[Math] Area inside outer curve of limacon $r=1+2\cos\theta$

Question

I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$

Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta=\frac{r^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=2\frac{(1+2\cos\theta)^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=0
$$
so, 0 area is clearly wrong. Where was my mistake?

Answer 1

The outer curve of the limacon lies between $\theta=\pm\frac{2\pi}{3}$ but using symmetry it is twice the area between $\theta=0$ and $\theta=\frac{2\pi}{3}$

\begin{eqnarray} A&=&2\int_0^{2\pi/3}\frac{1}{2}r^2d\theta\\ &=&\int_0^{2\pi/3}(1+2\cos\theta)^2d\theta\\ &=&\int_0^{2\pi/3}3+4\cos\theta+2\cos(2\theta)\,d\theta =[3\theta+4\sin\theta+\sin(2\theta)]_0^{2\pi/3}\\ &=&2\pi+\frac{3\sqrt{3}}{2} \end{eqnarray}