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.

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