62. The volume of a sphere. In some works on solid geometry and in books on mensuration it is shown that the volume of a pyramid is equal to one third the product its base and altitude. Now suppose that a polyedron (i.e. a solid bounded by plane faces) is circumscribed about a sphere, each of the faces of the polyedron, accordingly, touching the sphere. This polyedron may be regarded as made up of pyramids which have a common vertex (namely, the centre of the sphere), and a common altitude (namely, the radius of the sphere), and which have the faces of the polyedron as bases. Then, R being the radius of the sphere,

Vol. of polyedron = Rx (sum of faces of polyedron). (1) If the number of faces of the polyedron be increased and the area of each face be decreased, then the sum of the faces becomes more nearly equal to the area of the surface of the sphere, and the volume of the polyedron becomes more nearly equal to the volume of the sphere. By increasing the number of faces and decreasing the area of each face, the difference between the sum of the faces of the polyedron and the area of the sphere can be made as small as one please; and, likewise, the difference between the volume of the polyedron and the volume of the sphere can be made as small as one please. In other words:

The area of the surface of the sphere is the limit of the area of the surface of the polyedron, and the volume of the sphere is the limit of the volume of the polyedron, when the faces of the latter are increased without limit, and each face is made to approach zero in area.

Hence, from (1), Vol. of sphere = Rx surface of sphere

63. Definitions. A spherical pyramid is a portion of a sphere bounded by a spherical polygon and the planes of the sides of the polygon. The polygon is called the base of the pyramid.

*For a note concerning the measurement of the circle and the sphere see Plane Trigonometry, Art. 72, and Note C, p. 171. For the proofs of Archimedes, see T. L. Heath, The Works of Archimedes edited in modern notation, with introductory chapters (Cambridge, University Press), pp. 39, 41, 93.

For example, in Fig. 11, Art. 12, 0-ABCD, 0-ABC, 0-ABD, are spherical pyramids; their bases are ABCD, ABC, ABD.

A spherical sector is the portion of a sphere generated by the revolution of a sector of a circle about any diameter of the circle as axis. For example, in Fig. 47, Art. 53, when the semicircle ATB revolves about AB, each of the circular sectors AOL, LOT, LOK, etc., describes a spherical sector.

(One of

A spherical segment is the portion of a sphere bounded by two parallel planes and the zone intercepted between them. the planes may be tangent to the sphere.)

64. Volume of a spherical pyramid; of a spherical sector. By reasoning analogous to that in Art. 62, it can be shown that, in a sphere of radius R,

vol. of a spherical pyramid = R x area of its base; Rx area of its zone.

vol. of a spherical sector

=

Since the area of a zone of height h = 2 Rh (Art. 53),

#R2h.

vol. O-ABCD = } OA × area ABCD;

1. Find the volumes of the spherical pyramids whose bases are the triangles described in Art. 57, Exs. 1-6.

2. Find the volumes of the following spherical sectors:

(a) The sector whose base is a zone of height 2 inches on a sphere of radius 18 inches.

(b) The sector whose base is a zone of height 3 feet on a sphere of radius 12 feet.

65. Volume of a spherical segment. Let AB be an arc of a semicircle of radius R having the diameter DD'. From A, B, draw Aa, Bb, at right angles to DD'. It is required to find the volume of the spherical segment generated by the revolution of ABba about DD'.

b

D

D'
FIG. 51

Let h denote the height of the segment, and Pu P, the lengths of the perpendiculars from the centre to the parallel bases of the segment. On making the revolution of the semicircle DAD', it is seen that

segment generated by ABba = cone generated by BObspherical sector generated

by AOB cone generated by AOa.

vol. cone generated by BOb = {} πr1⁄22P2;

vol. sector generated by AOB = }πR2h ; (Art. 64)

vol. cone generated by AOα = }πr12P1•

.. vol segment = (r2p2+ 2 R2h — ri2p1).

(1)

The result (1) can be reduced to various forms. For example,

then vol. segment = TR2 (p2 - P1) + } πр2(R2 − р22) — § πî1(R2 — p12)

-

On substituting the last result in (3), expressing p12 and p22 in terms of R, rı, r2, and reducing, the following formula is obtained, viz. :

h2

vol. segment ==2 (m2 + n2 + 12).

πλ
2

(4)

EXAMPLES.

1. Show that if (in Fig. 51) angle AOD = α, then the volume of the spherical sector generated by AOD is πR3(1 — cos α).

2. Show that if angle AOD=α, then the volume of the segment generated by the revolution of ADa is πR3 sin1 ¦ α(1 + 2 cos2 } α).

SUGGESTION. Segment generated by ADa : = sector generated by AODcone generated by AOa.

3. Find the volume of a spherical segment, the diameters of its ends being 10 and 12 inches, and its height 2 inches.

4. The diameters of the ends of a spherical segment are 8 and 12 inches, and its height is 10 inches. Find its volume.

N.B. For questions and exercises on Chapter VI., see page 108.

CHAPTER VII.

PRACTICAL APPLICATIONS.

66. Geographical problem. To find the distance between two places and the bearing (i.e. the direction) of each from the other, when their latitudes and longitudes are known. An interesting application of spherical trigonometry can be made in solving this problem. In the following examples the earth is regarded as spherical, and its radius is taken to be 3960 miles.

EXAMPLES.

1. Find the shortest distance along the earth's surface between Baltimore (lat. 39° 17' N., long. 76° 37' W.) and Cape Town (lat. 33° 56' S., long. 18° 26' E.).

In Fig. 52 B and C represent Baltimore and Cape Town; EQ is the earth's

equator; NGS, NBS, NCS are the meridians of Greenwich, Baltimore, and Cape Town respectively; BC is the great circle arc whose length is required.

In the spherical triangle BNC, NB, NC, and BNC are known. For

Hence, BC can be determined in degrees by Art. 44; then, the radius of the sphere being given, BC can be determined in miles. The angles NBC, NCB, can also be found.

=

Answers: BC (65° 47' 48') = 4685.8 miles; NBC = 115° 1' 35'; NCB = 57° 42′ 23".

NOTE 1. The bearing of one place from a second place is the angle which the great circle arc joining the two places makes with the meridian of the second place. Thus, in Fig. 52 the bearing of Cape Town from Baltimore is the angle NBC, and the bearing of Baltimore from Cape Town is NCB.