Find the area of the region.one petal of r = 3sin(2𝜃)

To find the area of one petal of the polar curve r = 3sin(2θ), we use the formula for the area A of a polar curve: A = 1/2 ∫[a,b] (r(θ))^2 dθ.

Since we're looking for the area of one petal, we need to find the limits of integration a and b. For the curve r = 3sin(2θ), a petal starts at θ = 0 and ends when sin(2θ) = 0 again. This happens at θ = π/2.

So, the area A of one petal is given by:

A = 1/2 ∫[0,π/2] (3sin(2θ))^2 dθ = 1/2 ∫[0,π/2] 9sin^2(2θ) dθ.

To solve this integral, we use the power-reduction identity sin^2(x) = 1/2(1 - cos(2x)). So, the integral becomes:

A = 1/2 ∫[0,π/2] 9/2(1 - cos(4θ)) dθ = 9/4 ∫[0,π/2] (1 - cos(4θ)) dθ = 9/4 [θ - 1/4 sin(4θ)] evaluated from 0 to π/2.

Evaluating this gives:

A = 9/4 [π/2 - 1/4 sin(2π) - (0 - 1/4 sin(0))] = 9/4 [π/2 - 0] = 9/8 π.

So, the area of one petal of the curve r = 3sin(2θ) is 9/8 π square units.