What is the surface area of the solid created by revolving #f(x)=-2x^3-3x^2+6x-12# over #x in [2,3]# around the x-axis?

1 Answer
Jan 20, 2016

Use the surface area of rotation integral:
#SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx#
Integrate the following:
#SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx#

Explanation:

#SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx#
may look bad it is not really. Take the derivative of f(x)
#(df(x))/dx = -6x^2 -6x+6 #
now integrate

#SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx#