included between parallel planes, if every plane cross-section parallel to the given planes has areas in the same ratio in both solids, then the volumes of the solids are in that ratio.
In particular, if the areas of the sections are always equal, then so are the volumes.
The principle is deceptively powerful and before we discuss Torricelli’s use of it with the trumpet we will acclimatize ourselves to it by looking at a famous example of its use: computing the volume of a sphere (knowing the volume of a circular cone).
On the left of figure 8.2 is a hemisphere of radius r with a horizontal section at height h above the base of the hemisphere. On the right is a cylinder of radius r and of height r with another horizontal section cut at height h above its base. Within the cylinder is inscribed a circular cone with base the top of the cylinder and vertex at the centre of the cylinder’s base. The area of the horizontal circular section within the hemisphere is π ( r 2 − h 2 ). Since the height of the cone at this level is h , so must be its base radius. This means that the area of the annular region withinthe cylinder is πr 2 − πh 2 = π ( r 2 − h 2 ) also. Using Cavalieri’s Principle it must be that the volume of the hemisphere is equal to the volume of the cylinder minus the volume of the cone. We have, then, that the volume of the hemisphere is
Figure 8.2. Cavalieri’s Principle at work.
Figure 8.3. Torricelli’s use of Cavalieri’s Principle which demonstrated the infinite volume.
and the volume of the sphere is
Torricelli used an extended form of the principle to show that the volume of the infinite trumpet is itself finite. To appreciate his proof we must imagine the trumpet to have a ‘lip’ at its open end and be made up of an infinite number of concentric, horizontal cylinders. figure 8.3 shows a particular example of such a cylinder and let us suppose that the lip begins at the point (1,1).
Figure 8.4.
Suppose now that the right end of a particular cylinder touches the hyperbola in the xy -plane at the point ( x 0 , y 0 ), then the cylinder has curved surface area (2 πy 0 ) x 0 = 2 πx 0 y 0 = 2 π since y 0 = 1/ x 0 . This area is, therefore, constant as the point of contact varies. Now unravel the cylinder so that it becomes a horizontal rectangle located at height y 0 , as shown in figure 8.4 (a). Finally, construct a vertical cylinder made up of horizontal discs each of area 2 π at height y 0 from its base, as shown in figure 8.4 (b). The height of this cylinder will be 1. Cavalieri’s Principle is itself a limiting argument, if that is now extended to allow x 0 → ∞ and so approach the base of the vertical cylinder, the volume of the trumpet is the same as the volume (2 π ) of that cylinder.
Of course, this argument is special. It works because xy = 1 and could not be easily extended to other cases, but it does work and it did shock. Torricelli himself said
It may seem incredible that although this solid has an infinite length, nevertheless none of the cylindrical surfaces we considered has an infinite length but all of them are finite.
If we look at the problem through modern eyes, taking the finite trumpet from x = 1 to x = N (forgetting the lip) and then allowing N to become arbitrarily large, we have that the volume of the trumpet is given by
and as N → ∞ the volume approaches π .
The simple calculation
shows that we have a solid of infinite cross-sectional area but finite volume.
To calculate the surface area of the finite solid and so prove that it is also infinite requires more e3ort.
The Trumpet’s Surface Area
Using the standard formula for the surface area of a volume of revolution, as described in appendix C:
Here we have y = 1/ x and so
so the formula becomes
This clearly diverges since
but it is pleasing, if a little messy, to find an exact form for the integral
We will attack it in two stages: first using integration by parts and then
In particular, if the areas of the sections are always equal, then so are the volumes.
The principle is deceptively powerful and before we discuss Torricelli’s use of it with the trumpet we will acclimatize ourselves to it by looking at a famous example of its use: computing the volume of a sphere (knowing the volume of a circular cone).
On the left of figure 8.2 is a hemisphere of radius r with a horizontal section at height h above the base of the hemisphere. On the right is a cylinder of radius r and of height r with another horizontal section cut at height h above its base. Within the cylinder is inscribed a circular cone with base the top of the cylinder and vertex at the centre of the cylinder’s base. The area of the horizontal circular section within the hemisphere is π ( r 2 − h 2 ). Since the height of the cone at this level is h , so must be its base radius. This means that the area of the annular region withinthe cylinder is πr 2 − πh 2 = π ( r 2 − h 2 ) also. Using Cavalieri’s Principle it must be that the volume of the hemisphere is equal to the volume of the cylinder minus the volume of the cone. We have, then, that the volume of the hemisphere is
Figure 8.2. Cavalieri’s Principle at work.
Figure 8.3. Torricelli’s use of Cavalieri’s Principle which demonstrated the infinite volume.
and the volume of the sphere is
Torricelli used an extended form of the principle to show that the volume of the infinite trumpet is itself finite. To appreciate his proof we must imagine the trumpet to have a ‘lip’ at its open end and be made up of an infinite number of concentric, horizontal cylinders. figure 8.3 shows a particular example of such a cylinder and let us suppose that the lip begins at the point (1,1).
Figure 8.4.
Suppose now that the right end of a particular cylinder touches the hyperbola in the xy -plane at the point ( x 0 , y 0 ), then the cylinder has curved surface area (2 πy 0 ) x 0 = 2 πx 0 y 0 = 2 π since y 0 = 1/ x 0 . This area is, therefore, constant as the point of contact varies. Now unravel the cylinder so that it becomes a horizontal rectangle located at height y 0 , as shown in figure 8.4 (a). Finally, construct a vertical cylinder made up of horizontal discs each of area 2 π at height y 0 from its base, as shown in figure 8.4 (b). The height of this cylinder will be 1. Cavalieri’s Principle is itself a limiting argument, if that is now extended to allow x 0 → ∞ and so approach the base of the vertical cylinder, the volume of the trumpet is the same as the volume (2 π ) of that cylinder.
Of course, this argument is special. It works because xy = 1 and could not be easily extended to other cases, but it does work and it did shock. Torricelli himself said
It may seem incredible that although this solid has an infinite length, nevertheless none of the cylindrical surfaces we considered has an infinite length but all of them are finite.
If we look at the problem through modern eyes, taking the finite trumpet from x = 1 to x = N (forgetting the lip) and then allowing N to become arbitrarily large, we have that the volume of the trumpet is given by
and as N → ∞ the volume approaches π .
The simple calculation
shows that we have a solid of infinite cross-sectional area but finite volume.
To calculate the surface area of the finite solid and so prove that it is also infinite requires more e3ort.
The Trumpet’s Surface Area
Using the standard formula for the surface area of a volume of revolution, as described in appendix C:
Here we have y = 1/ x and so
so the formula becomes
This clearly diverges since
but it is pleasing, if a little messy, to find an exact form for the integral
We will attack it in two stages: first using integration by parts and then
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