Answer: To find the area of one loop of the rose r = 3 cos(3θ), we can use the formula:
A = 1/2 ∫θ2 θ1 (f(θ))^2 dθ
where f(θ) is the function that defines the curve, and θ1 and θ2 are the angles that define one loop of the curve.
In this case, the curve completes one loop when θ goes from 0 to π/6 (or from π/6 to π, since the curve is symmetric about the y-axis). Therefore, we can compute the area as:
A = 1/2 ∫0^(π/6) (3cos(3θ))^2 dθ
A = 9/2 ∫0^(π/6) cos^2(3θ) dθ
Using the identity cos^2(θ) = (1 + cos(2θ))/2, we can simplify this to:
A = 9/4 ∫0^(π/6) (1 + cos(6θ)) dθ
A = 9/4 (θ + sin(6θ)/6) ∣∣0^(π/6)
A = 9/4 (π/6 + sin(π)/6)
A = 3π/8 - 3√3/8
Therefore, the area of one loop of the rose r = 3 cos(3θ) is 3π/8 - 3√3/8.