# Area between polar curves

1. Oct 12, 2014

### TheRedDevil18

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

Find the area inside one loop of r = 2cos(3 theta) and outside the circle r = 1

2. Relevant equations

3. The attempt at a solution

I need to clarify something about the limits of integration. I found the intersection of the two curves to be at an angle of pi/9. This is how I setup my integral

A = 2*integral from 0 to pi/9 of 1/2(2cos(3 theta))^2 d theta

Is it correct ?

2. Oct 12, 2014

### vela

Staff Emeritus
Not quite. You haven't used the fact that the area is outside the circle r=1.

3. Oct 12, 2014

### TheRedDevil18

So my expression is for the area inside the circle ?, I'm confused

4. Oct 12, 2014

### vela

Staff Emeritus
No. Why do you think the circle has do to anything with your expression at all?

5. Oct 12, 2014

### TheRedDevil18

Do I have to subtract away the circle ?, I thought by integrating to the point where the two graphs intersect I would get the area outside the circle

6. Oct 12, 2014

### vela

Staff Emeritus
Think about this. Suppose the question asked you to calculate the area inside the circle. How would the integral change? The two curves still intersect at the same points, so using your logic, you'd end up with same integral. Obviously, that can't be right. There's no reason to believe the area inside and outside the circle are the same.

I recommend you rethink the problem starting from the more general formula for the area
$$A = \iint r\,dr\,d\theta,$$ with the appropriate limits, and try to understand where the formula
$$A = \int \frac 12 r^2 \,d\theta$$ comes from. The latter is a special case of the first one, and you need to understand when you can actually use it.

7. Oct 12, 2014

### TheRedDevil18

Ok, so is this expression right ?

A = 2*integral from 0 to pi/9 of (1/2(2cos(3 theta))^2) - 1/2(1)^2 d theta

8. Oct 12, 2014

### vela

Staff Emeritus
How'd you come up with it?

9. Oct 12, 2014

### TheRedDevil18

At the point where they intersect by integrating the flower petal it includes part of the circle, therefore I have to subtract away that part

10. Oct 12, 2014

### LCKurtz

Yes.

11. Oct 13, 2014

Ok thanks