Consider the polar curve: r=3cos(3ϴ)
a) Find the area of the first leaf of the graph.
-Thank you.
taking advantage of symmetry, and knowing that
r(0)=3
r(π/6)=0
all we have to do is double the area for that domain.
a = 2∫[0,π/6] 1/2 r^2 dθ = ∫[0,π/6] 9cos^2(3θ) dθ = 3π/4
To find the area of the first leaf of the polar curve r = 3cos(3θ), you can use the formula for finding the area enclosed by a polar curve:
A = (1/2)∫[a,b] r^2 dθ
In this case, the "first leaf" refers to one complete petal of the polar curve, which corresponds to one complete revolution of θ from its starting point to the next point where r = 0.
To determine the limits of integration, you need to find the points where r = 0 (where the petal starts and ends). In this case, r = 0 when 3cos(3θ) = 0, which occurs at θ = π/6 and θ = 7π/6.
Now you can calculate the area using the formula:
A = (1/2)∫[π/6,7π/6] (3cos(3θ))^2 dθ
A = (9/2)∫[π/6,7π/6] cos^2(3θ) dθ
Since cos(2θ) = cos^2(θ) - sin^2(θ), you can rewrite cos^2(3θ) as:
cos^2(3θ) = (cos(6θ) + 1)/2
Therefore, the area integral becomes:
A = (9/2)∫[π/6,7π/6] (cos(6θ) + 1)/2 dθ
Now you can solve the integral:
A = (9/2) * [(1/6)sin(6θ) + (θ/2)] evaluated from π/6 to 7π/6
A = (9/2) * [(1/6)sin(6(7π/6)) + (7π/12) - ((1/6)sin(6(π/6)) + (π/12))]
A = (9/2) * [(1/6)sin(7π/2) + (7π/12) - (1/6)sin(π/2) + (π/12)]
A = (9/2) * [(-1/6) + (7π/12) - (1/6) + (π/12)]
Simplifying further:
A = (9/2) * [(7π/12) + (π/12) - (1/6) - (1/6)]
A = (9/2) * [(8π/12) - (2/6)]
A = (9/2) * [(2π/3) - (1/3)]
A = (9/2) * (π/3)
A = (9π/6)
A = (3π/2)
Therefore, the area of the first leaf of the graph defined by the polar curve r = 3cos(3θ) is (3π/2) square units.