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## AakashSudhakar one year ago [Calculus 3] Can anyone help me with the following line integral? Thanks!

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1. AakashSudhakar

Use Stoke's Theorem to evaluate $\int\limits_{C}^{ } F dr$ where F is given by $F = <e ^{-2x}-3yz, e ^{-4y}+3xz, e ^{-2z}>$ and C is the circle $x^2 + y^2 = 9$ on the plane z = 2 having traversed counterclockwise orientation when viewed from above.

2. AakashSudhakar

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.

3. ganeshie8

The question explicitly asks you to use stokes theorem right ?

4. AakashSudhakar

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.

5. ganeshie8

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

6. ganeshie8

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 $$C$$

7. AakashSudhakar

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.

8. ganeshie8

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

9. ganeshie8

|dw:1438227332463:dw|

10. AakashSudhakar

The normal vector would always just be <0, 0, 1>, right? Since it lies on the plane of z=2?

11. ganeshie8

thats right, it doesn't change over the region

12. AakashSudhakar

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?

13. ganeshie8

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/

14. ganeshie8

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..

15. AakashSudhakar

What do I substitute in for z in the integrand? Just 2?

16. ganeshie8

the surface is entirely in plane z=2, so we can replace z by 2

17. AakashSudhakar

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.

18. ganeshie8

np, i hope you have used the geometry to evaluate that double integral

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