As a simple example, consider the graph of the function for

If we take the region between the graph and the x-axis and revolve it about the x-axis, we obtain a solid

To help you in plotting surfaces of revolution, A Maple procedure
called `revolve` has been written. The commands used to produce
the graphs are shown below. The `revolve` procedure, as well as
the `RevInt`, `LeftInt`, and `LeftDisk` procedures
described below are all part of the `CalcP7` package, which must
be loaded first. The last line in the example below shows the
optional argument for revolving the graph of about the line
instead of the default .

> with(CalcP7): > f := x -> x^2+1; > plot(f(x),x=-2..2,y=0..5); > revolve(f(x),x=-2..2); > revolve(f(x),x=-2..2,y=-2);

The `revolve` command has other options that you should read about
in the help screen. For example, you can speed the command up by only
plotting the surface generated by revolving the curve with the ` nocap` argument, and you can also plot a solid of revolution formed
by revolving the area between two functions. Try the following
examples. (Note: The last example shows how to use `revolve` with
a piecewise defined function using the `piecewise` command.)

> revolve({f(x),0.5},x=-2..2,y=-1); > revolve(cos(x),x=0..4*Pi,y=-2,nocap); > revolve({5,x^2+1},x=-2..2); > g := x-> piecewise(x<0,-x+1/2,x^2-x+1/2); > revolve(g(x),x=-1..2);

It turns out that the volume of the solid obtained by revolving the
region between the graph and the -axis about the -axis can be
determined from the integral

to have the value . More generally, if you revolve the area under the graph of for about the -axis, the volume is given by

Where does this formula come from? The integral formula given above for the volume of a solid of revolution comes, as usual, from a limit process. Recall the rectangular approximations we used for plane regions. If you think of taking one of the rectangles and revolving it about the x-axis, you get a disk whose radius is the height of the rectangle and thickness is , the width of the rectangle. The volume of this disk is . If you revolve all of the rectangles in the rectangular approximation about the x-axis, you get a solid made up of disks that approximates the volume of the solid of revolution obtained by revolving the plane region about the x-axis.

To help you visualize this approximation of the volume by disks, the
`LeftDisk` procedure has been written. The syntax for this
command is similar to that for `revolve`, except that the number
of
subintervals must be specified. The examples below produce
approximations with five and ten disks. The latter approximation is
shown in the graph below. The `LeftInt` command adds the disks.

> LeftDisk(f(x),x=-2..2,5); > LeftInt(f(x),x=-2..2,5); > LeftDisk(f(x),x=-2..2,10); > LeftInt(f(x),x=-2..2,10);

In order to calculate the volume of a solid of revolution use the `int` command implementing the formula above.

> Pi*int(f(x)^2,x=-2..2); > evalf(Pi*int(f(x)^2,x=-2..2));

- For the function
over the interval
,
**A**- Plot over the given interval.
**B**- Plot the approximation of the solid of revolution using
`LeftDisk`with 12 disks. **C**- Plot the solid formed by revolving about the -axis using the
`revolve`command. **D**- Find the exact volume of the solid of revolution and label your output
`exact`. **E**- Find the number of subintervals needed to approximate the
volume of the solid of revolution about the -axis using
`LeftInt`with error no greater than 0.1.

- Consider the funnel formed by revolving the curve
over the interval .
- A)
- Plot the solid of revolution.
- B)
- Consider the same funnel over the interval . Without plotting, find the volume of this funnel as
. Use the command
`value`instead of`evalf`.

- A doughnut is to be made by revolving a circle of radius centered at the origin about the line.
- A)
- Write the equation of the circle using two functions for the top and bottom halves.
- B)
- Revolve your two functions about the line. (To make it look like a donut right-click on the graph and click on scaling=constrained)
- C)
- To find the volume we need to shift the functions so the picture revolves about the x-axis. Rewrite your two functions shifting each 2 units.
- D)
- Revolve your new functions to show that it is the same donut just shifted.
- E)
- Find the volume of the donut.

2013-11-20