Question

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

Answer 1

First let's have a look at our areas:

Basically we want the area of the two leafs like shapes at `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(theta)=1-sin(theta)` and `theta=pi/6`

In general the area in polar form is:
`1/2int_(theta_1)^(theta_2)r^2d(theta)` (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= `1/2int_0^(pi/6)(sin^2(theta))d(theta)+1/2int_(pi/6)^(pi/2)[1-sin(theta)]^2d(theta)`

(Please, check my calculations)