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