Volumes+of+solids+of+revolution

=﻿__Volume of Solids of Revolution__= Up to this point, we have used integration to find the areas under curves. So how can these methods be applied for use on 3-dimensional shapes? One of the simplest applications of integrals in 3D is finding the area of Solids of Revolution: a shape constructed by revolving a known shape around an axis.

To do this, we will take the integral of the areas of the circles that will be created when the original curve is revolved. When the areas of these circles are added, we will have the Volume of the solid. The radius of these circles is given by the curve that bounds the solid.

The region bounded by the curves x=0, x=5, and y=2 is revolved about the x-axis. What is the volume of the solid that is created?
 * Example One **

Solution: The area of each circle is A=p r 2 =p (2) 2 =4p. (With a radius of 2 due to the y=2 line) The region goes from x=0 to x=5, so we will add up the areas of the circles from x=0 to 5: 

The region bounded by the curves: x=0, x=7 and y=(1/7)x 2 +2 is revolved about the x-axis. What is the volume of the resulting solid?
 * Example Two **

Solution: In this problem, the radius is defined by the y=(1/7)x 2 +2 curve. Note that in this example, the radius will be changing continuously. We have the boundaries of 0<x<7 and the radius, so we can now sum the areas of the circles to find the Volume (here we use a calculator):

//Solids with Holes//
If the region that is being revolved does not border the axis of revolution, then a hole will be created in the 3D figure. In order to correctly find the Volume, we will find the integral of the areas just as we have up to this point, but we will then subtract out the integral of the area of the hole. In other words, we will find the Volume of the solid created by the outermost function, and subtract the Volume of the innermost solid.

This can be written mathematically as:

Find the Volume of the solid created by revolving the region bounded by x=0, x=5, y=1 and y=2 about the x-axis.
 * Example Three **

Solution: The region is the area between y=1 and y=2 for the interval 0<x<5. To we will find the volume of the outer solid (bounded by y=2) and subtract the Volume of the inner solid (y=1). (We will be using the answer from Example One for the Volume of the outer solid)

//Beyond the x-axis//
These techniques can be applied to any axis of revolution, not just the x-axis. For example, if we were to revolve Example Three around the line y=4, we would use the same method, but the radii will be changed to reflect the new axis of revolution: The outer radius would be of length 3 (from line y=1) and the inner radius would be of length 2 (from line y=2).

The axis of revolution need not be of the form y=k either. Lets try an example of revolving a region about the y-axis:

**Example Four** What is the Volume of the region created by revolving the region bounded by y=1, y=4, y=x, and y=2x about the y-axis?

Solution: For this problem (and any problem where the axis of revolution is of the from x=k), we want every equation to have x as the dependent variable, so we will rewrite the last equation to x=(1/2)y. Notice that the line x=(1/2)y is closer to the y-axis than x=y, so x=(1/2)y will be our inner radius function and x=y will be the outer radius function. Now we will take the integral on the interval 1<y<4: