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So far we have used the integral mainly to to compute areas of plane regions.
It turns out that the definite integral can also be used to calculate
the volumes of certain types of three-dimensional solids. The class of
solids we will consider in this lab are called *Solids of
Revolution* because they can be obtained by revolving a plane region
about an axis.

As a simple example, consider the graph of the function *f*(*x*) = *x ^{2}*+1
for , which appears in Figure 1.

If we take the region between the graph and the x-axis and revolve it about the x-axis, we obtain the solid pictured in Figure 2.

To help you in plotting surfaces of revolution, A
Maple procedure called `revolve` has been written. The
command used to produce the graphs in Figures 1 and
2 is shown below. The `revolve` procedure, as well
as the `RevInt`, `LeftInt`, and
`LeftDisk` procedures described below are all part of the `
CalcP` package, which must be loaded first. The last line in the
example below shows the
optional argument for revolving the graph of *f*(*x*) about the line
*y*=-2 instead of the default *y*=0.

> with(CalcP):

> f := x -> x^2+1;

> plot(f(x),x=-2..2);

> 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 function defined piecewise.)

> 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 -> if x < 0 then -x +1/2 else x^2-x+1/2 fi ;

g := proc(x) options operator,arrow; if x < 0 then -x+1/2 else x^2-x+1/2 fi end

> revolve('g(x)',x=-1..2);

It turns out that the volume of the solid obtained by revolving the
region in Figure 1 between the graph and the *x*-axis
about the *x*-axis can
be determined from the integral

Where does this formula come from? To help you understand it, Two more
Maple procedures, `RevInt` and `LeftDisk`, have been written.
The procedure `RevInt` sets up the integral for the volume of a
solid of revolution, as shown below. The Maple commands `evalf`
and `value` can
be used to obtain a numerical or analytical value.

> RevInt(f(x),x=-2..2);

> value(RevInt(f(x),x=-2..2));

> evalf(RevInt(f(x),x=-2..2));

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 *h* 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 procedure 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 Figure 3.

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

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

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

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

The two `LeftInt` commands above add up the volumes in the disk
approximations of the solid of revolution.

> f:= x-> sqrt(x) +1;

> vol:= int(Pi*f(x)^2, x=0..9);

> evalf(vol);

> S:= int(2*Pi*f(x)*sqrt(1+D(f)(x)^2), x= 0..9);

> evalf(S);

to find the surface area of the solid obtained by rotating

- 1.
- Let and consider the interval . Plot the solid of revolution obtained by revolving the
graph of this function around the
*x*axis. Then compute the volume of the solid of revolution you obtained. Finally, use the`LeftInt`command and determine the number of subintervals needed to approximate the volume to within 0.1. - 2.
- Repeat the first exercise, this time using the function
*g*(*x*) = 1 +*x*/4. The number of subintervals required should be larger than in the previous exercise. Try to explain why this happens. Some things that might help you are to compare the graphs of the two functions, to look at plots with`LeftDisk`, and to examine the integrals that give the volumes. - 3.
- Compute the volume of the solid generated by revolving the
region bounded by the
*x*-axis, the graph of the function ,*x*=0, and*x*=10. - 4.
- Two years ago, Chris Zannella and Eric Pauly (both
class of '98) were asked to design a drinking glass by revolving a
suitable function about the
*x*axis. Here is the function they came up with.*x*axis over the interval [-0.9,4.5]. The Maple command they used to define this function is given below.> f := x -> if x<0 then x^4+0.088 elif x<3 then 0.088 else sin(x-3)+0.088 fi;

Plot this function (without revolving it) over the interval [-0.9,4.5] and identify the formula for each part of the graph. Then, revolve this function about the

*x*axis over the same interval and comment on the glass Eric and Chris designed. Finally, compute the volume of the part of this glass that could be filled with liquid, assuming the stem is solid. (Hint - your integral will involve only one of the formulas used to define the function.) - 5.
- Suppose that a tank for holding molasses is to be constructed as
a right circular
cone of height 10 meters and base radius 5 meters. It is to be
set up so that the point of the cone is at the bottom and the circular
base is at the top. If
*x*is the height of molasses in the cone, find and plot a function*v*(*h*) (where h is the height) which gives the volume of molasses in the tank.

9/14/1999