Consider axisymmetric motion in two dimensions, so that the flow is confined to a plane. Use cylindrical coordinates \( (r, \phi, z) \), where \( z \) is the direction perpendicular to the plane, with velocity components \( (u^r, u^\phi, u^z) \). For axisymmetric flow \( u^z = u^r = 0 \) but \( u^\phi \neq 0 \).
\(\because\) \( \boldsymbol{\omega} = \nabla \times \mathbf{v} \), \( \nabla \times \mathbf{v} = \begin{vmatrix} \hat{\mathbf{r}} & \hat{\boldsymbol{\phi}} & \hat{\mathbf{z}} \\ \frac{\partial}{\partial r} & \frac{1}{r} \frac{\partial}{\partial \phi} & \frac{\partial}{\partial z} \\ u^r & u^\phi & u^z \end{vmatrix} \)
\(\because\) \( u^r = 0, u^z = 0, u^\phi = \Omega r \) \( \therefore \mathbf{v} = (0, \Omega r, 0) \)
\(\because\) \( \omega^r = 0, \omega^\phi = 0 \), \( \omega^z = \frac{1}{r} \frac{\partial}{\partial r} (r u^\phi) \)
\(\because\) \(\frac{1}{r} \frac{\partial}{\partial r} (r u^\phi) = \frac{1}{r} \left( u^\phi + r \frac{\partial u^\phi}{\partial r} \right)\)
\(\because\) \( u^\phi = \Omega r \), \( \therefore r u^\phi = \Omega r^2 \)
\(\therefore\) \( \frac{\partial}{\partial r} (\Omega r^2) = 2 \Omega r \), \(\frac{1}{r} \frac{\partial}{\partial r} (r^2 \Omega) = \frac{1}{r} (2 r \Omega) = 2\Omega\)
\(\therefore \omega^z = \frac{1}{r} (2 \Omega r) = 2 \Omega \)
1Vladimir Zorich (2016). Mathematical Analysis II.