Question

Solve the following :

Find the area between the parabolas y² = 7x and x² = 7y.

Answer 1

Given equations of the parabolas are y² = 7x  ...(i)
and x² = 7y
∴ y = `x^2/(7)`        ...(ii)
From (i), we get
y = `sqrt(7)x`          ...(iii) [∵ In first quadrant, y > 0]
Find the points of intersection of y² = 7x and x² = 7y.
Subbstituting (ii) in (i) we get
`(x^2/7)^2` = 7x
∴ x⁴ = 343x
∴ x⁴ – 34x = 0
∴ x(x³ – 343) = 0
∴ x = 0 or x³ = 343 = 7³
∴ x = 0 or x = 7

When x = 0, y = 0 and when x = 7, y = 7
∴ The points of intersection of y2 = 7x and x2 = 7y are O(0, 0) and B(7, 7).
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 = 7x – area under the parabola x² = 7y

= `int_0^7 sqrt(7x)*dx - int_0^7 x^2/(7)*dx` ...[from (iii) and (ii)]

= `sqrt(7) int_0^7 x^(1/2) *dx - (1)/(7) int_0^7 x^2*dx`

= `sqrt(7)[x^(3/2)/(3/2)]_0^7 - (1)/(7)[x^3/3]_0^7`

= `(2sqrt(7))/(3)[97)^(3/2) - 0] - (1)/(21)[(7)^3 - 0]`

= `(2sqrt(7))/(3) (7sqrt(7)) - (1)/(21)(343)`

= `(98)/(3) - (49)/(3)`

= `(49)/(3)"sq. units"`.