Question

What is the surface area of the solid created by revolving `f(x) = 2x^2-6x+18 , x in [2,3]` around the x axis?

Answer 1

`A=94pi/3`

Explanation:

Given:

`f(x)=2x^2-6x+18`
The surface of solid is made up of number of rings with centers lying on x axis and having radius as ordinate of the function.
The area of the ring when advanced through an increment of dx is given by the product of the perimeter with the advancement dx

`dA=2pif(x)dx=2pi(2x^2-6x+18)dx`

The area enclosed between the limits x=2, and x=3 can be obtained by integrating the above expression between the limits x=2 to x=3

`intdA=int2pi(2x^2-6x+18)dx`between x=2 and 3

`=2pi(2/3x^3-6/2x^2+18x)`

`2pi(2/3(3^3-2^3)-3(3^2-2^2)+18(3-2))`

`=pi(4/3(27-8)-6(9-4)+36(3-2))`

`=pi(4/3xx19-6xx5+36xx1)`

`=pi/3xx(4xx19-30xx3+36xx3)`

`=pi/3xx(76-90+108)`94
`=pi/3xx94`
`A=94pi/3`