2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $-24\pi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.