# Thread: Area Between Polar Curves

1. ## Area Between Polar Curves

I am having troubles finding the area between two curves from 0 to 2$\displaystyle \pi$:

The inside of r=1+cosθ and the outside of r=2cosθ

Is it possible to simply subtract 2cosθ from 1+cosθ and take the integral of that so it looks like:

$\displaystyle \frac{1}{2} \int_0^{2\pi} (1-cos\theta)^2d\theta$

Thanks ahead of time for the help

2. Hi

Look at the sketch below
The area is twice the one for $\displaystyle \theta$ between 0 and $\displaystyle \pi$ and the latter can be split into 2 parts : $\displaystyle \theta$ between 0 and $\displaystyle \frac{\pi}{2}$ and $\displaystyle \theta$ between $\displaystyle \frac{\pi}{2}$ and $\displaystyle \pi$

For $\displaystyle \theta$ between 0 and $\displaystyle \frac{\pi}{2}$ the radius varies between $\displaystyle 2 \cos \theta$ and $\displaystyle 1 + \cos \theta$

For $\displaystyle \theta$ between $\displaystyle \frac{\pi}{2}$ and $\displaystyle \pi$ the radius varies between 0 and $\displaystyle 1 + \cos \theta$

Therefore

$\displaystyle A = 2 \left(\int_{0}^{\frac{\pi}{2}} \int_{2 \cos \theta}^{1 + \cos \theta} r dr d\theta + \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{1 + \cos \theta} r dr d\theta\right)$

I can find $\displaystyle A = \frac{\pi}{2}$ which seems possible when looking at the figure

3. Thank you very much, that helped out tremendously