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