Question

Using integration, find the area of the region bounded by the curve y² = 4x and x² = 4y.

Answer 1

Given that the curves are y² = 4x and x² = 4y.

Now, the graph of the provided curves is as follows:

The given equations are:

y² = 4x   ...(i)

And x² = 4y

`y = x^2/4`  ...(ii)

Put the value of (ii) in (i), we get

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

`\implies x^4/16 = 4x`

`\implies` x⁴ = 4 × 16x

`\implies` x⁴ – 64x = 0

`\implies` x(x³ – 64) = 0

`\implies` x = 0 or x = 4

The curve is rewritten as follows:

y² = 4x

`\implies y = sqrt(4x) = 2sqrt(x)`

`\implies` x² = 4y

`\implies y = x^2/4`

Now, the area of the bounded region is given as:

A = `int_0^4 (2sqrt(x) - x^2/4)dx`

= `[2 xx x^(3/2)/(3/2) - x^3/12]_0^4`

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

= `[(4 xx 8)/3 - 64/12]`

= `(128 - 64)/12`

= `64/12`

= `16/3` sq.units