# Homework Help: Finding the area between Polar Curves

1. May 10, 2014

### jojo13

1. The problem statement, all variables and given/known data

Find the area of the region that consists of all points that lie within the circle r = 1 but outside the polar equation r = cos(2θ)

2. Relevant equations

A = ∫ 1/2 (r2^2 - r1^2) dθ, where r2 is outer curve and r1 is inner curve.

3. The attempt at a solution

Here is what the graph looks like if you want to see it: http://www.wolframalpha.com/input/?i=r+=+cos(2x)+polar

Ok so first I had to find out the limits for my integral. I set: 1 = cos(2θ) and got that θ = ∏.

Now I made my integral, but instead of going from -∏ to ∏ I went from 0 to ∏ and multiplied it by 2

A = 2 ∫ 1/2 [(1^2) - (cos(2θ)^2)] dθ, from 0 to ∏

A = ∫ [ (1) - (1/2 cos(4θ) +1)] dθ, from 0 to ∏

I solved it and received: ∏ - ∏/2

Which equals ∏/2.

Just want to know if I did it right. The main part I was worried about was the limits for my integral.

2. May 10, 2014

### epenguin

I think it's right but IMHO would benefit from being done more simply.

I guess you could take the area of a circle as known.

For the area enclosed by cos(2θ) you could equally well integrate 0 to π/4 and multiply by 8. But simpler still...

In your polar co-ordinate figure it looks quite plausible that the area enclosed by it is half that of the circle, but is not self-evident.

Whereas it is self-evident in. Cartesian co-ordinates. And you could make the argument purely by symmetry without any integrations.

Last edited: May 11, 2014
3. May 10, 2014

### LCKurtz

If you look at the graph for $0<\theta<\pi$ of the two figures, it is a bit of a stretch to describe what you have as an outer curve and an inner curve. Of course, squaring makes it all work out OK. I would have just gone from $0$ to $\frac \pi 4$ and multiplied by $8$.

4. May 10, 2014

### jojo13

I see. That's and interesting way of looking at the problem. Thanks for that.

Also, if I'm not mistaken both give the same answer (∏/2) right?

5. May 10, 2014

### LCKurtz

Yes, $\frac \pi 2$ is correct.