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Green's Theorem

  1. May 5, 2005 #1
    I'm having trouble on a line integral.

    Assuming that the closed curve C is taken in the counterclockwise sense. Use Green's Theorem.

    [tex] \int_C F\bullet dR [/tex]
    where F=([tex]x^2 + y^2[/tex])i + 3x[tex]y^2[/tex]j
    and C is the circle
    [tex]x^2 + y^2 = 9[/tex]

    This is what I have done so far....

    [tex]\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta[/tex]

    [tex]\int_0^{2\Pi} \int_0^3 \-r^3 drd\theta[/tex]

    [tex]\int_0^{2\Pi} \frac{-r^4}{4} \\]_0^3d\theta[/tex]

    [tex]\int_0^{2\Pi} \frac{-81}{4} d\theta[/tex]

    [tex]\frac{-81}{4} \theta\\]_0^{2\Pi}[/tex]

    [tex]\frac{-81\Pi}{2}[/tex]

    The book gives the answer as [tex]\frac{243\Pi}{4}[/tex]

    I have no idea where I went wrong.
     
  2. jcsd
  3. May 5, 2005 #2
    Theres another form of Green's theorem that might be more appropriate for this problem, try looking for it in your textbook.

    Also the integral of [itex] r^3 [/itex] is not [itex] \frac{-r^4}{4} [/tex]. No negative.

    You've also managed to neglect the vector field your integrating over.
     
  4. May 5, 2005 #3
    Can you show the work as to how you got

    [tex]\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta[/tex]

    ?
     
  5. May 5, 2005 #4

    quasar987

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    You've done something strange. Applying Green theorem should yield the equality

    [tex] \int_C F\bullet dR = \iint_D \left(\frac{\partial (x^2+y^2)}{\partial y} - \frac{\partial (3xy^2)}{\partial x} \right) dxdy [/tex]

    and that's not

    [tex]\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta[/tex]
     
  6. May 5, 2005 #5
    Sorry guys, I was looking at another problem and I took the derivative of the wrong force. I redid this problem and I got the correct answer. Thanks for the attempted help though.
     
  7. May 5, 2005 #6

    quasar987

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    Np.. happens to me all the time :uhh:
     
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