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

##### 1 Answer

Aug 26, 2017

#### Explanation:

The first step is to check to make sure that

The surface area of a curve on

#S = 2piint_1^2 1/x^2sqrt(1 + 4/x^6)dx#

#S = 2piint_1^2 1/x^2sqrt((x^6 + 4)/x^6)dx#

#S = 2piint_1^2 1/x^5sqrt(x^6 + 4) dx#

This integral has no elementary solution.

But according to Wolfram Alpha, the surface area is approximately

Hopefully this helps!