Problem 7 - Entrance Test

The inner loop of a limaçon occurs when r becomes negative and returns to zero. To find the angles that define the inner loop, set r = 0: 1 + 2sin(θ) = 0 sin(θ) = -1/2 For 0 ≤ θ < 2π, the solutions are θ = 7π/6 and θ = 11π/6. These are the limits of integration for the inner loop. The formula for the area of a polar region is A = (1/2) ∫ r² dθ. Area = (1/2) ∫ from 7π/6 to 11π/6 (1 + 2sin(θ))² dθ. = (1/2) ∫ from 7π/6 to 11π/6 (1 + 4sin(θ) + 4sin²(θ)) dθ. Use the identity sin²(θ) = (1 - cos(2θ))/2: = (1/2) ∫ from 7π/6 to 11π/6 (1 + 4sin(θ) + 4 * (1 - cos(2θ))/2) dθ. = (1/2) ∫ from 7π/6 to 11π/6 (1 + 4sin(θ) + 2 - 2cos(2θ)) dθ. = (1/2) ∫ from 7π/6 to 11π/6 (3 + 4sin(θ) - 2cos(2θ)) dθ. Now, integrate: [3θ - 4cos(θ) - sin(2θ)] from 7π/6 to 11π/6. Evaluate at 11π/6: 3(11π/6) - 4cos(11π/6) - sin(2 * 11π/6) = 11π/2 - 4(√3/2) - sin(11π/3) = 11π/2 - 2√3 - (-√3/2) = 11π/2 - 4√3/2 + √3/2 = 11π/2 - 3√3/2. Evaluate at 7π/6: 3(7π/6) - 4cos(7π/6) - sin(2 * 7π/6) = 7π/2 - 4(-√3/2) - sin(7π/3) = 7π/2 + 2√3 - (√3/2) = 7π/2 + 4√3/2 - √3/2 = 7π/2 + 3√3/2. Subtract the lower limit from the upper limit: (11π/2 - 3√3/2) - (7π/2 + 3√3/2) = (11π - 7π)/2 - (3√3/2 + 3√3/2) = 4π/2 - 6√3/2 = 2π - 3√3. Finally, multiply by 1/2: Area = (1/2) * (2π - 3√3) = π - 3√3/2. The final answer is A.