The area (in sq. units) of the smaller portion enclosed between the curves, x 2 + y 2 = 4 and y 2 = 3 x , is:

Step-by-step solution

To find point of intersection, eliminate y from both the equation, we get x 2 + 3 x - 4 = 0 ⇒   x + 4   x - 1 = 0 x = - 4 ,   x = 1 Area = ∫ 0 1 3 · x · d x + ∫ 1 2 4 - x 2 · d x × 2 = 3 x 3 2 3 2 0 1 + x 2 4 - x 2 + 2 sin - 1 ⁡ x 2 1 2 × 2 = 3 2 3 + 2 · π 2 - 3 2 + π 3 × 2 = 2 3 - 3 2 + 2 π 3 × 2 = 1 2 3 + 2 π 3 × 2 = 1 3 + 4 π 3 sq. units