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Homework Help: Line integral around an ellipse

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data
    What is [tex]\int_{\gamma} xy dx + x^2 dy[/tex] in each of the following cases?

    1. [tex]\gamma[/tex] is the lower half of the curve [tex]2x^2 + 3y^2 = 8[/tex], traveled from [tex](2,0)[/tex] to [tex](-2,0)[/tex].

    2. [tex]\gamma[/tex] is the full curve [tex]2x^2 + 3y^2 = 8[/tex], traveled counterclockwise.


    2. Relevant equations

    The line integral formula, I suppose. The fact that the integral can be expressed as the dot product of the vector field [tex](xy, x^2)[/tex] with the unit tangent vector to the curve can also be helpful.



    3. The attempt at a solution

    I parametrized the curves for (1) and (2) in different ways, viz.

    1. [tex]x = t, y = -2\sqrt{\frac{2}{3}\left(1 - \frac{t^2}{4}\right)}[/tex].

    2. [tex]x = 2\cos{\theta}, y = 2\sqrt{2/3}\sin{\theta}[/tex].

    Then standard integration rules, but I came up with 0 for both the answers. Am I correct?
     
  2. jcsd
  3. Jan 9, 2010 #2
    For no.1, the xy dx part is equal to 0, by symmetry: for every point (x,y) [which contributes xy dx] on the curve, there is another point (-x,y) [which contributes -xy dx] on the opposite side of the y axis. dx is always positive on this curve, so they cancel each other out.

    We can also see that the x^2 dy part is equal to 0: because for the first half of the curve, dy is negative, for the second half, dy is positive, while x^2 is positive and mirrored. so the first half cancels out the second half again.


    So 0 seems right.

    2nd question follows easily from the first by splitting it up into 2 integrals: first around the bottom half and then around the top half of the ellipse. We already know from the first question that the bottom half = 0, and by symmetry the top half must be 0 too. 0+0=0.


    Symmetry arguments are good for checking your work.
     
  4. Jan 9, 2010 #3

    I checked your answers and they are correct. Here is a graphical analysis of the problem in both cases:
    http://img30.imageshack.us/img30/8343/intego.jpg [Broken]

    AB
     
    Last edited by a moderator: May 4, 2017
  5. Jan 9, 2010 #4
    A brilliant technical argument, but less mathematical. :wink:
     
  6. Jan 9, 2010 #5

    vela

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    [STRIKE]Note that [tex]d\Phi(x,y)=xydx+x^2dy[/tex] is an exact differential, so the integrals depend only on the endpoints. (I'll leave it to you to find [tex]\Phi(x,y)[/tex].) The second integral is then trivially zero because the start and end points are the same. The first integral is equal to [tex]\Phi(-2,0)-\Phi(2,0)=0[/tex].[/STRIKE]
     
    Last edited: Jan 9, 2010
  7. Jan 9, 2010 #6
    many thanks, guys! i can't find [tex]\Phi(x)[/tex] though, i keep just missing it by a scalar factor. it would seem that [tex]\Phi = x^2y[/tex] would work, but it just misses it. poincare's lemma doesn't seem to help either. am i missing something obvious? :(
     
  8. Jan 9, 2010 #7

    vela

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    Why does [tex]\Phi(x,y)=x^2y[/tex] not work? You could add on a constant, obviously, but that wouldn't make a difference in evaluating the integral.
     
  9. Jan 9, 2010 #8
    oh, i thought because [tex]d(x^2y) = 2xy dx + x^2 dy[/tex]. how do we get rid of the scalar multiple of 2? it isn't an additive constant, right?
     
  10. Jan 9, 2010 #9

    vela

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    Oh, I'm sorry. I misled you. For some reason I thought there was a two there in the original problem. You're right. It's not exact, so you have to do the integral by hand. (And they were both zero when I did them earlier.)
     
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