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
Noah G
Aug 26, 2017

#S~~ 4.46#

Explanation:

The first step is to check to make sure that #f(x)# is defined on the interval #[1, 2]#. The only discontinuity of the the function is at #x= 0#, so we do not have to concern ourselves with this.

The surface area of a curve on #[a, b]# rotated around the x-axis is defined by #S = 2piint_a^b f(x)sqrt(1 + (f'(x))^2)dx#. The derivative of #f(x)# is #f'(x) = -2/x^3#. This means that #(f'(x))^2 = (-2/x^3)^2 = 4/x^6#.

#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 #4.46# square units.

Hopefully this helps!

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