Verifying Stokes' Theorem on two intersecting cylinders 
Let $S\subset \mathbb{R}^3$ be the portion of the cylinder $y^2 + z^2 = 4$ with $z>0$ and $x^2 + y^2 ≤ 1$. Let $\mathbf{f}(x,y,z) = (zx-y)\mathbf{i}$, where $\mathbf{i}$ is the usual unit vector in the $x$-direction.
Verify that Stokes' Theorem is satisfied in this case.

So this question is in two parts: I first have to evaluate $\iint_S \nabla \times \mathbf{f} \cdot \mathrm{d}\mathbf{S}$, then $\oint_{\partial S}\mathbf{f} \cdot \mathrm{d} \mathbf{r}$, and finally show that they are equal.
My Attempt
I first calculated $\nabla \times \mathbf{f}$ to be $\begin{vmatrix}\
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\
(zx-y) & 0 & 0\\
\end{vmatrix} = (0,x,1).$
I then parameterised the cylinder by $\mathbf{r}(\theta,z) = (\cos\theta,\sin\theta,z)$ , with my normal $\frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial z} = (\cos\theta,\sin\theta,0)$ which is outward pointing as required. And my integral becomes $$\begin{align} \int_{\theta = 0} ^{2\pi}\int_{z=0}^{\sqrt{4-\sin^2\theta}}(0,\cos\theta,1)\cdot(\cos\theta,\sin\theta,0)\mathrm{d}z\mathrm{d}\theta & = \int_{\theta = 0} ^{2\pi}\int_{z=0}^{\sqrt{4-\sin^2\theta}}\sin\theta\cos\theta\mathrm{d}z\mathrm{d}\theta\\
& = \frac{1}{2}\int_{\theta = 0} ^{2\pi}\sin2\theta\sqrt{4-\sin^2\theta}\mathrm{d}\theta\end{align}$$ at which point I am stuck.
I think perhaps my parameterisation is wrong, or maybe I have to calculate the surface integrals in two steps, but the latter method didn't seem to work either and I can't find any better parameterisations. 
I also am having trouble on how to parameterise the line integral I need to compute. Should there be two boundaries which I need to combine to find the whole boundary $\partial S$?
 A: Observe that due to the periodicity of the integrand we have 
$$\int_0^{2\pi} \sin \theta \cos \theta \sqrt{4-\sin^2 \theta} \, d\theta=\int_{-\pi}^{\pi} \sin \theta \cos \theta \sqrt{4-\sin^2 \theta} \, d\theta$$
Further note that the integrand is an odd function of $\theta$ and therefore the integral is identically zero.

The integral of interest should be 
$$\int_0^{2\pi} \int_0^1 \nabla \times \vec f(\rho, \theta,z(\rho,\theta)) \cdot\hat n \frac{\rho d\rho d\theta}{(\hat n \cdot \hat z)}$$
The curl of $\vec f$ is $\nabla \times \vec f=\hat yx +\hat z$ while the unit normal $\hat n$ to the surface of the cylinder is $\hat n=\frac12(\hat yy+\hat zz)$ since $\sqrt{y^2+z^2}=2$.  Finally, note that 
$$dS=\frac{dxdy}{(\hat z \cdot \hat n)}=\frac{dxdy}{\frac12 z}$$
Thus, we have 
$$\begin{align}
\hat n \cdot \nabla \times \vec f dS&= \frac12(xy+z) \frac{dxdy}{\frac12 z}\\\\
&=\left(\frac{xy}{\sqrt{4-y^2}}+1\right)dxdy\\\\
&=\left(\frac{\rho^2 \sin \theta \cos \theta}{\sqrt{1-\rho^2 \sin^2 \theta}}+1\right)\rho d\rho d\theta
\end{align}$$
Thus we have 
$$\int_0^{2\pi} \int_0^1 \nabla \times \vec f(\rho, \theta,z(\rho,\theta)) \cdot\hat n \frac{\rho d\rho d\theta}{(\hat n  \cdot \hat z)}=\int_0^{2\pi} \int_0^1 \left(\frac{\rho^2 \sin \theta \cos \theta}{\sqrt{1-\rho^2 \sin^2 \theta}}+1\right) \rho d\rho d\theta$$
The first term integrates to zero since it has period $2\pi$ in $\theta$ and is an odd function of $\theta$.  The second term is trivial to integrate and yields $\pi$.
Let's verify this by carrying out the line integral.  The parameterization is $x=\cos \theta$, $y =\sin \theta$, and $z=\sqrt{4-\sin^2 \theta}$ with $\theta$ going from $-\pi$ to $\pi$.  On the contour of interest, $d\vec l=\hat \theta d\theta$.  Thus, 
$$\begin{align}
\int_C \vec f \cdot d\vec l &=\int_{-\pi}^{\pi} \left(zx-y\right)\hat x\cdot \hat \theta d\theta\\\\
&=\int_{-\pi}^{\pi} \left(\sqrt{4-\sin^2 \theta}\cos \theta-\sin \theta\right)(-\sin \theta) d\theta\\\\
&=\int_{-\pi}^{\pi} \left(-\sin \theta \cos \theta \sqrt{4-\sin^2 \theta}+\sin^2 \theta\right) d\theta\\\\
&=\pi
\end{align}$$
since the first integral is an odd in $\theta$ and the second integral trivially integrates to $\pi$.
