Question

How do you find the area of the region bounded by the polar curve `r=3cos(theta)` ?

Answer 1

The area of the region is `9/4pi`.

Let us look at some details.

The region is a disk, which looks like this:

If you are allowed to use the formula for the area of a circle, then

`A=pir^2=pi(3/2)^2=9/4pi`

If you wish to use integration, then

`A=int_0^{pi}\int_0^{3cos theta}rdrd theta`

`=int_0^{pi}[r^2/2]_0^{3cos theta}d theta`

`=int_0^{pi}{9cos^2theta}/2 d theta`

by the trig identity: `cos^2theta=1/2(1+cos2theta)`,

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

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

`=9/4pi`

I hope that this was helpful.