The area (in sq. units) of the smaller portion enclosed between the curves x² + y² = 4 and y² = 3x is:

Here, x² + y² = 4 and y² = 3x
So finding the co-ordinates of x ⇒ x² + 3x = 0
(x + 4)(x - 1) = 0
x = -4, x = 1
Area = (∫₀¹ √3x dx + ∫₁² √(4 - x²) dx) × 2
= [√3(x³/²/³/₂)₀¹ + (x²/2 √(4 - x²) + 2 sin⁻¹(x/2))₁²] × 2
= (√3(2/3) + 2 × π/2 - (√3/2 + π/3)) × 2
= (2√3 - √3/2 + 2π/3) × 2
= (3√3/2 + 2π/3) × 2
= 3√3 + 4π/3
= √3 + 4π/3