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Applying Green's THM, Polar Coords substitution

  1. Apr 9, 2012 #1
    Use Green's THM to calculate the line integral ∫C(F<dot> dx), where C is the circle (x-2)2 + (y - 3)2=1 oriented counterclockwise, and F(x,y)=(y+ln(x2+y2), 2tan-1(x/y)).



    Green's THM
    ∂SF<dot>dx=∫∫S(∂F2/∂x) - ∂F1/∂y)




    I tried doing it by brute force. I took the partials and put them under the integral. I also computed the bounds of integration and split it into the multiple integral. However, the bounds were pretty messy:
    x from 1 to 3, y from 3 to √(1-(x-2)2)+3

    When I evaluated the first integral with respect to y, I got an intractible function to take an integral over.

    I feel like I should be using polar coordinates, but I am not sure how to substitute them in this case.
     
  2. jcsd
  3. Apr 9, 2012 #2

    tiny-tim

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    welcome to pf!

    hi dbkats! welcome to pf! :smile:
    same way as usual …

    x2 + y2 = r2

    x = rcosθ, y = rsinθ, dxdy = rsinθdrdθ :wink:

    show us your integrand in x and y :smile:
     
  4. Apr 9, 2012 #3
    Re: welcome to pf!

    I feel that the conventional polar coordinates will not do well because the disk is not centered at the origin. So when we transform the disk region, we do not get a nice square. Or did I misunderstand how that works?

    My integral with respect to x and y:

    2∫13dx∫3√(1-(x-2)2)+3{(2/(1 + x/y)2)-1-(2y/(x2+y2))}dy
     
    Last edited: Apr 9, 2012
  5. Apr 9, 2012 #4

    tiny-tim

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    Re: welcome to pf!

    first, let's simplify …
    :wink:
     
  6. Apr 9, 2012 #5
    Re: welcome to pf!

    So, I actually messed up writing it out here. It should actually be:
    Integrand = (2/(1 + (x/y)2))-1-(2y/(x2+y2))

    I feel very stupid for not seeing that it drastically simplifies when I work out the common denominator.

    Integrand = (y2 - 2y - x2)/(x2 + y2)

    Barring any other stupid arithmetic mistakes. Okay, so I have this very pretty integrand, but the bounds of integration are still really weird. Should I try to transition to polar?

    EDIT: As an aside, is there a way to enter my fractions here so they are human-readable?
     
  7. Apr 9, 2012 #6

    Dick

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    Re: welcome to pf!

    It will simplify even more if you get it right. You forgot to use the chain rule on arctan(x/y). It will simplify so much your coordinate problems will be over. And you could check out using LaTex https://www.physicsforums.com/showthread.php?t=8997
     
    Last edited: Apr 9, 2012
  8. Apr 9, 2012 #7
    Re: welcome to pf!

    >.< I feel silly. The integrand evaluates to -1, so I can now use whatever parametrization I feel like. I think that (x-2)=cosθ, (y-3)=sinθ would be a good bet.

    I guess once I saw the disk, all I could think about was "how am I going to parametrize this ugly thing"...

    Anyway, thanks so much for the help :smile:
     
  9. Apr 9, 2012 #8

    Dick

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    Re: welcome to pf!

    You're welcome. You don't need coordinates at all. To integrate a constant you just need to know the area of the disk.
     
  10. Apr 9, 2012 #9
    Re: welcome to pf!

    Or that, ya hahaha....
     
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