Prove that the curves y^{2 }= 4x and x^{2} = 4y divide the area of square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

#### Solution

A(OABC) = 4 × 4 = 16 sq. units

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

`(x^2/4)^2 = 4x`

or `x^4/16 = 4x`

or x^{4} - 64x = 0

or x(x^{3} - 64) = 0

or x = 0 or x = 4

when x = 0, y = 0

x = 4, y = 4

Point of intersection of the two parabolas is (0, 0) and (4, 4).

Area of part III = `int_0^4 y dx` (parabola x^{2} = 4y)

= `int_0^4 (x^2)/4 dx = [1/4 x^3/3]_0^4`

= `1/12(64 - 0)`

= `64/12`

= `16/3` sq.units

Area of I = Area of square - Area of II and III

= `16 - int_0^4 sqrt(4x) dx`

= `16 - (2 × 2)/3 [x^(3/2)]_0^4`

= `16 - 32/3` sq. units

= `16/3` sq. units

Area of II = Area of square - Area of I - Area of III

= 16 - `16/3 - 16/3` sq. units

= `16/3` sq. units

The two curves divide the square into three equal parts.