Question

What is the surface area of the solid created by revolving `f(x) = 6x^2-3x+22 , x in [2,3]` around the x axis?

Answer 1

`14086/5 pi`

Explanation:

around the x axis, the volume of revolution is

`V = pi int_{x=2}^3 y^2 \ dx`

that looks horrendous and might be made simpler by a shift in axis. completing the square should ultimately make the algebra a little bit easier to bear

so `y(x) = 6 (x^2 - 1/2 x + 22/6)`
`= 6 ((x - 1/4)^2 - 1/16 + 22/6)`
`= 6 ((x - 1/4)^2 + 173/48)`
so `y(u) = 6 u^2 + 173/8` where `u = x - 1/4, du = dx`

and the integral becomes

`pi int_{u=7/4}^{11/4} y^2 \ du`
`= pi int_{u=7/4}^{11/4} ( 6 u^2 + 173/8)^2 \ du`
`= pi int_{u=7/4}^{11/4} 36 u^4 + 519/2 u^2+ 29929/64 \ du`
`= pi ( 36/5 u^5 + 519/6 u^3+ 29929/64 u)_{u=7/4}^{11/4}`
`= 14086/5 pi`