Question

Calculate the volume of the solid of revolution generated by revolving the region bounded by the parabolas y 2 = 2 (x − 3) and y 2 = x about y = 0.

Answer 1

`9\pi`

Explanation:

given are two parabolas with vertex as (3,0) and (0.0)

`y^2 =2(x-3)\\y^2 =x`

These two intersect at x=6

Volume of II curve rotated about x axis - volume of I curve rotated about x axis = Volume of solid of revolution

For the second curve limits for x are from 0 to 6 and for I curve it is from 3 to 6

V2 =\pi
`\int\limits^6_0 {y^2} \, dx \\=\pi\int\limits^6_0 {x} \, dx\\= \pi(x^2)/(2) \\=18\pi`

V1 =
`\pi \int\limits^6_3 {y^2} \, dx\\\pi \int\limits^6_3 {2x-6} \, dx\\=\pi(x^2-6x)\\= \pi[(36-9)-6(6-3[)\\= (27-18)\pi\\=9\pi`

Volume of solid of revolutin = V2-V1 =
`9\pi`