# Integration of polar equation

1. Feb 21, 2009

### celeramo

1. The problem statement, all variables and given/known data
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 :)

2. Feb 22, 2009

### Unco

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. Feb 22, 2009

### HallsofIvy

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

4. Feb 22, 2009

### celeramo

Thanks very much, my mistake