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First supply the introductory scaling and geometry. The leading tangential Stokes equation in a thin gap balances $P/L$ against $\mu U/H^2$. The shear stress scale is $\tau \sim \mu U/H$, so lubrication pressure dominates shear stress:The outer circle, viewed along a ray $r(\cos \theta ,\sin \theta )$ from the inner centre, satisfies $r^2-2\alpha \Delta r\sin \theta +{\alpha }^2{\Delta }^2=(a+\Delta {)}^2$. Expanding its positive root to first order in $\Delta /a$ gives $r=a+\Delta +\alpha \Delta \sin \theta +O({\Delta }^2/a)$, hence $h=\Delta (1+\alpha \sin \theta )$.

Take positive $\Omega$ to mean increasing $\theta$, and put $s=a\theta$, $y=r-a$, $U=\Omega a$. With no-slip boundary conditions at $y=0,h$, lubrication theory givesSteady mass conservation makes $q$ independent of $\theta$. Single-valued pressure requires ${\int }_0^{2\pi }{p}_{\theta }d\theta =0$. Write $I_n={\int }_0^{2\pi }(1+\alpha \sin \theta {)}^{-n}d\theta$ and $D_{\alpha }=1+{\alpha }^2/2$. It follows thatThe tangential fluid tractions on the inner and outer solids, measured in the same increasing-$\theta$ direction, are respectivelyThis sign convention distinguishes the traction on a wall from the corresponding stress component in the fluid.

To leading order the holding force balances the pressure traction on the inner circle. Integration by parts givesUse $p_{\theta }=a{p}_s$ and $\int \sin \theta /(1+\alpha \sin \theta {)}^{n}d\theta =(I_{n-1}-I_n)/\alpha$, with the continuous limit at $\alpha =0$. Substituting the boxed flux yieldsThe zero vertical component also follows from reflection in mathematics in the vertical line: this leaves the geometry and vertical force unchanged, reverses $\Omega$, and Linearity of Stokes flow then requires $F_z=0$.

Pressure gives no moment about either circular wall's own centre. Integrating the viscous shear torque with lever arm $a$ to leading order gives the fluid couplesIndeed $C_i=a^2\int {\tau }_{i}d\theta$ and $C_o=a^2\int {\tau }_{o}d\theta$ at this order. Applied holding or driving couples are the negatives of these fluid couples. They need not cancel because they are moments about different axes. The fluid force on the outer solid equals the holding force $F_x$ on the inner one. Transferring its moment to the inner axis adds $-\alpha \Delta F_x$, so angular momentum balance about a common axis isFor $\alpha =0$ the usual equal and opposite concentric couples are recovered. All these formulas are leading thin-gap expressions at fixed $|\alpha |<1$; taking the contact limit requires maintaining the local slender-gap assumptions.

Rotation about the vertical axis gives the inner surface the azimuthal speed $\Omega a\sin \theta$. The leading Couette flow shear traction is opposing and has magnitudeIts moment arm about the vertical axis is $a\sin \theta$, and $dS=2\pi a^2\sin \theta d\theta$. ConsequentlyPutting $t=\lambda \cos \theta$ and using the supplied integral givesThis viscous shear torque opposes the rotation. Its concentric limit is $-8\pi \mu \Omega a^4/(3\Delta )$, agreeing with the thin-gap limit of Torque in rotational Stokes flow between concentric spheres.