Find the area of the region (x,y): y² ≤ 6ax and x² + y² ≤ 16a² using method of integration.

We have (x,y): y² ≤ 6ax and x² + y² ≤ 16a²
Consider y² = 6ax — (1) and x² + y² = 16a² — (2)
Curve (2) represents a circle centered at (0, 0) having radius of 4a.
Solving (1) and (2),
x² + 6ax - 16a² = 0 ⇒ (x + 8a)(x - 2a) = 0
Therefore, x = 2a and x = -8a
So, x = 2a, y = ±2√3a and point of intersection are (2a, ±2√3a).
Now required area = 2 × ar(OAPMO) = 2[∫₀²ᵃ y₁dx + ∫₂⁴ᵃ y₂dx]
= 2[√6a ∫₀²ᵃ √x dx + ∫₂⁴ᵃ √(4a)² - x² dx]
= 2(2/3√6a[x^(3/2)]₀²ᵃ + [x/2√(4a)² - x² + 16a²/2sin⁻¹(x/4a)]₄ᵃ₂ᵃ)
= 4/3√6a[(2a)^(3/2)] + [x/2√(4a)² - x² + 16a²/2sin⁻¹(x/4a)]₄ᵃ₂ᵃ
= 4/3√6a√2a(2a) + [(4a × 0 + 16a²/2sin⁻¹(1)) - (2a × 2√3a + 16a²/2sin⁻¹(1/2))]
= 16√3/3a² + [8πa² - (4√3a² + 16a²(π/6))]
= 16√3/3a² + [8πa² - 4√3a² - 8π/3a²]
= 4√3/3a² + 16π/3a²
= 4a²/3(4π + √3) Sq. units.