Circulation for the streamfunction $\psi= -C \ln(r/a)$ using Stokes theorem.

Question

Show, by using Stokes' theorem, that the circulation for $\psi= -C \ln(r/a)$ is the same for any simple curve once round the origin. What is the result if the curve does not enclose the origin, or goes twice round it?

This is a question from Paterson fluid dynamics. According to the solutions the answer should be $2n\pi C$, with $n$ the number of times round is.

I don't know how to solve this problem. I calculated the velocity field $ {\bf v} = (0, C/r),$ and hence the rotor of ${\bf v}$ is zero and thus the circulation is zero by Stokes Theorem. But this is a wrong according to the solution. Can someone help me? Thanks in advance.

Answer 1

\begin{align*} f(z) &= \phi(x,y)+i\psi(x,y) \\ \frac{C}{i}\ln z &= C\arg (x+yi)-iC\ln r \\ f'(z) &= \frac{\partial \phi}{\partial x}+i\frac{\partial \psi}{\partial x} \\ &= \frac{\partial \phi}{\partial x}-i\frac{\partial \phi}{\partial y} \\ &= u-vi \\ \Gamma &= \oint_{\gamma} \mathbf{v} \cdot d\mathbf{r} \\ &= \oint_{\gamma} (u\, dx+v\, dy) \\ &= \operatorname{Re} \left[ \oint_{\gamma} (u-vi)(dx+i\,dy) \right] \\ &= \operatorname{Re} \left[ \oint_{\gamma} f'(z) \, dz \right] \\ &= \operatorname{Re} \left[ \oint_{\gamma} \frac{C}{iz} \, dz \right] \\ &= 2\pi C \end{align*}