one petal has -π/6 ≤ θ ≤ π/6
Using the symmetry of the curve, the area is
A = 2∫[0,π/6] 1/2 r^2 dθ = ∫[0,π/6] 9 cos^2(3θ)
= 9/2 ∫[0,π/6] 1+cos(6θ) dθ
Now finish it off
Find the area of one petal of the polar curve r = 3 cos 3θ. (4 points)
A) 7 times pi over 12
B) 3 times pi over 4
C) 11 times pi over 12
D) 2π
2 answers
Can you please show the workings?