Subsections

• Introduction
• Background
• Finding Volumes of Revolution
• Exercises

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. The specific properties that we wish to study are their graphs and volume. Some common volume formulas that we have seen before are the volume of a sphere, cylinder, or a cone, but we have never seen these formulas proven.

Background

So far we have used the integral mainly 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 $-2\leq x \leq 2$, which appears in Figure 1.

 

Figure 1:  Plot of f(x)=x²+1. Note: y-axis doesn't cross x-axis at the origin

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)

 

Figure 2:  Solid generated by rotating f(x)=x²+1 about the x-axis.

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 to discover how some of the options work. (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

$$\pi \int_{-2}^2 (x^2+1)^2 \, dx$$

to have the value $\frac{412}{15} \pi$. More generally, if you revolve the area under the graph of g(x) for $a \leq x \leq b$ about the x-axis, the volume is given by

$$\pi \int_{a}^{b} (g(x))^2 \, dx.$$

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 $\Delta x$, the width of the rectangle. The volume of this disk is $\pi h^2 \Delta x$. 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.

 

Figure 3:  Cylindrical approximation with n=10 to the solid generated by rotating f(x)=x²+1 about the x-axis.

Finding Volumes of Revolution

Since this involves straightforward integration, one may simply issue commands such as:

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

  > vol:= RevInt(f(x),x=0..9);

  > evalf(vol);

Exercises

1.

For each function below,

(a)

Plot its surface of revolution using the command LeftDisk for n=4 and n=10 and then plot the surface of revolution using revolve.

(b)

Approximate the volume of the solid of revolution about the x-axis for each of the n-values given above and then find the exact value. Explain why it would be difficult to do these calculations by hand and tell why you are grateful to have Maple do it for you.

$\bullet \;\;f(x) = \ln (x^2+1),\;\;\;0 \leq x \leq 2$
$\bullet\;\; \displaystyle g(x) = 1 + \frac{\sin x}{e^x},\;\;\; -\pi \leq x \leq \pi$
$\bullet \;\; \displaystyle h(x) = \frac{x}{\sqrt{x^2+1}},\;\;\; 1 \leq x \leq 3$

2.

The formula for the volume of a cone, $$V=\frac{\pi r^2 h}{3}$$ can be obtained by revolving the line passing through the points (0,0) and (h,r) about the x-axis over the interval $0 \leq x \leq h$. Follow a similar approach to derive the formula for the volume of a frustrum of a cone of length h having r₁ as the radius of the top and r₂ as the radius of the base. (A frustrum is a cone with the top cut off parallel to the base.) Show that the formula can be expressed as $$V=\frac{\pi h}{3} (r_1^2 +r_1r_2 +r_2^2)$$.

3.

This exercise deals with custom-making your own wedding band by revolving a region shown in the figure below about the x-axis. The resulting wedding band will have inner radius r (usually between 6 and 12mm), minimum thickness t (usually between .5 and 1.5mm) and width w (usually between 4 and 10mm).

(a)

If the customer asks the price of a wedding band with given dimensions r, t, and w, the jewler must first calculate the volume of the desired band to determine how much gold will be required to make it. Derive a formula for the volume which will be given in cubic millimeters in factored form using the factor command. (Hint: you will need to find the radius of the circle such that the intersection of the circle with the line will be at $$x = \frac{w}{2}$$)

(b)

Suppose that the jeweler plans to charge $1000 per troy ounce of alloy used to make the ring. The inner radius r of the wedding band is determined by the measurement of the customer's finger (in millimeters). Suppose that the jewler makes all wedding bands with t=1(mm). Then, for a given acceptable cost C (in dollars), the customer wants to know the maximum width of the band he or she can afford. Calculate the cost of a wedding band for a ring with inner radius of 10mm and width 6mm. Use the following conversion for the cost per cubic millimeter:

$$\frac{\$1000}{troy~ounce}\frac{12~troy~ounces}{453.59~gm}\frac{18.4~gm}{cm^3}\frac{1cm^3}{1000mm^3}$$

(c)

Measure your own ring finger to determine r (you can measure its circumfrence $2\pi r$ in millimeters with a piece of string and then divide by $2\pi$). Then choose a cost in the $100 to $500 price range. Use your formula for volume with t=1 and the conversion formula to find the maximum width w for of a band that costs C dollars (at $1000/oz). Use the maple solve command to solve for the maximum width w. Plot the ring that you have designed for yourself.