# How do you find the area of the region that lies inside the curves r= 1+cos(theta) and r= 1-cos(theta)?

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Gió Share
Feb 26, 2015

First let us "see" our area:

Basically you want the area of the two loops enclosed by the two curves (vertically along the vertical axis).

In general the area in polar form is:
$\frac{1}{2} {\int}_{{\theta}_{1}}^{{\theta}_{2}} {r}^{2} d \left(\theta\right)$ (have a look to any maths book on calculus/analytical geometry).

In this case let us start with the upper loop; you have:

$0 \to \frac{\pi}{2}$ for the red line + $\frac{\pi}{2} \to \pi$ for the blue line

Area1= $\frac{1}{2} {\int}_{0}^{\frac{\pi}{2}} {\left(1 - \cos \left(\theta\right)\right)}^{2} d \left(\theta\right) + \frac{1}{2} {\int}_{\frac{\pi}{2}}^{\pi} {\left(1 + \cos \left(\theta\right)\right)}^{2} d \left(\theta\right)$
So you get:

Hope it helps (check my maths)

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