Is My Integral Correct for Finding the Enclosed Area of a Polar Equation?

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    Integration Polar
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Homework Help Overview

The discussion centers around finding the area enclosed by the polar equation r=2cos(3θ). The original poster expresses uncertainty about their integral setup, which yields a different area than what is stated in the reference material.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to validate their integral for the area calculation, questioning whether their approach is correct given the discrepancy with the book's answer. Another participant suggests sketching the curve to understand the behavior of the polar function and its implications for the integral.

Discussion Status

The discussion is active, with participants exploring the original poster's integral and its implications. There is a recognition of the need to consider the curve's behavior, as one participant indicates that the integral may be accounting for multiple loops of the curve.

Contextual Notes

Participants are considering the possibility of the reference material being incorrect, as well as the implications of the integral's limits and the nature of the polar function in question.

celeramo
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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 :)
 
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celeramo said:

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.
 
Therefore, Unco, your answer to his question "Is this integral incorrect", is "Yes"!
 
Thanks very much, my mistake
 

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