[Calculus 3] Can anyone help me with the following line integral? Thanks!

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[Calculus 3] Can anyone help me with the following line integral? Thanks!

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Use Stoke's Theorem to evaluate \[\int\limits_{C}^{ } F dr\] where F is given by \[F = \] and C is the circle \[x^2 + y^2 = 9\] on the plane z = 2 having traversed counterclockwise orientation when viewed from above.
I thought to convert it to \[\int\limits_{S}^{ }curl (F) dS\]but my friend thought to use \[\int\limits_{D}^{ } F'(r(t)) r'(t) dA\]I'm not sure what approach to use.
The question explicitly asks you to use stokes theorem right ?

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Right, but once I converted to curl(F)*dS, I wasn't sure what to do from there. Is it just a simple conversion to polar coordinates from that point? I still would have a vector though that I can't integrate.
Stokes theorem says that the closed loop line integral equals the flux through ANY surface that is bounded by the loop. so start by picking a simple surface
Stokes theorem : \[\large \oint\limits_{C} \vec{F}.d\vec{r}~~=~~\iint\limits_{S}\text{curl}(F)\cdot d\vec{S}\] where \(S\) is ANY surface bounded by \(C\)
So does it make sense that the region S is given by the enclosed region of the circle x^2 + y^2 = 9? In that case, I could quickly convert to polar coordinates, with r's bounds being from 0 to 3 and theta's bounds being from 0 to 2*pi.
Yes you may choose the surface as \(x^2+y^2\le 9\) in the plane \(z=2\) next, you need to find its normal vector
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The normal vector would always just be <0, 0, 1>, right? Since it lies on the plane of z=2?
thats right, it doesn't change over the region
Oh, so then a simple dot product between the normal vector and the curl of F will get be the scalar expression I can use to integrate, after I convert to polar coordinates?
Once you setup the flux integral, it is just a double integral, so you can evaluate it using any of the tricks that you know about double integrals/
It turns out that you don't need to use polar coordinates and all that to evaluate the double integral... work the curl first, you will see why..
What do I substitute in for z in the integrand? Just 2?
the surface is entirely in plane z=2, so we can replace z by 2
Which fits nicely in the integrand, giving me the integral:\[12\int\limits_{0}^{2\pi}\int\limits_{0}^{3}rdrd \theta\]which gets be the value 108*pi after solving. This is correct! Thanks for all your help, @ganeshie8.
np, i hope you have used the geometry to evaluate that double integral

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