Find the volume of the function y = e^(-x^2) , from x=-1 to x=1, rotated around the line x = 5, using the method of shells.

The method of shells is based on breaking up the solid obtained by the revolution of the curve into thin cylindrical shells, and integrating the surface area of the shells to obtain the volume of the solid. 
The general formula for the shell method is
V = int 2pirhdr
where r is the radius and h is the height of the shells.
Since the revolution of the function y = e^(-x^2)
 in this example occurs around vertical line x = 5, this line will be the axis of the cylindrical shells, which passes through the center of the base of the cylinders. Then, the radius of each shell is r = 5 - x, the distance between the center (at x = 5) and the x-coordinate. Since the given curve is bounded by the points with the x-coordinates -1 and 1, the bounds of the integral will be -1 and 1.
The height of the each cylinder is the y-coordinate of each point on the function:
h = y = e^(-x^2)
Therefore, the resultant volume will be equal to the integral
V = int_(-1) ^1 2pi(5-x)e^(-x^2)dx
This integral breaks up into two:
V = 10piint_(-1) ^ 1 e^(-x^2) dx - 2piint_(-1) ^ 1 xe^(-x^2) dx
The first integral can only be evaluated numerically. The second integral is 0 because it is an integral of an odd function will symmetrical bounds.
The value of int_(-1) ^ 1 e^(-x^2)dx
is approximately 0.683, so 
V = 2pi*0.683 = 4.29
The volume of revolution of the given function is approximately 4.29.
 
 
http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx