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Help with line integrals and greens theorem

  1. Dec 9, 2007 #1
    I get an answer for this problem, but its 0 and i think thats wrong. if someone could plz, help that'd be great.

    1. The problem statement, all variables and given/known data
    Find the work using the Line Integral Method:
    W = Integral of ( Vector F * dr)

    Vector Field: F(x,y) = (xy^2)i + (3yx^2)j

    C: semi circular region bounded by x axis and y = squareroot(4-x^2) where y = squareroot(4-x^2) is greater than 0.

    2. Relevant equations

    So where vector is <P,Q>,
    Work = Integral of (P dx + Q dy) over the region R.

    3. The attempt at a solution

    So I first parameterized the curve to get P,Q,dx,dy in terms of a common variable:
    x = 2cost, y = 2sint for 0 <= t <= pi which implies dx = -2sint and dy = 2cost

    but when I carry out the integration, the limits of integration end up making my answer go to 0 because they are between 0 and pi and I always end up with a sin(t) in the result of the integral.
  2. jcsd
  3. Dec 9, 2007 #2
    The entire benefit of using Green's Theorem is that you don't need to parameterize.

    Note that what you have written there is equivalent to

    [tex] \displaystyle \oint \vec{F} \cdot d\vec{x} = \iint_D \left( \frac{F_2}{dx} - \frac{F_1}{dy} \right) dx dy [/tex]

    Thus you can very easily figure out [itex] \frac{F_2}{dx}, \;\; \frac{F_1}{dy} [/itex]

    Furthermore, [itex] -2<x<2 [/itex] and [itex] 0<y<\sqrt{4-x^2} [/itex]
  4. Dec 9, 2007 #3
    ya, sorry, i dont know how to write in those fancy symbols. Anyway was my answer of 0 correcT?
  5. Dec 10, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Why ask for help if you ignore the advice? Do the integration over the half disk and see if you also get 0 there.
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