The area of the region {(x,y) : y² ≤ 4x, 4x² + 4y² ≤ 9} is:

The area of the region {(x,y) : y² ≤ 4x, 4x² + 4y² ≤ 9} is:

• A. (√2/3 + 9π/8 − 9/4 sin⁻¹(√8/3)) sq units
• B. (9π/4) sq units
• C. (9π/8) sq units
• D. (√2/3) sq units

Answer: A) (√2/3 + 9π/8 − 9/4 sin⁻¹(√8/3)) sq units

Explanation: Parabola y² = 4x and circle x² + y² = 9/4. Intersection: x² + 4x − 9/4 = 0 → 4x² + 16x − 9 = 0 → x = [−16 ± √(256+144)]/8 = [−16 ± 20]/8 → x = 1/2. Area = 2[ ∫[0 to 1/2] √(4x) dx + ∫[1/2 to 3/2] √(9/4 − x²) dx ] = computed to given option.