# What is the surface area of the solid created by revolving #f(x) = 2x^2-4x+8 , x in [1,2]# around the x axis?

##### 1 Answer

May 29, 2016

approximately

#### Explanation:

The surface area

#S=2piint_a^bf(x)sqrt(1+[f'(x)]^2)dx#

Using

#S=2piint_1^2(2x^2-4x+8)sqrt(1+(4x-4)^2)dx#

Actually integrating this is very complex and far beyond the scope of this problem, so put this into a calculator--be mindful of parentheses.

The answer you receive (don't forget to multiply by

#Sapprox100.896#