What is the surface area of the solid created by revolving #f(x) =2x^3-2 , x in [2,4]# around the x axis?

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Answer 1 · 2017-02-26T17:19:55

#2piint_2^4(2x^3-2)sqrt(1+36x^4)dxapprox49266.1936#

The surface area of a solid created by rotating a function #f# around the #x# axis on #x in[a,b]#is given by:

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

Here, #f(x)=2x^3-2# so #f'(x)=6x^2#. Then:

#S=2piint_2^4(2x^3-2)sqrt(1+(6x^2)^2)dx#

#S=2piint_2^4(2x^3-2)sqrt(1+36x^4)dx#

This cannot be easily integrated, so plug this into a calculator.

#S=49266.1936...#