# Area of polar curve

1. Jan 29, 2012

### aub

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

Find the area inside the inner loop of the limacon curve : r = 1 + 2cos(θ)

2. Relevant equations

A = ∫$\stackrel{α}{β}$($\frac{1}{2}$r2)dθ

3. The attempt at a solution

i have the solution, my question is : how do you find α and β ?
here α = 2π/3 and β = π

A = 2∫$\stackrel{2π/3}{π}$($\frac{1}{2}$r2)dθ

2. Jan 29, 2012

### SammyS

Staff Emeritus
Have done the polar graph of r = 1 + 2cos(θ)?

Have you done the Cartesian graph ?

3. Jan 29, 2012

### aub

i did the polar one, i can see that its tangent to 2∏/3 on the origin but i cant get the point if there is any

and for the cartesian one, we never did that in class for those curves..

thanks

4. Jan 29, 2012

### SammyS

Staff Emeritus
You need to find the range of θ values that correspond to the inner loop, correct?

Looking at both graphs might help. Also, solving
1+2cos(θ) = 0 ​
for θ might help.

What is unusual about the values of r on the inner loop?

5. Jan 29, 2012

### aub

yeah, i thought what i need to do is solving for θ, it gave me α = 2∏/3 and β = 4∏/3 and i here i don't need to multiply the integral by 2 anymore.. im lost

do you mean the symmetry?

6. Jan 29, 2012

### aub

also, if anyone can help me to find the area of the region R between the inner loop and outer loop
what would be α and β ?

7. Jan 29, 2012

### aub

i guess i got it just tell me if its right

so for the area of the inner loop A1, α = 2∏/3 and β = 4∏/3 and i apply the formula

for the area between the inner loop, i find the whole area with α = 0 and β = 2π then substract A1

8. Jan 29, 2012

### SammyS

Staff Emeritus
No. For $\displaystyle \frac{2\pi}{3}<\theta<\frac{4\pi}{3}\,,$ 1 + 2cos(θ) is negative, so r is negative.

Last edited: Jan 29, 2012