The surface area #S# is given by
#S=2piint_1^3xe^-x-e^xdx=2pi[int_1^3xe^-xdx-int_1^3e^xdx]#.
#intxe^-x=-e^-x(x+1)+C#
which can be obtained by integration by parts, so
#S=2pi[-e^-x(x+1)-e^x]# evaluated from 1 to 3.
#S=2pi[-e^-3*4-e^3+e^-1*2+e]~~-105.75#
To make this physically meaningful, we would just take the absolute value of this and say that
#S~~105.75#.
The volume, #V#, can be calculated from #piint_1^3(xe^-x-e^x)^2dx#.
#piint_1^3(xe^-x-e^x)^2dx=piint_1^3x^2e^(-2x)-2x+e^(2x)dx#
#=pi(int_1^3x^2e^(-2x)dx-2int_1^3xdx+int_1^3e^(2x)dx)#
The tough one here is #intx^2e^(-2x)dx#. This is done by integration by parts. If you need me to show you the steps for you, just indicate in the comments. For now, I will simply note that
#intx^2e^(-2x)dx=-1/4e^(-2x)(2x^2+2x+1)+C#
And so we have
#V=pi[-1/4e^(-2x)(2x^2+2x+1)-x^2+e^(2x)/2]# evaluated from 1 to 3.
#V=pi[-e^(-6)/4*25-9+e^6/2+e^(-2)/4*5+1-e^2/2]~~597.45#
