Verified answer

One petal of this flower has beginning and end point for consecutive values of θ when r = 0:

6 cos(3θ) = 0

==> cos(3θ) = 0

==> 3θ = π/2 + πk for any integer k

==> θ = π/6 + πk/3.

Letting k = -1 and 0 yields θ = ±π/6.

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So, the area in one petal equals

∫ (1/2) r^2 dθ

= ∫(θ = -π/6 to π/6) (1/2) (6 cos(3θ))^2 dθ

= ∫(θ = -π/6 to π/6) 18 cos^2(3θ) dθ

= 2 * ∫(θ = 0 to π/6) 18 cos^2(3θ) dθ, since the integrand is even

= ∫(θ = 0 to π/6) 18 (1 + cos(6θ)) dθ, via half angle identity

= 18 (θ + sin(6θ)/6) {for θ = 0 to π/6}

= 3π.

I hope this helps!