How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #y=e^x# on the interval #0<=x<=1# ?

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Answer 1 · 2014-09-16T06:09:54

The answer is #pi/2[e^2-1]#.

Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method. The disk method is:

#V=int_a^b Adx#
#=int_a^b pi r^2dx#
#=int_a^b pi [f(x)]^2dx#

We have the known values:

#f(x)=e^x#
#a=0#
#b=1#

And now we can substitute:

#V=int_0^1 pi (e^x)^2dx#
#=pi int_0^1 e^(2x)dx#
#=pi (e^(2x))/2|_0^1#
#=pi/2[e^2-e^0]#
#=pi/2[e^2-1]#