Find the volume of the function y=e^(-x^2) , from x = -1 to x = 1, rotated about y = -3, using Disk/Washers method.

The method of disk or washers is based on breaking up the solid obtained by the revolution of the curve into either thin disks or thin washers (rings), and integrating their area to obtain the volume of the solid. When the function is rotated around a horizontal line, as in the given problem, the plane of the disks will be perpendicular the x-axis and the integrating will be over dx.
The general formula for the disk/washer method, when they are perpendicular to the x-axis, is
V = int A(x)dx , where A(x) is the area of the disks/washers. In case of disks, A(x) = pir^2 , where r is the radius of the disk.
Because in the given problem the function is rotated around y = -3, the line y = -3 will be the central axis of the disks and the radius of each disk will be
r = |-3 - y(x)| = |-3 - e^(-x^2)| = 3 + e^(-x^2)
Then, the integral becomes
V = int_(-1) ^1 pi(3+e^(-x^2))^2dx .
Simplifying this results in
V = 9pi int_(-1) ^1dx + 6pi int_(-1) ^1 e^(-x^2) dx + pi int_(-1) ^1 e^(-2x^2) dx
The first integral equals
9pi int_(-1) ^1 dx = 9pix | _(-1) ^ 1 = 18 pi
The other two integrals can only be taken numerically. The second integral is approximately
6pi*0.683 = 12.9
To take the third integral, first use substitution u = sqrt2x , du = sqrt(2)dx , which will turn it into
pi/sqrt2 int_(-sqrt2) ^ sqrt2 e^(-u^2)du = pi/sqrt2*1.692 = 3.76
Adding the values of the three integrals results in approximately 73.2
The volume of revolution of the given function is approximately 73.2.
http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx