Calculating Area of Lemniscate Polar Coordinates | Integral Method

[future_phd]

Homework Statement

Find the area inside the lemniscate r = 2sqrt(sin(2theta))

Homework Equations

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

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.

 

[Mark44]

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.

 

[future_phd]

Ahh that makes sense, thank you!