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

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Answer 1 · 2017-01-20T12:51:04

#S = sqrt(2)pi #

If you consider the interval #(x,x+dx)# the length of the arc element of the curve #y=f(x)# is:

#dl = sqrt(dx^2+dy^2) = sqrt(dx^2 + f'(x)^2dx^2) = dx sqrt(1+f'(x)^2)#

so the area of the surface element generated by the rotation around the #x#-axis is:

#dS = 2piydl = 2pi f(x) sqrt(1+f'(x)^2) dx#

Integrating over the interval, the surface is then calculated as:

#S = 2pi int_1^2 f(x) sqrt(1+f'(x)^2) dx #

As #f'(x) = 1#:

#S = 2sqrt(2) pi int_1^2 (x-1)dx = sqrt2pi [(x-1)^2]_1^2 = sqrt(2)pi #