What is the surface area of the solid created by revolving f(x) = xe^-x-e^(x) , x in [1,3] around the x axis?

1 Answer
Mar 29, 2018

The surface area is approximately 105.75.

Explanation:

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