42. COROLLARY I. The result of this proposition may be formulated,

V = R3.

43. COROLLARY II. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters.

PROPOSITION XII.-THEOREM.

44. The volume of a spherical sector is equal to the area of the zone which forms its base multiplied by one-third the radius of the sphere.

The proof is analogous to the proof of Proposition XI. The form of the circumscribed polyedron is, however, somewhat more complicated, as it will be bounded by a surface made up of plane faces tangent to the zone of the spherical sector, and by two pyramidal faces tangent to, or inscribed in, the two conical surfaces of the spherical sector.

45. Definition. A spherical pyramid is a solid bounded by a spherical polygon and the planes of the sides of the polygon; as O-ABCD. The centre of the sphere is the vertex of the pyramid; the spherical polygon is its base.

B

ע

EXERCISE.

Theorem.-The volume of a spherical pyramid is equal to the area of its base multiplied by one-third of the radius of the sphere.

PROPOSITION XIII.-PROBLEM.

46. To find the volume of a spherical segment.

Any spherical segment may be obtained from a spherical sector by adding to it, or subtracting from it, cones having as bases the bases of the segment.

A

For example, let us consider a segment of two bases which does not contain the centre of the sphere. The segment generated by the revolution of ABCD about OC may be obtained by taking the cone generated by OAD from the sum of the cone generated by OBC and the spherical sector generated by OAB.

B

D

Call OC p', OD p, DC h, AD r, BC r', and OA R, and the volume of the segment V. Then we have the simple relations

h = p' —\p,

pr2 + p2 = R2, r'2 + p22 = R2.

The area of the zone of the segment is 27R.h (Proposition IX., Corollary). Hence,

VhR2 + ƒπр‍r2 + ƒπрr2 (Proposition XII., and Proposition VII., Corollary I.),

V = ƒ ̃(p' — p)R2 + 3¬р′(R2 — p′2) — 1ñр(R2 — р2),

√ = (p' — p)ñR2 — 3ñ(р′3 — p3),

[1]

a convenient formula when the distances of the bases of the

segment from the centre of the sphere are given.

Another convenient formula can be obtained by introducing in [1] h, r, and in place of p and p'. We have

V = (p' — p)Ţ[3R1 — (p'2 + p′p + p2)].

This formula is convenient when the areas of the bases of the segment are given, and it may be put into words as follows:

The volume of a spherical segment is equal to the half sum of its bases multiplied by its altitude plus the volume of a sphere of which that altitude is the diameter.

EXERCISES ON BOOK IX.

THEOREMS.

1. GIVE a strict proof of Proposition I. and Proposition IV. for the volumes of cylinder and cone, by showing that the difference between the volumes of the inscribed and circumscribed figures can be decreased at pleasure.

2. Assuming that if a solid has a plane face the area of that face is less than the rest of the surface of the solid, prove, first, that if two convex solids have a plane face in common, and one solid is wholly included by the other, its surface is less than that of the other (v. V., 13), and then give a strict proof of Proposition I. and Proposition IV. for the surfaces of cylinder and cone.

3. The volumes of a cone of revolution, a sphere, and a cylinder of revolution are proportional to the numbers 1, 2, 3 if the bases of the cone and cylinder are each equal to a great circle of the sphere, and their altitudes are each equal to a diameter of the sphere.

4. An equilateral cylinder (of revolution) is one a section of which through the axis is a square. An equilateral cone (of revolution) is one a section of which through the axis is an equi、 lateral triangle. These definitions premised, prove the following theorems:

I. The total area of the equilateral cylinder inscribed in a sphere is a mean proportional between the area of the sphere and the total area of the inscribed equilateral cone. The same is true of the volumes of these three bodies.

II. The total area of the equilateral cylinder circumscribed about a sphere is a mean proportional between the area of the sphere and the total area of the circumscribed equilateral cone. The same is true of the volumes of these three bodies.

5. If h is the altitude of a segment of one base in a sphere whose radius is r, the volume of the segment is equal to πh2(R — }h).

6. The volumes of polyedrons circumscribed about the same sphere are proportional to their surfaces.

MISCELLANEOUS EXERCISES

ON THE

GEOMETRY OF SPACE.

1. A PERPENDICULAR let fall from the middle point of a line upon any plane not cutting the line is equal to one-half the sum of the perpendiculars let fall from the ends of the line upon the same plane.

2. The perpendicular let fall from the point of intersection of the medial lines of a given triangle upon any plane not cutting the triangle is equal to one-third the sum of the perpendiculars from the vertices of the triangle upon the same plane.

3. The perpendicular from the centre of gravity of a tetraedron upon any plane not cutting the tetraedron is equal to one-fourth the sum of the perpendiculars from the vertices of the tetraedron upon the same plane.

4. The volume of a truncated triangular prism is equal to the product of the area of its lower base by the perpendicular upon the lower base let fall from the intersection of the medial lines of the upper base.

5. The volume of a truncated parallelopiped is equal to the product of the area of its lower base by the perpendicular from the centre of the upper base upon the lower base.

6. If ABCD is any tetraedron, and O any point within it, and if the straight lines AO, BO, CO, DO, are produced to meet the faces in the points a, b, c, d, respectively, then