Line integral (not using Stokes theorem) Evaluate 
$$\int_C Fdr$$
$$F=<-y^2,x,z^2>$$
$C$ is the curve of intersection of the plane $y+z=2$ and the cylinder $x^2+y^2=1$
I can parametrize the curve using cylindrical coordinates but I don't know exactly how I would do that when $x$ is missing from the plane equation. Any ideas?
 A: Parametrize it useing $x=\cos\theta$, $y=\sin\theta$ and $z=2-\sin\theta$.
A: To answer your question about the equation for the plane $z+y=2$ ... $x$ spans from $-\infty$ to $+\infty$. 
Parametrically, let $u$ and $v$ span the plane.  Then, $x=u$, $y=v$, and $z=2-v$ is a parametric description that fits here.  Another is $x=u$, $y=2-v$, and $z=v$. Can you think of others?
Think about a plane that lies in the $x-y$ plane.  Here, $z=0$ defines the plane, while $x$ and $y$ span it.  So, $x=u$, $y=v$, and $z=0$ is a valid parameterization.
Now, on to the line integral itself.

Use of cylindrical coordinates would be a natural choice.  But I will go brute force.  This will provide a general outline for tackling similar problems for which there is no well-know curvilinear coordinate system to which you can appeal.

To that end, let $t$ be a parameter.
Then, there are two "pieces."  
On "Piece 1," $x=\sqrt{1-t^2}$, $y=t$, and $z=2-t$, and $t$ starts at $-1$ and ends at $1$.
On "Piece 2," $x=-\sqrt{1-t^2}$, $y=t$, and $z=2-t$, and $t$ starts at $1$ and ends at $-1$. 
Remember, that the differential line segment $d\vec r$ is given by
$$d\vec r =\hat x \frac{dx}{dt}+\hat y \frac{dy}{dt}+\hat z \frac{dz}{dt} dt$$
where
$$\frac{dx}{dt}=\mp \frac{t}{\sqrt{1-t^2}}$$ with the upper sign on "Piece 1," and the lower sign on "Piece 2." 
$$\frac{dy}{dt}=1$$
$$\frac{dz}{dt}=-1$$
It won't be pretty, but it is tenable.

For the corresponding surface integral, let $u$ and $v$ be parameters.
Then, $\vec r=\hat x u+\hat yv+\hat z (2-v)$ with $-1\le u\le 1$ and $-\sqrt{1-u^2}\le v\le \sqrt{1-u^2}$ defining the surface.
The unit vector normal to the surface is 
$$\hat n= \frac{\frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}}{|\frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}|}=\frac{\hat x \times (\hat y-\hat z)}{|\hat x \times (\hat y-\hat z)|}=\frac{\hat y+\hat z}{\sqrt{2}}$$
Again, a more natural setting for all of this would be a cylindrical coordinate system.  But, this does give you a way to proceed in cases for which there is no suitable well-known curvilinear coordinate system to use.
