# Polar Coordinates Area

1. Mar 12, 2009

### future_phd

1. The problem statement, all variables and given/known data
Find the area inside the lemniscate r = 2sqrt(sin(2theta))

2. Relevant equations
Integral from a to b of (1/2)[f(theta)]^2 d(theta)

3. The attempt at a solution
I tried integrating from 0 to 2pi and got an area of 0. Then I tried integrating from 0 to pi and still got an area of 0. I looked at the answer and they integrated from 0 to pi/2 and then multiplied the integral by 2. I don't understand why they chose to integrate from 0 to pi/2 or why they multiplied the integral by 2? Any help would be greatly appreciated.

2. Mar 12, 2009

### Staff: Mentor

y = sin(2x) has a complete cycle between 0 and pi. If you wanted the area between this curve and the x-axis, it wouldn't do you any good to integrate between 0 and pi -- you would get 0. So instead you would integrate between 0 and pi/2 to get the area under one arch, and then double it, to get the area of both regions.

A similar thing is happening with your polar curve.

3. Mar 12, 2009

### future_phd

Ahh that makes sense, thank you!