# How do you find the area inside of the Cardioid r=3+2cos(theta) for 0 <= (theta) <= 2pi?

Sep 26, 2015

$11 \pi$

#### Explanation:

$A = \frac{1}{2} \int {r}^{2} d \theta = \frac{1}{2} {\int}_{o}^{2 \pi} {\left(3 + 2 \cos \theta\right)}^{2} d \theta$

$A = \frac{1}{2} {\int}_{o}^{2 \pi} \left(9 + 12 \cos \theta + 4 {\cos}^{2} \theta\right) d \theta$

$A = \frac{1}{2} \left(9 \theta + 12 \sin \theta + 4 {\int}_{0}^{2 \pi} \frac{1 + \cos 2 \theta}{2} d \theta\right)$

$A = \frac{1}{2} \left(9 \theta + 12 \sin \theta + 4 \left(\frac{\theta}{2} + \frac{1}{4} \sin 2 \theta\right)\right) {|}_{0}^{2 \pi}$

$A = \frac{1}{2} \left(18 \pi + 4 \pi\right) = 11 \pi$