# How do you find the area of the region shared by the cardioid r=2(1+cos(theta)) and the circle r=2?

Sep 26, 2015

$5 \pi - 8$

#### Explanation:

Let's find the points of intersection:

$2 \left(1 + \cos \theta\right) = 2$
$1 + \cos \theta = 1$
$\cos \theta = 0$
$\theta = \frac{\pi}{2} \vee \theta = \frac{3 \pi}{2}$

We can see from the graph that (considering the half of the area due to the symmetry and doubling the integrals):

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

${A}_{1} = 4 \theta {|}_{0}^{\frac{\pi}{2}} = 2 \pi$

${A}_{2} = 4 {\int}_{\frac{\pi}{2}}^{\pi} \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right) d \theta$

${A}_{2} = 4 \left(\theta + 2 \sin \theta + {\int}_{\frac{\pi}{2}}^{\pi} \frac{1 + \cos 2 \theta}{2} d \theta\right)$

${A}_{2} = 4 \left(\theta + 2 \sin \theta + \frac{\theta}{2} + \frac{1}{4} \sin 2 \theta\right) {|}_{\frac{\pi}{2}}^{\pi}$

${A}_{2} = 4 \left(\frac{3 \pi}{2}\right) - 4 \left(\frac{3 \pi}{4} + 2\right) = 6 \pi - 3 \pi - 8 = 3 \pi - 8$

$A = 2 \pi + 3 \pi - 8 = 5 \pi - 8$