Finding area between polar curves

[Sentience]

Homework Statement

Find the area of the region inside the lemniscate r^2 = 2sin(2\theta) 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(\theta)^2 - g(\theta)^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\theta)^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?

 

[epenguin]

You said your classmate's figure was approximate and I do get approximately that, 0.68485.

Your approach as expressed in the integral appears perfectly OK.

Clue to where you have gone wrong is that your π/3 actually is the second part of your integral that you have to subtract from the first. I understand your g(θ) = 1. You don't need to do any integration of this part, it is just a third of a circle. Oh that's 2π/3. Well you work out the details. The pi factor should alert you to where error is because a trigonometric function of any fraction of pi wouldn't itself be any multiple of π.

Personally I would just integrate r²/2 between your limits, the whole shaded area in fig I will try to put up shortly, and subtract π/3 from that.

 

[epenguin]

Hope you can make out the circles of radii 1 and 2 there.

In the formula should have been cos 2θ not cosθ

 

[flyingpig]

You did it right, I think you probably pressed the wrong buttons on your calculator.

Here is how I did it

2\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}}} \int_{1}^{\sqrt{2\sin2\theta}} r dr d\theta