Find the area between the parabolas y2 = 5x and x2 = 5y - Mathematics and Statistics

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Sum

Find the area between the parabolas y2 = 5x and x2 = 5y

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Solution


Given equations of the parabolas are y2 = 5x   ......(i)

and x2 = 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 y2 = 5x and x2 = 5y.

Substituting (ii) in (i), we get

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

∴ x4 = 125x

∴ x4 – 125x = 0

∴ x(x3 – 125) = 0 

∴ x = 0 or x3 = 125 = 53

∴ x = 0 or x = 5

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

∴ The points of intersection of y2 = 5x and x2 = 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 y2 = 5x – area under the parabola x2 = 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

Concept: Standard Forms of Parabola and Their Shapes
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Chapter 1.7: Application of Definite Integration - Q.3
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