Find the area of the region lying between the parabolas 4y2 = 9x and 3x2 = 16y - Mathematics and Statistics

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Sum

Find the area of the region lying between the parabolas 4y2 = 9x and 3x2 = 16y

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Solution

Given equations of the parabolas are

4y2 = 9x       .......(i)

and 3x2 = 16y 

∴ y = `(3x^2)/16`    ......(ii)

From (i), we get

y2 = `9/4x`

∴ y = `3/2 sqrt(x)`  ....(iii)  ......[∵ In first quadrant, y > 0]

  Find the points of intersection of 4y2 = 9x and 3x2 = 16y.

Substituting (ii) in (i), we get

`4((3x^2)/16)^2` = 9x

∴ x4 = 64x

∴ x4 – 64x = 0

∴ x(x3 – 64) = 0

∴ x = 0 or x3 = 64 = 43

∴ x = 0 or x = 4

When x = 0, y = 0 and when x = 4, y = 3

∴ The points of intersection are O(0, 0) and B(4, 3).

Draw BD ⊥ OX.

Required area = area of the region OABCO

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

= area under the parabola 4y2 = 9x – area under the parabola 3x2 = 16y

= `int_0^4 3/2 sqrt(x)  "d"x - int_0^4  (3x^2)/16  "d"x`

= `3/2 int_0^4 x^(1/2)  "d"x - 3/16  int_0^4  x^2  "d"x`

= `3/2[(x^(3/2))/(3/2)]_0^4 - 3/16[x^3/3]_0^4`

= `[(4)^(3/2) - 0] - 1/16[(4)^3 - 0]`

= `8 - 1/16 (64)`

= 8 – 4

= 4 sq.units

Concept: Area Between Two Curves
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