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

Sep 20, 2015

$6 \pi$

#### Explanation:

$A = \int {\int}_{D} r \mathrm{dr} d \theta$

A=int_(0)^(2pi)d thetaint_(0)^(2+2costheta)rdr=1/2int_(0)^(2pi)d theta r^2|_(0)^(2+2costheta)

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

$A = \frac{1}{2} {\int}_{0}^{2 \pi} \left(4 + 8 \cos \theta + 4 {\cos}^{2} \theta\right) d \theta$

${\cos}^{2} \theta = \frac{1 + \cos 2 \theta}{2}$

$A = \frac{1}{2} {\int}_{0}^{2 \pi} \left(4 + 8 \cos \theta + 2 + 2 \cos 2 \theta\right) d \theta$

$A = \frac{1}{2} \left(4 \theta + 8 \sin \theta + 2 \theta + \sin 2 \theta\right) {|}_{0}^{2 \pi}$

$A = 6 \pi$