification prevents treating a further near-contact limit as uniformly covered.
On the sphere the leading <shear stress> is $\sigma_{xy}(h)=-2\mu U/h+4\mu Uh_0/h^2$, so
$$
\boxed{\int_{-\infty}^{\infty}\sigma_{xy}(h)dx
=-2\mu UJ_1+4\mu Uh_0J_2=0.}
$$
Positive and negative tangential stresses cancel within each angular strip. Thus the leading tangential resultant, and its leading torque contribution, vanish: negligible rotation is consistent with the <torque-free> condition. The shear is not zero pointwise, and the pressure forces are not zero. Pressure on a sphere cannot produce a torque about its centre because its normal is radial; its force contributions and the matched far pressure jump remain essential to the <force-free transport of a nearly occluding sphere>.
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