Question

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

Answer 1

The areas of both regions are `pi/2`.

The graph of `r=sin(2theta)`, `0leq theta <2pi` looks like this:

Since the area element in polar coordinates is `r dr d theta`, we can find the area of the four leaves above by
`A=int_0^{2pi}int_0^{sin(2 theta)}rdrd theta`.

Let us evaluate the inside integral first,
`A=int_0^{2pi}[{r^2}/2]_0^{sin(2 theta)}d theta =int_0^{2pi}{sin^2(2 theta)}/2 d theta`

By the double-angle idenitity `sin^2(2 theta)={1-cos(4 theta)}/2`,
`A=1/4int_0^{2pi}[1-cos(4 theta)]d theta =1/4[theta-{sin(4 theta)}/{4}]_0^{2pi} =1/4[2pi-{sin(8pi)}/4-(0-{sin(0)}/{4})]=1/4(2pi)={pi}/2`

Hence, the area is `{pi}/2`.

For `r=cos(2 theta)`, the area can be found by
`A=int_0^{2pi}int_0^{cos(2 theta)}rdrd theta`