Stokes' Theorem - cylindrical coordinates I'm currently having an issue with verifying the validity of Stokes' Theorem on a particular problem. I can solve the problem by using Stokes' theorem to turn a surface integral of the curl of a vector into a line integral of a vector. However, when I try to check the validity of this by explicitly computing the surface integral of the curl, I am apparently faced with integrating a zero vector. We have the following where we use the cylindrical coordinate system $(\rho,\varphi,z)$ and $R,\Gamma$ are constants:
\begin{align*} 
\boldsymbol{u} &= u_\varphi(\rho)\hat{\boldsymbol{\varphi}} 
\end{align*}
\begin{align*}
&u_\varphi(\rho) = \begin{cases}
\frac{\Gamma \rho}{2 \pi R^2} &\rho \leq R \\
\frac{\Gamma}{2 \pi \rho} & \rho > R \\
\end{cases}
\end{align*}
\begin{align}
\boldsymbol{\omega} &= \nabla \times \boldsymbol{u} = (0,0,\omega(\rho))
\end{align}
\begin{align*} 
&\omega(\rho) = \begin{cases}
\frac{\Gamma}{2 \pi \rho} &\rho \leq R \\
0 & \rho > R \\
\end{cases}
\end{align*}
We are given
\begin{align}
\Omega(\rho) = \int_{D(0,\rho)} \boldsymbol{\omega} \cdot \ dS
\end{align}
where $D(0,\rho)$ is a disc of radius $\rho$ centered at the origin. We are asked to show that:
\begin{align*} 
&\Omega(\rho) = \begin{cases}
\Gamma(\frac{\rho}{R})^2 &\rho \leq R \\
\Gamma & \rho > R \\
\end{cases}
\end{align*}
where $D(0,\rho)$ is a disc of radius $\rho$ centered at the origin.
While I can find $\Omega$ for $\rho \leq R$, I'm having an issue directly computing it when $\rho > R$. I have used Stokes' theorem to change the integral to $\int_{0}^{2 \pi} \boldsymbol{u} \cdot d\boldsymbol{x}$ which then gives me:
\begin{align*}
\Omega(\rho) &= \int_{0}^{2 \pi} (0, \frac{\Gamma }{2 \pi \rho},0) \cdot d \boldsymbol{x} \\
 &= \int_{0}^{2 \pi} \frac{\Gamma }{2 \pi \rho} \rho d \varphi \\
&= \frac{\Gamma }{2 \pi } [ \varphi ]_{0}^{2 \pi} = \Gamma
\end{align*}
However, when I try to compute this directly using:
\begin{align*}
\Omega(\rho) = \int_{D(0,\rho)} \boldsymbol{\omega} \cdot \ dS = \int_{D(0,\rho)} (\nabla \times\boldsymbol{u}) \cdot \ dS
\end{align*}
I get $(\nabla \times \boldsymbol{u})=(0,0,0)$ which would give $\Omega(\rho)= 0 \quad \forall \ \rho > R$
We are also provided with a formula for computing the curl of a vector field expressed in cylindral polar coordiantes $(\rho, \varphi, z)$:
\begin{align*} 
 \nabla \times \boldsymbol{A} = \frac{1}{\rho} \begin{vmatrix}
\hat{\boldsymbol{\rho}} & \rho \hat{\boldsymbol{\varphi}} & \hat{\boldsymbol{z}} \\
\partial_{\rho} & \partial_{\varphi} & \partial_{z} \\
A_{\rho} & \rho A_{\varphi} & A_{z}
\end{vmatrix}
\end{align*}
Any help on how I can get the two methods to agree or observations on where I have gone wrong would be much appreciated, thank you!
 A: $u_\varphi = \frac{\Gamma \rho}{2 \pi R^2}, \rho \leq R $
$u_\varphi = \frac{\Gamma}{2 \pi \rho},  \rho > R$
There is a mistake in your calculation of curl. It is easy to see that $\hat \rho$ and $\hat \varphi$ components of the curl are zero and you have that right. $\hat z$ component is
$\displaystyle \omega_z = \frac{1}{\rho} \bigg(\frac{\partial (\rho u_{\varphi})}{\partial \rho} - \frac{\partial u_{\rho}}{\partial \varphi}\bigg) = \frac{1}{\rho} \frac{\partial (\rho u_{\varphi})}{\partial \rho}$
For $\rho \leq R, \ \displaystyle \omega_z = \frac{1}{\rho} \frac{\partial}{\partial \rho} \bigg(\frac{\Gamma \rho^2}{2 \pi R^2}\bigg) = \frac{\Gamma}{\pi R^2}$
For $\rho \gt R, \ \omega_z = 0 \ $ as you mentioned.
So for disk with radius $\rho \ (\leq R),$ centered at the origin, surface integral is
$\displaystyle \int_0^{2\pi} \int_0^{\rho} \frac{\Gamma}{\pi R^2} \ \rho \ d\rho \ d\varphi = \frac{\Gamma \rho^2}{R^2}$.
For $\rho \gt R$, it is
$\displaystyle \int_0^{2\pi} \int_0^{R} \frac{\Gamma}{\pi R^2} \ \rho \ d\rho \ d\varphi + \int_0^{2\pi} \int_{R}^{\rho} (0) \ \rho \ d\rho \ d\varphi = \Gamma + 0 = \Gamma$
