<s>[INST] How to parametrise intersection of two cylinders I have a question that asks me to verify Stokes Theorem on 
$S:=$portion of the cylinder $ y^2+z^2=4$ with $ z>0$ and $x^2+y^2\leq1$
How would I parametrise the curve and surface so that I can apply Stokes Theorem. I have plotted this surface and can imagine what it looks like but I am finding it difficult to parametrise it. Any help will be appreciated. I can complete the question once I have found a suitable parametrisation.
The parametrisation I came up with is $r(\theta,x)=(x,2cos\theta,2sin\theta)$ with $0<\theta<\pi$, $-\sqrt{1-(2cos\theta)^2}\leq x \leq\sqrt{1-(2cos\theta)^2}$ If this is incorrect what is wrong with this?
Or would a better parametrisation be the usual cylinderical coordinates with $0\leq\theta\leq2\pi$ and $0<z<\sqrt{4-sin^2\theta}$ If either of these are correct, why is the other one incorrect?
 [/INST] Based on the fact that $z = \sqrt{4-y^2}$ :
$$
\sigma(r,\theta) = (x,y,z) = (r\cos\theta, r\sin\theta, \sqrt{4-(r\sin\theta)^2}) \\
r \in [0,1],\ \theta \in [0,2\pi[
$$
According to WA, (And my understanding of the surface), this seems right.

The cross product is painful, I agree...
$$
\sigma_r = (\cos\theta, \sin\theta, -\frac{r\sin^2\theta}{\sqrt{4-r^2\sin^2\theta}}) \\
\sigma_\theta = (-r\sin\theta, r\cos\theta, -\frac{r^2\sin\theta\cos\theta}{\sqrt{4-r^2\sin^2\theta}}) \\
\sigma_r \times \sigma_\theta = (0,\frac{r^2(\cos^2\theta\sin\theta  + \sin^3\theta)}{\sqrt{4-r^2\sin^2\theta}},r)
$$
Regarding your parametrization it doesn't work for $\theta < \pi/3, \theta > 2\pi/3$, the square roots in the bounds give complex numbers but I see what you meant to do. The problem is that the bound for $x$ doesn't define a circle of radius $1$ as it should be. Here you are mixing the cylinder of radius $2$ (which gives you the $y=2\cos\theta$) and the equations $x = \pm \sqrt{1-y^2}$ and to make this equation work $y$ needs to be between $-1$ and $1$ but by its nature its range is $[-2,2]$.

Checking Stokes theorem with this surface:
$$
A= \iint_S rot \vec{F} \cdot d\vec{n} = \iint_S (0,x,1) \cdot d\vec{n} =
\iint_S\left( r\cos\theta \cdot \frac{r^2(\cos^2\theta\sin\theta  + \sin^3\theta)}{\sqrt{4-r^2\sin^2\theta}} + r \right)dS = \\
\underbrace{\int_0^1 dr \int_0^{2\pi} d\theta \frac{r^3(\cos^3\theta\sin\theta  + \sin^3\theta\cos\theta)}{\sqrt{4-r^2\sin^2\theta}}}_{=I_1} + \int_0^1 dr \int_0^{2\pi} d\theta\ r = 0 + \pi
$$
Then
$$
\partial S(\theta) = (\cos\theta,\sin\theta, \sqrt{4-\sin^2\theta}), \ \partial S'(\theta) = (-\sin\theta,\cos\theta, -\frac{\sin\theta \cos\theta}{\sqrt{4-\sin^2\theta}}),\ \theta\in [0,2\pi[ \\
B = \oint_{\partial S} \vec{F}\cdot d\vec{l} = \oint_{\partial S} (zx-y,0,0) \cdot d \vec{l} = \int_0^{2\pi}  (\cos\theta \sqrt{4-\sin^2\theta} -\sin\theta) (-\sin\theta)\; d\theta = \pi
$$
Therefore $A = B$. However I agree that the integrals are not be that simple but clearly doable which may indicate that the parametrization I chose was maybe not the best, but at least it worked !

$$
I_1 = \int_0^1 dr\ r^3 \int_0^{2\pi}d\theta \cos\theta \frac{(1-\sin^2\theta)\sin\theta + \sin^3\theta}{\sqrt{4-r^2\sin^2\theta}} \overset{w = \sin\theta, dw = \cos\theta d\theta}{=} \int_0^1 dr\ r^3\int_0^0 dw \frac{w}{\sqrt{4-r^2 w^2}} = 0
$$
</s>