In like manner, it may be proved that the surface described by CD is equal to the altitude HK multiplied by the circumference of the inscribed circle; and the same may be proved of the other sides. Hence the entire surface described by ABCDEF is equal to the circumference of the inscribed circle multiplied by the sum of the altitudes AG, GH, HK, KL, and LF; that is, the axis of the polygon.

Let, now, the arcs AB, BC, etc., be bisected, and the number of sides of the polygon be indefinitely increased, its perimeter will coincide with the circumference of the semicircle, and the perpendicular IM will become equal to the radius of the sphere; that is, the circumference of the inscribed circle will become the circumference of a great circle. Hence the surface of a sphere is equal to the product of its diameter by the circumference of a great circle.

Cor. 1. The area of a zone is equal to the product of its altitude by the circumference of a great circle.

For the surface described by the lines BC, CD is equal to the altitude GK multiplied by the circumference of the inscribed circle. But when the number of sides of the polygon is indefinitely increased, the perimeter BC+CD becomes the arc BCD, and the inscribed circle becomes a great circle. Hence the area of the zone produced by the revolution of BCD is equal to the product of its altitude GK by the circumference of a great circle.

Cor. 2. The area of a great circle is equal to the product of its circumference by half the radius (B. VI., Pr. 12), or one fourth of the diameter; hence the surface of a sphere is equivalent to four of its great circles.

Cor. 3. The surface of a sphere is equal to the convex surface of the circumscribed cylinder.

For the latter is equal to the product of its altitude by the circumference of its base. But its base is equal to a great circle of the sphere, and its altitude to the diameter; hence the convex surface of the cylinder is equal to the product of its diameter by the circumference of a great circle, which is also the measure of the surface of a sphere.

Cor. 4. Two zones upon equal spheres are to each other as their altitudes, and any zone is to the surface of its sphere as the altitude of the zone is to the diameter of the sphere.

Cor. 5. Let R denote the radius of a sphere, D its diameter, C

the circumference of a great circle, and S the surface of a sphere; then we shall have

Also

C=2πR, or TD (B. VI., Pr. 13, Cor. 2).

S=2TRX2R=4πR2, or #D2.

If H represents the altitude of a zone, its area will be

The volume of a sphere is equal to one third the product of its surface by the radius.

D

B

Let ACEG be the semicircle by the revolution K of which the sphere is described. Inscribe in the semicircle a regular semi-polygon ABCDEFG, and draw the radii BO, CO, DO, etc.

The solid described by the revolution of the ° polygon ABCDEFG about AG is composed of the solids formed by the revolution of the triangles ABO, BCO, CDO, etc., about AG.

First. To find the value of the solid formed by the revolution of the triangle ABO.

From O draw OH perpendicular to AB, and from B draw BK perpendicular to AO. The two triangles ABK, BKO, in their revolution about AO, will describe two cones having a common base, viz., the circle whose radius is BK.

The solid described by the triangle ABO will then be repre- . sented by R2H, or BK2 × AO (Prop. 5, Cor. 2).

But, by similar triangles,

therefore

BK: BA::HO: AO;

BKxAO=HO×AB;

or, multiplying by BK, we have

BK2 × AO=HO×TAB× BK.

But the surface described by AB=TAB× BK (Prop. 3, Cor.). Hence the solid described by the triangle ABO is equal to HO× the surface described by AB.

Secondly. To find the value of the solid formed by the revolution of the triangle BCO.

Produce BC until it meets AG produced in L.

It is evident, from the preceding demonstration, that the solid described by the triangle LCO is equal to

OM

surface described by LC;

and the solid described by the triangle LBO is equal to

OMX surface described by LB; hence the solid described by the triangle BCO is equal to

40MX surface described by BC.

In the same manner, it may be proved that N the solid described by the triangle CDO is equal D ON × surface described by CD,

to

and so on for the other triangles. But the perpendiculars OH, OM, ON, etc., are all equal; hence the solid described by the polygon ABC DEFG is equal to the surface described by the perimeter of the polygon multiplied by JOH.

Let, now, the number of sides of the polygon be indefinitely increased, the perpendicular OH will become the radius OA, the perimeter ACEG will become the semi-circumference ADG, and the solid described by the polygon becomes a sphere; hence the volume of a sphere is equal to one third of the product of its surface by the radius.

Cor. 1. The volume of a spherical sector is equal to the product of the zone which forms its base by one third of the radius of the sphere.

For the solid described by the revolution of BCDO is equal to the surface described by BC+CD multiplied by JOM.

