Integration of polar equation

  • Thread starter celeramo
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  • #1
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Homework Statement


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 :)
 

Answers and Replies

  • #2
156
0

Homework Statement


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?
No; sketch the curve. Beginning at theta=0, it returns to itself at theta=pi. Hence your integral is going around the curve twice.
 
  • #3
HallsofIvy
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Therefore, Unco, your answer to his question "Is this integral incorrect", is "Yes"!
 
  • #4
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Thanks very much, my mistake
 

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