Answer:
[tex]9\pi[/tex]
Step-by-step explanation:
given are two parabolas with vertex as (3,0) and (0.0)
[tex]y^2 =2(x-3)\\y^2 =x[/tex]
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 [tex]\int\limits^6_0 {y^2} \, dx \\=\pi\int\limits^6_0 {x} \, dx\\= \pi\frac{x^2}{2} \\=18\pi[/tex]
V1 =[tex]\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[/tex]
Volume of solid of revolutin = V2-V1 = [tex]9\pi[/tex]
