# Homework Help: Polar curve area question

1. Mar 29, 2013

### steel1

1. The problem statement, all variables and given/known data
Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside r=2

2. Relevant equations
area of polar curves = .5∫R^2(outside)-r^2(inside) dθ

3. The attempt at a solution
r^2=8cos(2θ) and r=2, so....

4=8cos(2θ)
.5=cos(2θ) since .5 is positive, we need the angles in the first and fourth quadrant
2θ=∏/3 and -∏/3
θ=∏/6 and -∏/6

.5∫8cos(2θ)-4 from -pi/6 to pi/6
=difference of .5[4sin(2θ)-4θ] evaluated at pi/6 and -pi/6. but when i plug the values in, i cant seem to get the right answer

2. Mar 29, 2013

### SammyS

Staff Emeritus
Consider what happens when cos(2θ) < 0 .

Ignore this post. I misread the problem. DUH !

Last edited: Mar 30, 2013
3. Mar 29, 2013

### steel1

not sure what your trying to get at. :<

4. Mar 29, 2013

### Stimpon

What answer should you be getting?

5. Mar 29, 2013

### steel1

2.739

6. Mar 29, 2013

### Stimpon

im so dumb

Last edited: Mar 29, 2013
7. Mar 29, 2013

### LCKurtz

Are you getting exactly half the desired answer? Are you aware there is a symmetric area on the left side of the y axis?

8. Mar 29, 2013

### steel1

yes, i was getting exactly half the right answer. forgot about the symmetry, thanks!