Problem 2 - Entrance Test

The area between two polar curves r_outer and r_inner is given by A = (1/2) ∫ (r_outer² - r_inner²) dθ. First, find the intersection points by setting the two radii equal: 3 + 2sin(θ) = 2 2sin(θ) = -1 sin(θ) = -1/2 This occurs at θ = 7π/6 and θ = 11π/6. Over the interval from 7π/6 to 11π/6, the curve r = 3 + 2sin(θ) is outside r = 2. So, r_outer = 3 + 2sin(θ) and r_inner = 2. Area = (1/2) ∫ ( (3 + 2sin(θ))² - 2² ) dθ from 7π/6 to 11π/6. = (1/2) ∫ ( 9 + 12sin(θ) + 4sin²(θ) - 4 ) dθ = (1/2) ∫ ( 5 + 12sin(θ) + 4(1 - cos(2θ))/2 ) dθ (using sin²(θ) = (1 - cos(2θ))/2) = (1/2) ∫ ( 5 + 12sin(θ) + 2 - 2cos(2θ) ) dθ = (1/2) ∫ ( 7 + 12sin(θ) - 2cos(2θ) ) dθ Now, integrate: [7θ - 12cos(θ) - sin(2θ)] from 7π/6 to 11π/6. Evaluate at 11π/6: 7(11π/6) - 12cos(11π/6) - sin(2 * 11π/6) = 77π/6 - 12(√3/2) - sin(11π/3) = 77π/6 - 6√3 - sin(2π + 5π/3) = 77π/6 - 6√3 - (-√3/2) = 77π/6 - 12√3/2 + √3/2 = 77π/6 - 11√3/2 Evaluate at 7π/6: 7(7π/6) - 12cos(7π/6) - sin(2 * 7π/6) = 49π/6 - 12(-√3/2) - sin(7π/3) = 49π/6 + 6√3 - sin(2π + π/3) = 49π/6 + 6√3 - (√3/2) = 49π/6 + 12√3/2 - √3/2 = 49π/6 + 11√3/2 Subtract the lower limit from the upper limit: (77π/6 - 11√3/2) - (49π/6 + 11√3/2) = (77π - 49π)/6 - (11√3/2 + 11√3/2) = 28π/6 - 22√3/2 = 14π/3 - 11√3. Finally, multiply by 1/2: Area = (1/2) * (14π/3 - 11√3) = 7π/3 - 11√3/2. The final answer is A.