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Greens theorem homework help

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the [tex]\int[/tex] xyi - xj dot dr over the curve y = 1 - x2 from 1,0 to 0,1

    2. Relevant equations



    3. The attempt at a solution

    I used greens theorem

    [tex]\int[/tex] -1 - x dy dx

    dy is from 0 to 1-x2

    dx is from 0 to 1

    [tex]\int[/tex]-x3 + x2 - x - 1 dx = -1.4166666666666666666

    is this correct
     
  2. jcsd
  3. May 15, 2009 #2

    Dick

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    Re: Integrals

    How can you use Green's theorem? The curve isn't closed. Just do the line integral.
     
  4. May 15, 2009 #3
    Re: Integrals

    Well, I suppose he could use Green's theorem by completing the loop through the axes and subtracting their contribution which seems to be zero.

    Are you sure the sign of the x3 term is correct though?
     
  5. May 15, 2009 #4
    Re: Integrals

    "just do the line integral"

    [tex]\int[/tex]gradf dot dr = [tex]\int[/tex] xyi - xj dot dr

    Is this correct to say...I can say that xyi -xj is grad f as long as i can find the f, and then i would plug the two points into f to get the answer
     
  6. May 15, 2009 #5
    Re: Integrals

    No, this only applies to conservative vector fields, that is fields whose curl is 0. The vector field you have (xyi -xj) is not conservative and therefore the function you describe doesn't exist. Try finding it and you will get contradicting terms from different derivatives. You can't just plug the end points in because the integral depends on which path you take between those points.

    But you were going in the right direction in your first post, just remember that Green's Theorem applies to closed curves.
     
  7. May 15, 2009 #6
    Re: Integrals

    ok

    the -x3 should be positive and thus the answer is -11/12

    But how do i make it a closed curve and what do i subtract
     
  8. May 15, 2009 #7
    Re: Integrals

    You have a curve that takes you from (1,0) to (0,1), which means that you need to add another curve that takes you from (0,1) to (1,0), preferably one that would be easier to integrate over than the original. Then you'd have a closed loop and you could use Green's Theorem. But the answer you get won't correspond to your original integral so you'll have to subtract the portion that you added in the beginning. What would be the easiest way to close the curve in your example?

    Note that by calculating the area the way you did you already chose the curve that completes the loop.

    Of course you can just calculate the original integral by itself as Dick suggested. For that you would need to parametrize x and y using a single variable so that your integral becomes one-dimensional.
     
  9. May 15, 2009 #8
    Re: Integrals

    I think i should do it as a line integral over a paramaterized curve

    y = 1-x2

    r(t) = ti + (-t2 +1)j
    r'(t) = i + -2tj

    xyi - xj

    F(r(t) = (-t3 + t)i - tj
    F(r(t) dot r'(t) = -t3 + 2t2 + t

    -[tex]\int[/tex]-t3 + 2t2 + t = -11/12

    is this correct
     
  10. May 15, 2009 #9

    Dick

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    Re: Integrals

    Yes, that's the easy way to do it. But why did you change the sign?? Oh, I see, x goes from 1 to 0, not 0 to 1. That's why isn't it?
     
    Last edited: May 15, 2009
  11. May 15, 2009 #10
    Re: Integrals

    t goes from 1 to 0
     
  12. May 15, 2009 #11
    Re: Integrals

    is that just a cusidence that i got the same answer when i used greens theorem incorectly
     
  13. May 15, 2009 #12
    Re: Integrals

    Yes, it so happens that the contribution of the curve that you should have used to complete the loop is zero. Basically what you calculated using GT was an integral of xyi - xj first along y = 1 - x2 from (1,0) to (0,1), then from (0,1) to (0,0) along the y-axis, then from (0,0) to (1,0) along the x-axis thus coming to the point you started from. However, the latter two integrals are zero so the only contribution comes from the first one.
     
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