Intersection curve between a circle and a plane - Stokes theorem What is the intersection curve between the circle
$$x^2+y^2=1$$
and the plane
$$x+y+z=0$$
If i am not wrong, I should solve the equation system
\begin{align}
x^2+y^2-1=0 \\
x+y+z=0
\end{align}
But I don't get the right curve. If i solve x in the first equation, then i get the system\begin{align}
x=- \sqrt{1-y^2} \\
x+y+z=0
\end{align}
and
\begin{align}
x= \sqrt{1-y^2} \\
x+y+z=0
\end{align}.
And this gives me
$$x=- \sqrt{1-y^2} => z= -y + \sqrt{1-y^2}$$
$$x=\sqrt{1-y^2} => z= -y - \sqrt{1-y^2}$$But it's wrong. Where am i doing wrong?
EDIT: If I have to calculate the work of a field are doing along the curve p, do I need to calculate the curve or should I use the plane equation to get the normal vector??
That's what this guy here
Stokes's Theorem on a Curve of Intersection did. But I just wan't to know if I am always going to do it so  I don't make any mistakes. And i would appreciate if you could link any good sites to learn more about this, thanks!
If i do parametrize the plane equation as
\begin{align}
x=s \\
y=t \\
z=-s-t \\
\end{align}
then i will get the normal vector
$$r_s x  r_t =(1,1,1)$$
and with this I can use stokes theorem with x^2+y^2<=1, right?
 A: From $x+y+z=0$ we have $x=-(y+z)$, so $$x^2+y^2=(y+z)^2+y^2=1$$You can parametrize this equation by setting $y=\cos\theta$ and $y+z=\sin\theta$, i.e. $z=\sin\theta-\cos\theta$. Then our intersection is $$\boxed{\langle-\sin\theta,\cos\theta,\sin\theta-\cos\theta\rangle,\,\theta\in[0,2\pi)}$$
You could also have parametrized $x=\cos\theta$ and $y=\sin\theta$ (so that $x^2+y^2=1$) and therefore $z=-(\cos\theta+\sin\theta)$, giving another parametrization $$\boxed{\langle\cos\theta,\sin\theta,-\cos\theta-\sin\theta\rangle,\,\theta\in[0,2\pi)}$$
The problem with your solution $z=-y\pm\sqrt{1-y^2}$ is that it makes no reference to $x$, and hence determines a surface. But the intersection of the infinite cylinder $x^2+y^2=1$ and the plane $x+y+z=0$ is a curve. Your surface does contain the curve, and you could turn it into a parametrization by $$\boxed{\langle\underbrace{\pm\sqrt{1-y^2}}_{=x},y,\underbrace{-y\mp\sqrt{1-y^2}}_{=z}\rangle,\,y\in[-1,1]}$$
EDIT
A good website for learning vector calculus is MathInsight.
I'm not entirely sure what you're asking in regard to Stokes's theorem. If you wanted to calculate the line integral of a force field around the (closed) intersection curve I computed above, you wouldn't necessarily have to compute the intersection curve; you could instead use Stokes's theorem, which would involve calculating the curl of your force field and then calculating the flux of that curl through any surface having the intersection curve as a boundary.
But if your question is whether you're "always going to do it," the answer is: it depends on which integral is easier to do. If the line integral is easier, do the line integral. But if the curve is hard to calculate, consider doing the surface integral of the curl, instead.
Here is a picture of the plane $x+y+z=0$ (in green), the intersection of the plane with the cylinder (in dark blue), and the easiest surface to use for Stokes's theorem (the checkerboard part of the plane contained by the blue boundary curve):

A: I think the intersection just two points.
$$
(\sqrt{2}/2, -\sqrt{2}/2, 0)
$$ 
and
$$
(-\sqrt{2}/2, \sqrt{2}/2, 0).
$$
