Question

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

(A) 1/2√3 + π/3

(B) 1/2√3 + 2π/3

(C) 1/√3 + 2π/3

(D) 1/√3 + 4π/3

Answer 1

Answer is (D) 1/√3 + 4π/3

The given equation x² + y² = 4 is equation of circle of radius 2 centred at origin and equation y² = 3x is the equation of parabola.

Substituting Eq. (2) in Eq. (1), we get 

x² + 3x – 4 = 0

⇒ x² + 4x – x – 4 = 0

⇒ x(x + 4) – 1(x + 4) = 0

⇒ (x – 1)(x + 4) = 0

⇒ (x – 1) = 0 and (x + 4) = 0

Therefore, x = 1, −4. Considering x = 1, then from Eq. (2), we get y = √3, -√3. 

Thus, (1, √3) and (1, -√3) are the points of intersection of parabola and circle. 

The required area (A) is the area of the shaded region shown in the figure. Therefore,

Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows: