I have a problem where I have to evaluate the line integral by evaluating the surface integral in Stokes's theorem with a choice $S$, assuming $C$ has a counterclockwise orientation.

The $F= \left< 2y,-z,x \right> $ and $C=x^2+y^2=12$ on $z=0$.

I get the $\operatorname{curl}(F)$ as $\left<-1,1,2 \right>$.

I keep messing up at finding the normal or the boundaries. I first tired making

$s: z=12-x^2-y^2$ and making the normal $\left<-dz/dx,-dz/dy,1 \right>$ so it became $n=\left<2x,2y,1\right>$ and I thought the limits would be $-\sqrt{12}<x<\sqrt{12}$ and $-\sqrt{12-x^2}<y<\sqrt{12-x^2}$ but that just ended with $0$ as its answer.

I then tried parametrizing $S$ and ended with the unit tangent normal being $\left< \cos u,\sin u,0 \right>$ but had no idea what i would make the bounds for $V$, anything I could come up with would not be the correct answer.

The correct answer is $-24\pi$ according to the book an I'm at a total loss how to reach it. Any help would be appreciated.


Take your surface as the disk with radius $\sqrt{12}$ and apply Stokes. Then your normal vector is $<0,0,1>$ and the z-component of your curl is actually $-2$ (we only care about the z-component because we're going to dot with the normal). The dot product of these is $-2$. Integrating $-2$ over the disk is simply multiplying $-2$ times the area of that disk which is $12\pi$, so the result is $-24\pi$.


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