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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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    The question explicitly asks you to use stokes theorem right ?

  4. AakashSudhakar
    • one year ago
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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.

  5. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    |dw:1438227332463:dw|

  10. AakashSudhakar
    • one year ago
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    The normal vector would always just be <0, 0, 1>, right? Since it lies on the plane of z=2?

  11. ganeshie8
    • one year ago
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    thats right, it doesn't change over the region

  12. AakashSudhakar
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    What do I substitute in for z in the integrand? Just 2?

  16. ganeshie8
    • one year ago
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    the surface is entirely in plane z=2, so we can replace z by 2

  17. AakashSudhakar
    • one year ago
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    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
    • one year ago
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    np, i hope you have used the geometry to evaluate that double integral

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