Question

How do you find the area of one petal of `r=cos2theta`?

Answer 1

`pi/8=0.3927` areal units, nearly.

Explanation:

Period of `r(theta)` is `(2pi)/2=pi`.

As `r = cos 2theta >= 0, 2theta in [-pi/2, pi/2] to theta in [-pi/4, pi/4]`,

for one petal. So, the area (by symmetry about #theta = 0)

`=2(1/2 int r^2 d theta)`, from `0 to pi/4`

`=int cos^ 2 2theta d theta`, for the limits

`=int (1+cos 4theta)/2 d theta`, for the limits

=`[1/2[theta+1/4sin 4theta]`, between `theta = 0 and theta = pi/4`

`=pi/8` areal units.

graph{(x^2+y^2)sqrt(x^2+y^2)=x^2-y^2 [-2.5, 2.5, -1.25, 1.25]}