# 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# ?

##### 1 Answer

Sep 16, 2014

The answer is

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]#