Question

Find the area of the region bounded by the curve y² = 2x and x² + y² = 4x.

Answer 1

Equations of the curves are given by x² + y² = 4x  ......(i)

And y² = 2x   ......(ii)

⇒ x² – 4x + y² = 0

⇒ x² – 4x + 4 – 4 + y² = 0

⇒ (x – 2)² + y² = 4

Clearly it is the equation of a circle having its centre (2, 0) and radius 2.

Solving x² + y² = 4x and y² = 2x

x² + 2x = 4x

⇒ x² + 2x – 4x = 0

⇒ x² – 2x = 0

⇒ x(x – 2) = 0

∴ x = 0, 2

Area of the required region

= `2[int_0^2 sqrt(4 - (x - 2)^2)  "d"x - int_0^2 sqrt(2x)  "d"x]`

∴ Parabola and circle both are symmetrical about x-axis.

= `2[(x - 2)/2 sqrt(4 - (x - 2)^2) + 4/2 sin^-1  (x - 2)/2]_0^2 - 2*sqrt(2)* 2/3 [x^(3/2)]_0^2`

= `2[(0 + 0) - (0 + 2 sin^-1 (-1)] - (4sqrt(2))/3 [2^(3/2) - 0]`

= `-2 xx 2 * (- pi/2) - (4sqrt(2))/3 * 2sqrt(2)`

= `2pi - 16/3`

= `2(pi - 8/3)` sq.units

Hence, the required area = `2(pi - 8/3)` sq.units