Question

Find the area of the region lying between the parabolas y² = 4x and x² = 4y.

Answer 1

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.

From the equation x² = 4y, `y = x^2/(4)`

∴ `y = x^4/(16)`

∴ `x^4/(16) = 4x`

∴ x⁴ – 64x = 0

∴ x(x³ – 64) = 0

∴ x = 0 or x³ = 64

i.e. x = 0 or x = 4

When x = 0, y = 0

When x = 4, y = `(4^2)/(4)` = 4

∴ The points of intersection are O(0, 0) and A(4, 4).

Required area = area of the region OBACO

= [area of the region ODACO] – [area of the region ODABO]

Now, area of the region ODACO

= area under the parabola y² = 4x,

i.e. `y = 2sqrt(x)` between x = 0 and x = 4

= `int_0^4 2sqrt(x)  dx`

= `[2 (x^(3/2))/(3/2)]_0^4`

= `2 xx (2)/(3) xx 4^(3/2) - 0`

= `(4)/(3) xx (2^3)`

= `(32)/(3)`

Area of the region ODABO

= area under the parabola x² = 4y,

i.e. `y = x^2/(4)` between x = 0 and x = 4

= `int_0^4 (1)/(4)x^2 dx`

= `(1)/(4)[x^3/(3)]_0^4`

= `(1)/(4)(64/3 - 0)`

= `(16)/(3)`

∴ Required area = `(32)/(3) - (16)/(3) = (16)/(3)"sq units"`.