But when the number of sides of the polygon is indefinitely increased, the perpendicular OM becomes the radius OB, the quadrilateral BCDO becomes the sector BDO, and the solid described by the revolution of BCDO becomes a spherical sector. Hence the volume of a spherical sector is equal to the product of the zone which forms its base by one third of the radius of the sphere. Cor. 2. Let R represent the radius of a sphere, D its diameter, S its surface, and V its volume; then we shall have

Also

S=4πR2, or D2 (Pr. 7, Cor. 5).
V=RXS=TR3, or πD';

hence the volumes of spheres are to each other as the cubes of their radii.

If we put H to represent the altitude of the zone which forms the base of a sector, then the volume of the sector will be represented by 2πRHXR=R2H.

Cor. 3. Every sphere is two thirds of the circumscribed cylinder.

For, since the base of the circumscribed cylinder is equal to a great circle, and its altitude to a diameter, the volume of the cylinder is equal to a great circle multiplied by the diameter (Pr. 2).

But the volume of a sphere is equal to four great circles multiplied by one third of the radius, or one great circle multiplied by of the radius, or of the diameter. Hence a sphere is two thirds of the circumscribed cylinder.

A spherical segment with one base is equivalent to half of a cylinder having the same base and altitude, plus a sphere whose diameter is the altitude of the segment.

D

Let BD be the radius of the base of the segment, AD its altitude, and let the segment be generated by the revolution of the circular half segment AEBD about the axis AC. Join CB, and from the centre C draw CF perpendicular to AB.

The solid generated by the revolution of the segment AEB is equal to the difference of the solids generated by the sector ACBE and the triangle ACB. Now the solid generated by the sector ACBE is equal to CB2 x AD (Pr. 8, Cor. 2).

And the solid generated by the triangle ACB, by Pr. 8, is equal to CF multiplied by the convex surface described by AB, which is 27CF ×AD (Pr. 7), making, for the solid generated by the triangle ACB, TCF2X AD.

Therefore the solid generated by the segment AEB is equal to

kor

that is,

πAD × (CB2-CF2),

TAD × BF2;

TAD XAB2,

because CB2-CF2 is equal to BF2, and BF2 is equal to one fourth of AB2.

Now the cone generated by the triangle ABD is equal to πAD × BD2 (Pr. 5, Cor. 2).

Therefore the spherical segment in question, which is the sum of the solids described by AEB and ABD, is equal to

that is,

AD(2BD2+AB2);

AD(3BD2+AD2),

because AB2 is equal to BD2+AD2.

This expression may be separated into the two parts

ADX BD2, and

AD3.

The first part represents the volume of a cylinder having the same base with the segment and half its altitude (Pr. 2); the other part represents a sphere, of which AD is the diameter (Pr. 8, Cor. 2). Therefore a spherical segment, etc.

Cor. The volume of the spherical segment of two bases generated by the revolution of BCED about c the axis AE may be found by subtracting that of the segment of one base generated by ABD from that of the segment of one base generated by ACE.

B

B

EXERCISES ON THE PRECEDING PRINCIPLES. 1. What is the entire surface of a triangular prism whose base is an equilateral triangle, having each of its sides equal to 17 inches, and its altitude 5 feet?

2. What is the entire surface of a regular triangular pyramid whose slant height is 15 feet, and each side of the base 4 feet?

3. What is the convex surface of the frustum of a square pyramid whose slant height is 14 feet, each side of the lower base being 3 feet, and each side of the upper base 2 feet?

4. What is the volume of a triangular prism whose height is 12 feet, and the three sides of its base 4, 5, and 6 feet?

5. What is the volume of a triangular pyramid whose altitude is 25 feet, and each side of the base 4 feet?

6. What is the volume of a piece of timber whose bases are squares, each side of the lower base being 14 inches, and each side of the upper base 12 inches, the altitude being 25 feet?

7. What is the entire surface of a cylinder whose altitude is 17 feet, and the diameter of its base 3 feet?

8. What is the entire surface of a cone whose side is 24 feet, and the diameter of its base 5 feet?

9. What is the entire surface of a frustum of a cone whose side is 18 feet, and the radii of the bases 5 feet and 4 feet?

10. What is the volume of a cylinder whose altitude is 16 feet, and the circumference of its base 5 feet?

11. What is the volume of a cone whose altitude is 13 feet, and the circumference of its base 7 feet?

12. What is the volume of a frustum of a cone whose altitude is 22 feet, the circumference of its lower base 18 feet, and that of the upper base 14 feet?