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Problem using Green's Theorum

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    F = [−y^3, x^3], C the circle x^2 + y^2 = 25

    Book gives answer as Pi*1875*1/2, I get Pi*1875

    3. The attempt at a solution

    [tex]\int\int(3x^2 + 3y^2)dxdy[/tex]

    [tex]\int\int(75(cos^2\vartheta + sin^2\vartheta))rdrd\vartheta[/tex]

    [tex]75\int[1/2 r^2]^{5}_{0}d\vartheta[/tex]



    [tex]=1875\pi[/tex] Where did I go wrong?
  2. jcsd
  3. Nov 3, 2009 #2


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    Homework Helper

    When you have

    \int\int(3x^2 + 3y^2)dxdy

    and then convert to polar coordinates x=rcos(theta) and y=rsin(theta)

    r is not 5. r is a variable you are integrating.
  4. Nov 3, 2009 #3
    Of course!!! Thank you for the quick reply.
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