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Polar graph area

  1. Feb 1, 2013 #1
    Question is attached:


    [tex] r^2 = a^2 + 6acos(\theta) + 9cos^2(\theta) [/tex]

    using [tex] \frac{1}{2}\displaystyle\int^{2\pi}_0 r^2d\theta [/tex]

    using this I get a = 7

    are my limits right, as it says theta can't be 2pi?

    Attached Files:

  2. jcsd
  3. Feb 1, 2013 #2


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    a=7 looks ok. It doesn't really matter if they write the limit as <2pi or <=2pi. Including or excluding a single point doesn't change the integral.
    Last edited: Feb 1, 2013
  4. Feb 1, 2013 #3
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