# How do you find the area of the region bounded by the polar curve r=3(1+cos(theta)) ?

Sep 7, 2014

The area of the region enclosed by the curve $r = 3 \left(1 = \cos \theta\right)$ can be found by the double integral
${\int}_{0}^{2 \pi} {\int}_{0}^{3 \left(1 + \cos \theta\right)} r \mathrm{dr} d \theta = \frac{27}{2} \pi$

Let us evaluate the double integral.
${\int}_{0}^{2 \pi} {\int}_{0}^{3 \left(1 + \cos \theta\right)} r \mathrm{dr} d \theta$
by Power Rule,
$= {\int}_{0}^{2 \pi} {\left[{r}^{2} / 2\right]}_{0}^{3 \left(1 + \cos \theta\right)} d \theta$
$= {\int}_{0}^{2 \pi} \frac{{\left[3 \left(1 + \cos \theta\right)\right]}^{2}}{2} d \theta$
$= \frac{9}{2} {\int}_{0}^{2 \pi} \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right) d \theta$
by ${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$,
$= \frac{9}{2} {\int}_{0}^{2 \pi} \left(\frac{3}{2} + 2 \cos \theta + \frac{1}{2} \cos 2 \theta\right) d \theta$
$= \frac{9}{2} {\left[\frac{3}{2} \theta + 2 \sin \theta + \frac{1}{4} \sin 2 \theta\right]}_{0}^{2 \pi}$
$= \frac{9}{2} \cdot \frac{3}{2} \left(2 \pi\right) = \frac{27}{2} \pi$