Polar coordinates and overlapping areas

Question

The area of the region that lies inside both curves $r = 1 - \cos \theta$ and $r = \frac{1}{2},$ when simplified, can be expressed in the form $\frac{a \pi}{b} - \frac{c \sqrt{d}}{e}$

where $a,$ $b,$ $c,$ $d,$ and $e$ are all positive integers. Find $a + b + c + d + e.$

How do I find the proper points of intersection? I think I will need to do that, then find out what angle sector is cut out.

Answer 1

The curves intersect when they have the same $r$ and $\theta$.

You can find $\theta$ by solving the equation $1-\cos(\theta)=\frac{1}{2}$.

This will give you two values of $\theta$ where the curves intersect.

Note, you need to make sure you take into account which curve has the lower radius so that you capture the region that lies inside both curves.

Answer 2

The two curves intersect at $(\frac12, \pm \frac\pi3)$. Then, the area integral is

$$\frac12\int_{-\pi/3}^{\pi/3} (1-\cos\theta)^2d\theta +\frac12\int_{\pi/3}^{5\pi/3} \frac14d\theta= (-\frac{7\sqrt3}8+\frac\pi2 )+ \frac\pi6= \frac{2{\pi}}3 -\frac{7\sqrt3}8$$

Thus, $a + b + c + d + e=23$.