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Green's Theorem

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Use Green's Theorum to evaluate the line integral c (x^2)y dx, where c is the unit circle centered at the origin.

    2. Relevant equations


    3. The attempt at a solution
    Taking the partial derivative with respect to y and subtracting it from zero(I'm taking the dy in the original problem to be zero because there wasn't one), I set up the double integral:

    ∫y = -1 to y =1 ∫x = -sqrt(1 - (y^2)) to x = sqrt(1 - ( y^2)) -(x^2) dxdy

    I'm just kind of confused because there is no dy in the problem, and I'm not sure why the answer was -π/4.
    I might not have set it up right because when I put the x bounds into -(x^3)/3, I'm not sure how to then integrate with respect to y.
     
  2. jcsd
  3. Dec 10, 2014 #2

    jedishrfu

    Staff: Mentor

    Since the path is a circle about the origin have you thought of using polar coordinates?
     
  4. Dec 10, 2014 #3

    Zondrina

    User Avatar
    Homework Helper

    You should first apply Green's theorem properly.

    $$\oint_C \vec F \cdot d \vec r = \oint_C x^2y \space dx + 0 \space dy = \iint_D Q_x - P_y \space dA$$

    Where ##\vec F = P \hat i + Q \hat j## and ##d \vec r = \vec r'(t) \space dt = x'(t) \hat i + y'(t) \hat j##.

    So what are ##Q_x## and ##P_y##?

    Now the curve ##C## is the unit circle ##x^2 + y^2 = 1## in a counter clockwise orientation. How do you parametrize a counter-clockwise circle around the origin? This should give you limits for ##r## and ##\theta##.
     
  5. Dec 10, 2014 #4
    Thank you...this helped!
     
  6. Dec 10, 2014 #5
    Thank you...that worked!
     
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