Question

Find the area of the region bounded by the curve y² = 4x, x² = 4y.

Answer 1

We have y² = 4x and x² = 4y.

y = `x^2/4`

⇒ `(x^2/4)^2` = 4x

⇒ `x^4/16` = 4x

⇒ x⁴ = 64x

⇒ x⁴ – 64x = 0

⇒ x(x³ – 64) = 0

∴ x = 0, x = 4

Required area = `int_0^4 sqrt(4x)  "d"x - int_0^4  x^2/4  "d"x`

= `2 int_0^4 sqrt(x)  "d"x - 1/4 int_0^4 x^2  "d"x`

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

= `4/3 [(4)^(3/2) - 0] - 1/12 [(4)^3 - 0]`

= `4/3 [8] - 1/12[64]`

= `32/2 - 16/3`

= `16/3` sq.units

Hence, the required area = `16/3` sq.units