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Green's Theorem

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data

    Use, using the result that for a simple closed curve C in the plane the area enclosed is:

    A = (1/2)∫(x dy - y dx) to find the area inside the curve x^(2/3) + y^(2/3) = 4

    2. Relevant equations
    Green's Theorem:
    ∫P dx + Q dy = ∫∫ dQ/dx - dP/dy

    3. The attempt at a solution
    I solved the equation of the curve for x:

    x = (4 - y^(2/3))^(3/2)

    Also, from the original curve equation x^(2/3) + y^(2/3) = 4, when x = 0, y = +/- 8 because 4^(3/2) = 8.

    But when I plug x = +/- (4 - y^(2/3))^(3/2) in for the x bounds and y = +/- 8 in for the y bounds in the resulting double integral

    (1/2)∫∫ 2 dxdy

    I have trouble integrating x = (4 - y^(2/3))^(3/2) it with respect to y.

    Does anybody happen to know if there is a more correct way to solve this problem?

    Thank you for your help!
     
  2. jcsd
  3. Feb 10, 2015 #2

    LCKurtz

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    The question does not want you to work the integral out by doing a double integral. It wants you to find a nice parameterization of the curve and do$$
    A = \frac 1 2 \int_C x~dy - y~dx$$ So your first job is to find a nice parameterization. As a hint think about a way to parameterize it so that you can use the identity ##(r\cos\theta)^2 + (r\sin\theta)^2 = r^2##.
     
  4. Feb 12, 2015 #3
    Thank you for your help!
     
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