# How do you find the area inside of the Cardioid r = 2+2cosθ and outside the circle r = 3?

##### 1 Answer
Jul 5, 2016

$= 8 - \frac{3 \pi}{2}$

#### Explanation:

area in polar is $\frac{1}{2} {\int}_{\theta 1}^{\theta 2} {r}^{2} \left(\theta\right) d \theta$ .

due to symmetry, the shaded area is $2 \times {\int}_{0}^{\frac{\pi}{2}}$

which means
$A = 2 \times \left(\frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} {\left(2 + 2 \cos \theta\right)}^{2} d \theta - \frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} {3}^{2} d \theta\right)$

$= {\int}_{0}^{\frac{\pi}{2}} d \theta q \quad {\left(2 + 2 \cos \theta\right)}^{2} - 9$

$= {\int}_{0}^{\frac{\pi}{2}} d \theta q \quad 4 {\cos}^{2} \theta + 8 \cos \theta - 5$

$= {\int}_{0}^{\frac{\pi}{2}} d \theta q \quad 2 \left(\cos 2 \theta + 1\right) + 8 \cos \theta - 5$

$= {\int}_{0}^{\frac{\pi}{2}} d \theta q \quad 2 \cos 2 \theta + 8 \cos \theta - 3$

$= {\left[\sin 2 \theta + 8 \sin \theta - 3 \theta\right]}_{0}^{\frac{\pi}{2}}$

$= 8 - \frac{3 \pi}{2}$