# Thread: area of polar coordinates

1. ## area of polar coordinates

I'm trying to figure out the the area inside the outer loop and above the x axis for the polar curve r = 1 - 2sin(theta).

i think the integral goes from pi to pi/6. I want to say that the integral for the area is 1/2 (1-2sin(theta))^2 d(theta) multiplied by 2 but I'm really not sure. Can someone clarify this for me?

2. Originally Posted by etha
I'm trying to figure out the the area inside the outer loop and above the x axis for the polar curve r = 1 - 2sin(theta).

i think the integral goes from pi to pi/6. I want to say that the integral for the area is 1/2 (1-2sin(theta))^2 d(theta) multiplied by 2 but I'm really not sure. Can someone clarify this for me?
I think you want 0 to pi/6, not pi to pi/6. Then it's all fine.

3. is it still supposed to be multiplied by 2? Or do I not need that

4. Originally Posted by etha
is it still supposed to be multiplied by 2? Or do I not need that
I only corrected the integral terminals. The rest I said was fine. So yes, it is still multiplied by 2.

A graph of the curve clearly shows why the factor of 2 is necessary.