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Subsections

# Solids of Revolution

## Introduction

The purpose of this lab is to use Maple to study solids of revolution. Solids of revolution are created by rotating curves in the x-y plane about an axis, generating a three dimensional object. They are discussed in Chapter 6 of Calculus by Varberg and Purcell (sections 2 and 3). The specific properties of them that we wish to study are their volume, surface area, and graph.

## Background

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) = x2+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 -> 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 in Figure 1 between the graph and the x-axis about the x-axis can be determined from the integral

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

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.

## Finding Volumes of Revolution

Since this involves straightforward integration, one may simply issue commands such as:
> f:= x-> sqrt(x) +1;

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

> evalf(vol);

## Surface Area

Assuming the function has been defined, one might issue:
> 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 f(x) about the x-axis.

## Exercises

1.
Compute the volume of the solid generated by revolving the region bounded by the x-axis and the graph of the function

for . Then find the surface area of the solid.
2.
Four years ago, Kevin Nordberg and James Rush (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.

They obtained the shape of their glass by revolving this function about the x axis over the interval [-1,2]. The Maple command they used to define this function is given below.

> f := x -> piecewise(x<-3/4,-2*x-1/2,x<0,1/6,x^(2/3)+1/6);

Plot this function (without revolving it) over the interval [-1,2] 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 Kevin and Jim 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.)

3.
A doughnut is to be made by revolving a circle of radius r centered at the point (0,a) about the x axis. You may assume that a > r. Find the volume of this doughnut as a function of the parameters r and a. Give reasonable values for r and a for a real doughnut and plot your doughnut. (Hint - the equation for the circle is x2 + (y-a)2 = r2. You will need to solve this equation for y, which will give you the two functions you need.

4.
(extra credit) Consider the function f that Nordberg and Rush used to define their glass. Is this function continuous? Your answer should be no. Can you find a function F(x) such that

where F is continuous and differentiable, except at one point? Furthermore, wherever F is differentiable, F'(x) = f(x). How is F influenced by f(x)=1/6 for ? Does the definition of f(x) matter over this subinterval? Why or why not? Graph F(x) corresponding to the given f(x). Then change the definition of f(x) by only changing the middle function. How does F(x) change?

Next: About this document ... Up: Labs and Projects for Previous: Labs and Projects for

Jane E Bouchard
11/16/1999