I am a little stuck on the following problem:
Use Stokes's Theorem to show that
$$\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2,$$
where $C$ is the suitably oriented intersection of the surfaces $x^2 + y^2 + z^2 = a^2$ and $x + y + z = 0$.
OK, so Stokes's Theorem tells me that:
$$\oint_{C}\vec{F} \cdot d\vec{r} = \iint_{S}\operatorname{curl} \vec{F} \cdot \vec{N} ~dS$$
I have calculated:
$$\operatorname{curl} \vec{F} = -\vec{i} - \vec{j} - \vec{k}.$$
I then figured that on the surface $S$ we must have:
$$\vec{N}dS = \vec{i} + \vec{j} + \vec{k}dxdy$$
since this follows from the equation of the given plane.
However this will then give me:
$$\operatorname{curl} \vec{F} \cdot \vec{N} = -1 -1 -1 = -3$$
And thus I would get, if project this onto the $xy$-plane:
$$\iint_{S} \operatorname{curl} \vec{F} \cdot \vec{N} ~dS = -3 \iint_{A} dA = -3 \pi a^2$$
which is obviously not correct.
I would greatly appreciate it if someone could help me with this. I actually had multivariable calculus a few years ago, and I know that I knew this stuff then. However, now that I need it again I notice that I've become quite rusty.
Thanks in advance :)