Area of a petal using Green's Theorem?

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.