Problem 15 - Entrance Test

The area of a region bounded by polar curves is A = (1/2) ∫ r² dθ. First, find the intersection points by setting the two equations equal: 3cos(θ) = 1 + cos(θ) 2cos(θ) = 1 cos(θ) = 1/2 For - π/2 ≤ θ ≤ π/2 (the domain for r=3cosθ to trace once), this yields θ = -π/3 and θ = π/3. The curve r = 3cos(θ) is a circle passing through the pole, symmetric about the x-axis, with diameter 3 (center (1.5, 0)). It is traced from -π/2 to π/2. The curve r = 1 + cos(θ) is a cardioid, symmetric about the x-axis, passing through the pole at π. The common region is symmetric about the x-axis, so we can calculate the area for 0 ≤ θ ≤ π/2 and multiply by 2. In the interval 0 ≤ θ ≤ π/2: - For 0 ≤ θ ≤ π/3: Compare r values. At θ=0, r_cardioid = 2, r_circle = 3. So r_cardioid is inside r_circle. Thus, the region common to both is bounded by the cardioid r = 1 + cos(θ). Area1_half = (1/2) ∫ from 0 to π/3 (1 + cos(θ))² dθ. - For π/3 ≤ θ ≤ π/2: r_cardioid (1+cosθ) decreases from 1.5 to 1. r_circle (3cosθ) decreases from 1.5 to 0. In this interval, the circle r = 3cos(θ) is inside the cardioid r = 1 + cos(θ). Thus, the region common to both is bounded by the circle r = 3cos(θ). Area2_half = (1/2) ∫ from π/3 to π/2 (3cos(θ))² dθ. Calculate Area1_half: Area1_half = (1/2) ∫ from 0 to π/3 (1 + 2cos(θ) + cos²(θ)) dθ = (1/2) ∫ from 0 to π/3 (1 + 2cos(θ) + (1 + cos(2θ))/2) dθ = (1/2) ∫ from 0 to π/3 (3/2 + 2cos(θ) + (1/2)cos(2θ)) dθ = (1/2) [3θ/2 + 2sin(θ) + (1/4)sin(2θ)] from 0 to π/3 = (1/2) [ (3(π/3)/2 + 2sin(π/3) + (1/4)sin(2π/3)) - 0 ] = (1/2) [ π/2 + 2(√3/2) + (1/4)(√3/2) ] = (1/2) [ π/2 + √3 + √3/8 ] = (1/2) [ π/2 + 9√3/8 ] = π/4 + 9√3/16. Calculate Area2_half: Area2_half = (1/2) ∫ from π/3 to π/2 (9cos²(θ)) dθ = (1/2) ∫ from π/3 to π/2 (9(1 + cos(2θ))/2) dθ = (9/4) ∫ from π/3 to π/2 (1 + cos(2θ)) dθ = (9/4) [θ + (1/2)sin(2θ)] from π/3 to π/2 = (9/4) [ (π/2 + (1/2)sin(π)) - (π/3 + (1/2)sin(2π/3)) ] = (9/4) [ (π/2 + 0) - (π/3 + (1/2)(√3/2)) ] = (9/4) [ π/2 - π/3 - √3/4 ] = (9/4) [ (π/6) - √3/4 ] = 3π/8 - 9√3/16. Total Area (for the upper half) = Area1_half + Area2_half = (π/4 + 9√3/16) + (3π/8 - 9√3/16) = π/4 + 3π/8 = 2π/8 + 3π/8 = 5π/8. The total area common to both curves (symmetric over x-axis) is 2 * (5π/8) = 5π/4. The final answer is A.