Answer:
3√3
Step-by-step explanation:
r = 3 cos θ
r = 2 - cos θ
First, find the intersections.
3 cos θ = 2 - cos θ
4 cos θ = 2
cos θ = 1/2
θ = -π/3, π/3
We want the area inside the first curve and outside the second curve. So R = 3 cos θ and r = 2 - cos θ, such that R > r.
Now that we have the limits, we can integrate.
A = ∫ ½ (R² - r²) dθ
A = ∫ ½ ((3 cos θ)² - (2 - cos θ)²) dθ
A = ∫ ½ (9 cos² θ - (4 - 4 cos θ + cos² θ)) dθ
A = ∫ ½ (9 cos² θ - 4 + 4 cos θ - cos² θ) dθ
A = ∫ ½ (8 cos² θ + 4 cos θ - 4) dθ
A = ∫ (4 cos² θ + 2 cos θ - 2) dθ
Using power reduction formula:
A = ∫ (2 + 2 cos(2θ) + 2 cos θ - 2) dθ
A = ∫ (2 cos(2θ) + 2 cos θ) dθ
Integrating:
A = (sin (2θ) + 2 sin θ) |-π/3 to π/3
A = (sin (2π/3) + 2 sin(π/3)) - (sin (-2π/3) + 2 sin(-π/3))
A = (½√3 + √3) - (-½√3 - √3)
A = 1.5√3 - (-1.5√3)
A = 3√3
The area inside of r = 3 cos θ and outside of r = 2 - cos θ is 3√3.
The graph of the curves is:
desmos.com/calculator/541zniwefe
