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

Nov 9, 2014

The region bounded by the polar curves looks like:

Since the region consists of two identical leaves that are symmetric about the $y$-axis, I will try to find a half of one leaf then multiply it by $4$.

$A = 4 {\int}_{0}^{\frac{\pi}{2}} {\int}_{0}^{1 - \cos \theta} r \mathrm{dr} d \theta$

$= 4 {\int}_{0}^{\frac{\pi}{2}} {\left[{r}^{2} / 2\right]}_{0}^{1 - \cos \theta} d \theta$

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

by ${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$,

$= {\int}_{0}^{\frac{\pi}{2}} \left(3 - 4 \cos \theta + \cos 2 \theta\right) d \theta$

$= {\left[3 \theta - 4 \sin \theta + \frac{1}{2} \sin 2 \theta\right]}_{0}^{\frac{\pi}{2}}$

$= \frac{3 \pi}{2} - 4$

I hope that this was helpful.