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Homework Help: Green's theorem for finding area.

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Use GT to find the area of one petal of the 8-leafed rose given by


    Recall that the area of a region D enclosed by a curve C can be found by
    [tex]A=1/2\int(xdy - ydx)[/tex]

    I calculated it using the parametrization

    [tex]x=rcos(\theta), y=rcos(\theta)[/tex]

    And I found a really long integral, evaluated it from 0 to pi/4, and got the correct answer.
    Here is my question: apparently, if x is defined as above, and I find

    [tex] dx = -rsin(\theta), dy = rcos(\theta)[/tex], then the integral

    [tex]A=1/2\int(xdy - ydx)[/tex] simplifies nicely to [tex]1/2\int(r^2)d\theta[/tex]. Evaluating this integral again from 0 to pi/4 gives the correct answer.

    So... why is it that I can pretend "r" is a constant when I'm evalutating dx and dy, when really, r is dependent on theta just as the x and y parametrizations are?
  2. jcsd
  3. Feb 19, 2010 #2
    Anyone, please?
  4. Feb 19, 2010 #3


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    Simple answer: You can't really. If you work it out properly, you'll see the terms proportional to dr cancel out. It's just a coincidence.
  5. Feb 19, 2010 #4
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