Acheson (1990). The fluid velocity is given (in our units) by $$\underline{u} = -\frac{U}{r}\frac{\partial \psi}{\partial \theta} \underline{e}_r + U \frac{\partial \psi}{\partial r} \underline{e}_{\theta}$$ On the cylinder the rate-of-strain tensor has components $e_{rr} = 0 = e_{\theta \theta}$, and $$2e_{r\theta} = U \frac{\partial^2 \psi}{\partial r^2}$$ The stress tensor $T_{ij} = -p \delta_{ij} + 2\mu e_{ij}$, so the force on the cylinder boundary is $$\underline{t} = T.\underline{e}_r =  -p \underline{e}_r  + \mu U \frac{\partial^2 \psi}{\partial r^2} \underline{e}_{\theta}$$ We project this into the $x$-direction and integrate over the circle to get the following formula for the drag $D_0$ (force per unit length on the cylinder): $$D_0 = - \int_0^{2\pi}\left[p \cos\theta + \mu U \frac{\partial^2 \psi}{\partial r^2} \sin\theta \right]d\theta$$ For low $R_e$ we estimate the pressure using the momentum equation in the form $$\del p = -\mu \del \wedge (\del \wedge \uvec )$$ and a short calculation gives $$\del p = \mu U \left\{-\frac{1}{r} \frac{\partial\ }{\partial \theta}(\delsq \psi)\ervec + \frac{\partial\ }{\partial r}(\delsq \psi)\ethvec \right\}$$ We need to integrate this in from infinity to a general point on the circle $r=1$. In general this is awkward without a global form of $\psi$. But if $\psi$ has the form $\psi = g(r) \sin\theta$ then we have 