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Independence of path

  1. Apr 25, 2008 #1
    Why is the given line integral independent of path in the entire xy-plane?

    Int((y2 + 2xy)dx + (x2 + 2xy)dy)[/i]
     
  2. jcsd
  3. Apr 25, 2008 #2

    Dick

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    Because it's a total derivative?? Because regarded as a line integral of a vector field it's curl is zero?? Because it's the contour integral of an analytic function?? Take your pick, depending on the course you are in. They are all the same. Show one of them.
     
  4. Apr 25, 2008 #3
    If you know Greens or Stokes theorem, you should check them out for this -- that is perhaps the quickest way to see the answer by just looking at the integral.
    If not, then have you talked about conservative fields? You should check out what the necessary conditions are for a conservative field and apply it here.
     
  5. Apr 26, 2008 #4

    HallsofIvy

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    Stoke's theorem clearly does not apply here. Green's theorem does but is far more general than needed here. As Dick said, your answer will depend upon the course, but the simplest thing to do is show that (y2 + 2xy)y= 2y+ 2x= (x2 + 2xy)x and so this is, as Dick said, a "total derivative" (I would say an "exact differential"): there exist a function F(x,y) having this as its differential. The integral along any curve is just the difference of F evaluated at the two endpoints.
     
    Last edited: Apr 26, 2008
  6. Apr 26, 2008 #5
    Stokes would deffinitely work. You just set your vector field equal to F=<y^2 + 2xy, x2 + 2xy,0> and then he has the integral F.dr. You use stokes theorem and get the same answer as greens would give you. But maybe I am missing something because I've only recently started studying Stokes.
     
  7. Apr 26, 2008 #6

    HallsofIvy

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    Stoke's theorem refers, in general, to an integration over a curved surface in R3. Green's theorem refers to an integeration over an area of R2. If you take your surface to be the xy-plane, then Stoke's theorem reduces to Green's theorem.
     
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