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

[celeramo]

Homework Statement

Find the area enclosed by r=2cos(3\theta)

I'm fairly confident how to do this but for some reason I am getting 2\pi rather than 1\pi, 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)\int^{2\pi}_{0} (2cos(3\theta))^2 d\theta

is this integral incorrect for the enclosed area?

Please and thank you very much :)

 

[Unco]

Homework Statement

Find the area enclosed by r=2cos(3\theta)

I'm fairly confident how to do this but for some reason I am getting 2\pi rather than 1\pi, 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)\int^{2\pi}_{0} (2cos(3\theta))^2 d\theta

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.

 

[HallsofIvy]

Therefore, Unco, your answer to his question "Is this integral incorrect", is "Yes"!

 

[celeramo]

Thanks very much, my mistake