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Find the area between the parabolas y^{2} = 5x and x^{2} = 5y

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#### Solution

Given equations of the parabolas are y^{2} = 5x ......(i)

and x^{2} = 5y.

∴ y = `x^2/5` ......(ii)

From (i), we get

y = `sqrt(5x)` ......(iii) [∵ In first quadrant, y > 0]

Find the points of intersection of y^{2} = 5x and x^{2} = 5y.

Substituting (ii) in (i), we get

`(x^2/5)^2` = 5x

∴ x^{4} = 125x

∴ x^{4} – 125x = 0

∴ x(x^{3} – 125) = 0

∴ x = 0 or x^{3} = 125 = 5^{3}

∴ x = 0 or x = 5

When x = 0, y = 0 and when x = 5, y = 5

∴ The points of intersection of y^{2} = 5x and x^{2} = 5y are O(0, 0) and B(5, 5).

Draw BD ⊥ OX

Required area = area of the region OABCO

= area of the region ODBCO – area of the region ODBAO

= area under the parabola y^{2} = 5x – area under the parabola x^{2} = 5y

= `int_0^5 sqrt(5x) "d"x - int_0^5 x^2/5 "d"x` ......[From (iii) and (ii)]

= `sqrt(5) int_0^5 x^(1/2) "d"x - 1/5 int_0^5 x^2 "d"x`

= `sqrt(5)[(x^(3/2))/(3/2)]_0^5 - 1/5[x^3/3]_0^5`

= `(2sqrt(5))/3 [(5)^(3/2) - 0] - 1/15 [(5)^3 - 0]`

= `(2sqrt(5))/3 (5sqrt(5)) - 1/15 (125)`

= `50/3 - 25/3`

= `25/3` sq.units

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