Polar Coordinates (Area enclosed)

Question

Find the common area enclosed by the curves $$r= 3 - 2 \cos \theta$$ $$r= 2$$

My attempt,

Area$$=4\pi-\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}\frac{1}{2}(4-(3-2\cos \theta)^2)d\theta$$

$$=36.753$$

Am I correct?

Answer 1

Area:

$$\begin{align} 2\left(\int_0^{\pi/3} \frac 12\big(\overbrace{3-2\cos\theta}^{\color{red}{r_1}}\big)^2 d\theta +\int_{\pi/3}^{\pi}\frac 12\cdot (\overbrace{\;\;2\;\;}^{\color{blue}{r_2}})^2 d\theta\right) &=2\left(\frac {11}2\bigg(\frac {\pi}3-\frac{\sqrt{3}}2\bigg)+ \frac {4\pi}3\right)\\ &=11\bigg(\frac {\pi}3-\frac{\sqrt{3}}2\bigg)+\frac {8\pi}3\\ &=\color{red}{\frac {19}3\pi-\frac {11\sqrt{3}}2}\\ &=\color{red}{10.37}\end{align}$$

Answer 2

As Rafa Budria pointed out, you made an error while calculating limits of $\theta$ which are obtained at the intersection points of two curves i.e.

$2=3-2cos\theta\implies cos\theta =\frac{1}{2}$ or $\theta=\pm π/3$