Question

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

Answer 1

The area of the enclosed region is `5/24 pi-sqrt{3}/4`.

It looks like this:

`A=int_0^{pi/3}int_0^{sin theta}rdrd theta +int_{pi/3}^{pi/2}int_0^{sqrt{3}cos theta}rdrd theta`

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

`=int_0^{pi/3} {sin^2theta}/2 d theta + int_{pi/3}^{pi/2}{3cos^2theta}/2 d theta`

`=1/4int_0^{pi/3} (1-sin2 theta) d theta + 3/4int_{pi/3}^{pi/2}(1+cos2 theta) d theta`

`=1/4[theta-{sin2theta}/2]_0^{pi/3} +3/4[theta+{sin2theta}/2]_{pi/3}^{pi/2}`

`=1/4(pi/3-sqrt{3}/4)+3/4(pi/6-sqrt{3}/4)`

`=5/24 pi-sqrt{3}/4`