So, the aim of this question is to show that Stokes' Theorem holds in this context, but I'd like some help understanding how this relates to cylindrical polar coordinates (something I struggle with still!)
$$ \text{Consider the flow} \; \textbf{v} = (v_r, v_{\theta},v_z) = (0, 2r^2cos^2\theta, 0) \; \text{and the closed path} \; C \;\text{given by the circle of radius}\; a \;\text{in the}\; z=0 \; \text{plane, centred at the origin and traversed anti-clockwise.}\\ \text{a)}\;\text{Evaluate the circulation}\; \Gamma = \oint_C \textbf{v}\cdot d\textbf{l}\\ \text{b)}\;\text{Evaluate the flux of}\; \boldsymbol\omega \;\text{through the surface}\; S \;\text{of constant}\; z\; \text{that is bounded by}\; C \\ \text{c)}\;\text{Compare the results obtained in parts (a) and (b). Is this what you expected?} $$
So, this is what I got (trying to save space by not including all steps):
$$\begin{align} \Gamma &= \oint_C \textbf{v}\cdot d\textbf{l}\\ &= \oint_C(0, 2r^2cos^2\theta, 0)\cdot(dr, rd\theta,dz)\\ &= \oint_C 2r^3cos^2\theta \;d\theta\\ &= \int_{0}^{2\pi} 2r^3cos^2\theta \;d\theta\\ &= 2\pi r^3\\ \\ \iint_S \boldsymbol\omega \cdot \boldsymbol{\hat{n}}\; dS &= \int_{0}^{2 \pi} \int_{0}^{a} 6r^2cos^2\theta \; drd\theta\\ &=\int_{0}^{2 \pi} 2a^3cos^2\theta \;d\theta\\ &=2 \pi a^3 \end{align}$$
So, by Stokes' Theorem $\Gamma = 2 \pi r^3 = 2 \pi a^3$. My question is then, is it valid for the first integral to write in the last step $2 \pi r^3 = 2 \pi a^3$? I.e. in cylindrical polar coordinates does it make sense to substitute the radius of the circle we are working with for r? As this would make this comparison more explicit for part c).