Q:

$C$ be the curve of intersection of sphere $x^2+y^2+z^2=a^2$ and plane $x+y+z=0$ ; to evaluate $\int_C ydx + z dy +x dz$ by Stoke's theorem?

Let $C$ be the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and the plane $x+y+z=0$ ; how to evaluate $\int_C ydx + z dy +x dz$ by Stoke's theorem ? 
$C$  is a great circle I think ; I am  not able to get the surface $S$ ; Please help . Thanks in advance 

A:

You are right, the curve $C$ is a great circle. This is rather obvious when you note that the center of the sphere is the origin and the plane passes through this point. For Stokes' theorem you need a surface $S$ such that $\partial S = C$. There are many choices but I would propose to the part of the plane that is "cut-out" by the sphere (which is a disk). 
We then have that
$$\int_C (y dx +z dy + x dz) =  -\int_S (dx\wedge dy + dy \wedge dz + dz \wedge dx) $$
by Stokes' theorem. The plane is parameterized by $u,v$ via 
$$ (x,y,z) = (u,v,-u-v).$$
We obtain
$$ \int_C (y dx +z dy + x dz) = - \int_S [du\wedge dv + dv \wedge(-du-dv) + (-du -dv)\wedge du ] = - 3 \int_S du \wedge dv .$$
The remaining integral $\int_S du \wedge dv$ is the projection of the disk $S$ onto the xy-plane. The disk has radius $a$ and is slanted at an angle of $ \phi=\arccos(1/\sqrt{3})$ with respect to the xy-plane. Elementary geometry yields $\int_S du \wedge dv = \pi a^2 /\sqrt{3}$ and thus
$$ \int_C (y dx +z dy + x dz) = - \sqrt{3} \pi a^2$$

