Problem 12 - Entrance Test

The region is in the first quadrant. The x-axis corresponds to θ=0. The curve r = 2cos(θ) is a circle centered at (1,0) with radius 1. The curve r = 1 is a circle centered at the pole with radius 1. Find the intersection point of r = 2cos(θ) and r = 1: 2cos(θ) = 1 ⇒ cos(θ) = 1/2. In the first quadrant, θ = π/3. Observe the region in the first quadrant: From θ = 0 to θ = π/3, the curve r = 2cos(θ) is outside r = 1. From θ = π/3 to θ = π/2, the curve r = 1 is outside r = 2cos(θ) (which drops to 0 at π/2). The area is the sum of two parts: Area1: from θ = 0 to θ = π/3, bounded by r = 2cos(θ). Area2: from θ = π/3 to θ = π/2, bounded by r = 1. Area1 = (1/2) ∫ from 0 to π/3 (2cos(θ))² dθ = (1/2) ∫ from 0 to π/3 (4cos²(θ)) dθ = 2 ∫ from 0 to π/3 ( (1 + cos(2θ))/2 ) dθ (using cos²(θ) = (1 + cos(2θ))/2) = ∫ from 0 to π/3 (1 + cos(2θ)) dθ = [θ + (1/2)sin(2θ)] from 0 to π/3 = (π/3 + (1/2)sin(2π/3)) - (0 + (1/2)sin(0)) = π/3 + (1/2)(√3/2) - 0 = π/3 + √3/4. Area2 = (1/2) ∫ from π/3 to π/2 (1)² dθ = (1/2) ∫ from π/3 to π/2 (1) dθ = (1/2)[θ] from π/3 to π/2 = (1/2)(π/2 - π/3) = (1/2)(3π/6 - 2π/6) = (1/2)(π/6) = π/12. Total Area = Area1 + Area2 = (π/3 + √3/4) + π/12 = (4π/12 + π/12) + √3/4 = 5π/12 + √3/4. The final answer is A.