Verifying Stokes theorem (i.e. prove both integrals are equal) I have that $\hat{F}=[-y,x,0]$ and I have a surface defined by $S: rz=1$ where $1\leq z \leq 2$.
I interpret this surface as a truncated cone with curved edges.
So I am trying to show that, indeed $\oint_{C} \mathbf{F} \cdot d \mathbf{r}=\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d \sigma$.
I have found that $\oint_{C} \mathbf{F} \cdot d \mathbf{r} = 3 \pi$, by parameterising the two boundary curves (the circles at top and bottom of S).
However, when calculating the surface integral I end up with $\frac{3}{ 2}\pi$.
Can anyone do this surface integral? Thank you!
 A: First, note that in cylindrical coordinates $\vec F(\vec r)$ is given by $\displaystyle \vec F(\vec r)=\hat \phi r$ and $\nabla \times \vec F(\vec r)=\hat z 2$.
Let the surface $S$ be the set defined by $S=\{(r,\phi,z)|rz=1, z\in [1,2], \phi\in[0,2\pi] \}$.
On $S$ at $z=1$, $r=1$.  On $S$ at $z=2$, $r=1/2$.  Hence, the line integral in the boundary of $S$ is
$$\begin{align}
\oint_{C_{z=1}}\vec F(\vec r)\cdot \,d\vec \ell+\oint_{C_{z=2}}\vec F(\vec r)\cdot \,d\vec \ell&=\int_0^{2\pi} \left(\hat \phi 1\right)\cdot \left(\hat \phi 1\right)\,d\phi+\int_0^{2\pi} \left(\hat \phi \frac12\right)\cdot \left(-\hat \phi \frac12\right)\,d\phi\\\\
&=2\pi -2\pi \left(\frac14\right)\\\\
&=3\pi/2\tag1
\end{align}$$
where the minus sign on $\hat \phi$ in the second integral on the right-hand side of $(1)$ is a consequence of the right-handed orientation of $S$.
Next, we calculate the surface integral of Stokes's Theorem using two methodologies.

METHODOLOGY $(1)$:  DIRECT COMPUTATION
We can parameterize the surface with $r$ and $\phi$ with
$$\vec r=\hat r(\phi) r +\hat z \frac1r $$
The surface differential vector is given
$$\begin{align}
\hat n d S&=\left(\frac{\partial \vec r}{\partial r}\times\frac{\partial \vec r}{\partial \phi}\right)\,dr\,d\phi\\\\
&=\left(\hat r\frac1r+\hat z r\right)\,dr\,d\phi
\end{align}$$
Hence, we find that
$$\begin{align}
\int_S \color{blue}{\nabla \times \vec F(\vec r)}\cdot \,\color{red}{\hat n \,dS}&=\int_0^{2\pi }\int_{1/2}^1 \color{blue}{\left(\hat z 2\right)}\cdot \color{red}{\left(\hat r\frac1r+\hat z r\right)\,dr\,d\phi}\\\\
&=3\pi/2
\end{align}$$
as was to be shown!


METHODOLOGY $(2)$:  USE OF THE DIVEREGENCE THEOREM
Let's close the surface $S$ with the surfaces $S_u=\{(r,\phi,z)|z=2, r\in[0,1/2], \phi\in[0,2\pi]\}$ and $S_l=\{(r,\phi,z)|z=1, r\in[0,1], \phi\in[0,2\pi]\}$.
Using $\nabla\cdot \nabla \times \vec F(\vec r)=0$ along with the Divergence Theorem, we have
$$\oint_{S+S_u+S_l}\nabla \times \vec F(\vec r)\cdot \hat n\,dS=0$$
Therefore we see that
$$\begin{align}
\oint_{S}\nabla \times \vec F(\vec r)\cdot \hat n\,dS&=-\oint_{S_u+S_l}\nabla \times \vec F(\vec r)\cdot \hat n\,dS\\\\
&=-\int_0^{2\pi}\int_0^{1/2} (\hat z 2)\cdot (\hat z)\,r\,dr\,d\phi-\int_0^{2\pi}\int_0^{1} (\hat z 2)\cdot (-\hat z)\,r\,dr\,d\phi\\\\
&=3\pi/2
\end{align}$$
as expected!
