# Homework Help: 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..

what were you hinting about?

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