Use Green's theorem to compute the area of one petal of the 4-leafed rose defined by r = 18 sin(2θ).
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as A = 1/2 ∫ xdy - ydx
Area of a petal using Green's Theorem?
I have an alternate solution that doesn't use Green's Theorem that you might or might not want to use.
One petal begins and ends at r = 0. So let's start by finding where θ is for r = 0
r = 0
18sin2θ = 0
sin2θ = 0
2θ = 0, ±π, ±2π, ...
θ = 0, ±π/2, ±π, ...
We choose two consecutive angles where the petal begins and ends. Let's choose a = 0 and b = π/2
The area for a polar curve is defined by:
A = ∫(a..b) 1/2 r^2 dθ
A = ∫(0..π/2) 1/2 (18sin2θ)^2 dθ
A = ∫(0..π/2) 1/2 * 324(sin2θ)^2 dθ
A = ∫(0..π/2) 162 (sin2θ)^2 dθ
Now you just use the half-angle to simply sin2θ and integrate from there to get your answer, which should be 81π/2.