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Integration of polar equation

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the area enclosed by r=2cos(3[tex]\theta[/tex])

    I'm fairly confident how to do this but for some reason I am getting 2[tex]\pi[/tex] rather than 1[tex]\pi[/tex], which the book claims is the answer. There is the possibility the book is wrong, but I want to make sure how to do this.

    I have the area=(1/2)[tex]\int[/tex][tex]^{2\pi}_{0}[/tex] (2cos(3[tex]\theta[/tex]))^2 d[tex]\theta[/tex]

    is this integral incorrect for the enclosed area?

    Please and thank you very much :)
     
  2. jcsd
  3. Feb 22, 2009 #2
    No; sketch the curve. Beginning at theta=0, it returns to itself at theta=pi. Hence your integral is going around the curve twice.
     
  4. Feb 22, 2009 #3

    HallsofIvy

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    Therefore, Unco, your answer to his question "Is this integral incorrect", is "Yes"!
     
  5. Feb 22, 2009 #4
    Thanks very much, my mistake
     
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