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Line Integral

  1. Apr 13, 2005 #1
    Starting from any one of the three corners of the path described here under,evaluate the line integral [tex]\int[/tex][tex]c[/tex] (2xy[tex]^3[/tex])dx + (4x[tex]^2[/tex]y[tex]^2[/tex])dy where C is the closed path forming the boundary of region in the first quadrant enclosed by x-axis,the line x=1 and the curve y=x[tex]^3[/tex]
  2. jcsd
  3. Apr 13, 2005 #2


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    Well, what have you got so far?
  4. Apr 13, 2005 #3


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    Any time you have a nice closed path with the line integral in that form, try and remember to use Green's Theorem:

    [tex]\oint_C Mdx+Ndy=\iint_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dA[/tex]

    Just plug it in right?
  5. Apr 13, 2005 #4


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    Not so fast saltydog - I dont think he is allowed to use Green or Stokes for now.

    just parametrize the curves, find derivative of that curve and then you can find individual line integrals of int(F dot dr) for all segments
  6. Apr 13, 2005 #5


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    It's easier to compute the double integral than the line integral...So i vote for Green as well.

  7. Apr 13, 2005 #6
    I might be embarrassing myself here but isn't there some sort of requirement that the function be analytic before you use Green's Theorem? And not all polynomials are analytic.
  8. Apr 13, 2005 #7


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    i don't think green has much hypotheses? maybe smoothness? try proving it and see what you need.

    it follows from fubini (repeated integration) plus ftc.

    of course all polynomials are analytic, but i am asuming you meant that you thought the form had to be closed, which is not necessary.
    Last edited: Apr 13, 2005
  9. Apr 13, 2005 #8


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    That domain should be described by x running from 0 to 1 and y from 0 to x^{3}...

  10. Apr 14, 2005 #9


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    All the polynomials I know are analytic!

    (And you don't need the function to be analytic for Green's theorem- "continuously differentiable" is enough.)

    The only question is whether kidia has had "Green's theorem" in Calculus class yet.

    It's not that hard to actually integrate along the path:
    1) from (0,0) to (1, 0) along the x- axis: take x= t, y= 0 so that dx= dt, dy= 0. Since y= 0 along that path, the integral on it is 0.

    2) from (1,0) to (1,1) along the line x= 1:take x= 1, y= t so that dx= 0, dy= dt. integrate
    4t2dt from 0 to 1.

    3) from (1,1) to (0,0) along the line y= x: take x= t, y= t so that dx= dt, dy= dt. integrate from t= 1 down to t= 0.
    Last edited: Apr 14, 2005
  11. Apr 14, 2005 #10
    I knew I was setting myself up for embarresment.
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