# area of polar coordinates

• May 7th 2008, 04:39 AM
etha
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?
• May 7th 2008, 05:12 AM
mr fantastic
Quote:

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.
• May 7th 2008, 05:16 AM
etha
is it still supposed to be multiplied by 2? Or do I not need that
• May 7th 2008, 05:38 AM
mr fantastic
Quote:

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.