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Homework Statement
Find the area of the region inside the lemniscate r^2 = 2sin(2[tex]\theta[/tex]) and outside the circle r = 1
It sucks because I wish I could post a graph, but the graph on my calculator looks like a circle around the origin with radius 1, with an infinity symbol going diagonally through it. (starting in quad 3 and pointing into quad 1.)
I believe the area they are after is the two ends of the lemniscate that don't quite fit inside the circle. (quads 1 and 3)
Homework Equations
1/2 * integral of (f([tex]\theta[/tex])^2 - g([tex]\theta[/tex])^2)
with your limits of integration being the angles where the two polar functions intersect.
The Attempt at a Solution
I set the equations equal to each other, and get the angles pi/12 and (5*pi)/12 as the limits of integration. Now, based on what the graph looks like, and the fact that I'm only after the ends of the lemniscate, it seems that:
2 * 1/2 * integral of ( 2sin(2[tex]\theta[/tex])^2 - (1)^2) with my limits of integration being pi/12 and (5 * pi)/12 would do the trick. I get an answer of pi/3. (approx. 1.05)
However, a classmate gave me the answer, an approximation, of .682. He said he went over it with the teacher, and I'm pretty sure he's right. Any tips on where I went wrong?