# How do you find the area of the region that lies inside the polar graphs, r = 1 - sin theta and r = sin theta?

Feb 26, 2015

First let's have a look at our areas:

Basically we want the area of the two leafs like shapes at $\frac{\pi}{6}$ inclination in the first and second quadrant.

By setting equal the two equations I get the inclination $\theta$ where they meet (green line):
$\sin \left(\theta\right) = 1 - \sin \left(\theta\right)$ and $\theta = \frac{\pi}{6}$

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).

To find the area of the first leaf (Area 1) you have:
(area from the red curve to the green line)+(area from the green line to the blue curve)=

Area 1= $\frac{1}{2} {\int}_{0}^{\frac{\pi}{6}} \left({\sin}^{2} \left(\theta\right)\right) d \left(\theta\right) + \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{\pi}{2}} {\left[1 - \sin \left(\theta\right)\right]}^{2} d \left(\theta\right)$