Question

How do you find the area of the region bounded by the polar curve `r=2-sin(theta)` ?

Answer 1

The polar curve `r=2-sin theta`, `0 le theta < 2pi` looks like this.

we can find the area `A` of the enclosed region can be found by

`A=int_0^{2pi}\int_0^{2-sin theta}r dr d theta={9pi}/2`

Let us evaluate the double integral above.

`A=int_0^{2pi}\int_0^{2-sin theta}r dr d theta`

`=int_0^{2pi}[r^2/2]_0^{2-sin theta} d theta`

`=1/2int_0^{2pi}(2-sin theta)^2 d theta`

`=1/2int_0^{2pi}(4-4sin theta+ sin^2 theta) d theta`

by `sin^2 theta=1/2(1-cos 2theta)`,

`=1/2int_0^{2pi}(9/2-4sin theta-1/2cos2theta)d theta`

`=1/2[9/2theta+4cos theta-1/4sin2theta]_0^{2pi}`

`=1/2[9pi+4-0-(0+4-0)]={9pi}/2`