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Green's Theorem and Line Integral

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\oint[/tex]xydx+x^2dy

    C is the rectangle with vertices (0,0),(0,1),(3,0), and (3,1)

    Evaluate the integral by two methods: (a) directly and (b) using green's theorem.



    2. Relevant equations





    3. The attempt at a solution
    Evaluating the integral directly:

    c1: y=0,x=t,dx=dt,dy=o {0[tex]\leq[/tex]t[tex]\leq[/tex]3}
    c2: x=3, y=t, dx=0, dy=dt {0[tex]\leq[/tex]t[tex]\leq[/tex]1}
    c3: y=1, dy=0, x=t, dx=dt {0[tex]\leq[/tex]t[tex]\leq[/tex]3}
    c4: x=0, dx=0, y=t, dy=dt {0[tex]\leq[/tex]t[tex]\leq[/tex]1}

    So I got c1 and c4 being the integral of zero which is just zero.

    Then for c2 and c3...

    [tex]\int[/tex]9dt {t:0[tex]\leq[/tex]t[tex]\leq[/tex]1} +[tex]\int[/tex]tdt {t:0[tex]\leq[/tex]t[tex]\leq[/tex]3}=27/2


    Then trying to use green's theorem:
    [tex]\int[/tex][tex]\int[/tex]xdydx=[tex]\int[/tex]xdx=9/2

    {y:0[tex]\leq[/tex]y[tex]\leq[/tex]1}
    {x:0[tex]\leq[/tex]x[tex]\leq[/tex]3}

    I am not sure where I messed up but I know I did because both of my answers should be the same. If someone could point me in the right direction, it would be much appreciated. Thank you for your time.
     
  2. jcsd
  3. Aug 2, 2010 #2
    Sorry if it looks a little confusing. I didn't intend on my function being where my top limit of integration would be.
     
  4. Aug 2, 2010 #3

    vela

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    You need to take into account the direction of the contour. You want to enclose the area going in the counter-clockwise direction.
     
  5. Aug 2, 2010 #4
    I don't really understand the positive orientation part of green's theorem. How do you even know what direction it is oriented?
     
  6. Aug 2, 2010 #5

    vela

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    If you have a region enclosed by a contour, the convention is that the positive orientation is when the contour encloses the region in the counter-clockwise direction. It's similar to the convention we use that a positive angle is measured going in the counter-clockwise direction from the x-axis. There's nothing inherently positive about it. It's just that everyone agrees that that's what they mean by "positive."

    In this problem, if you start at the origin, the positive sense of the contour would be going from (0,0) to (3,0) to (3,1) to (0,1) and finally back to the origin. So when you calculate C3, you have to go from (3,1) to (0,1), otherwise your answer for that leg comes out with the opposite sign and you don't get the cancellation you need.
     
  7. Aug 2, 2010 #6
    Ok. That makes perfect sense now. Thank you so much.
     
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