Question

What is the surface area of the solid created by revolving `f(x)=-2x^3-3x^2+6x-12` over `x in [2,3]` around the x-axis?

Answer 1

Use the surface area of rotation integral:
`SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx`
Integrate the following:
`SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx`

Explanation:

`SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx`
may look bad it is not really. Take the derivative of f(x)
`(df(x))/dx = -6x^2 -6x+6`
now integrate

`SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx`