**Solve the following :**

Find the area of the region lying between the parabolas :

y^{2} = 4x and x^{2} = 4y

#### Solution

For finding the points of intersection of the two parabolas, we equate the values of y^{2} from their equations.

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

∴ y = `x^4/(16)`

∴ `x^4/(16)` = 4x

∴ x^{4} – 64x = 0

∴ x(x^{3} – 64) = 0

∴ x = 0 or x^{3} = 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 y^{2} = 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 ofthe region ODABO

= area under the rabola x^{2} = 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"`.