DEFINITIONS.

1o. If a semi-circumference be divided into equal arcs, the chords of these arcs form half of the perimeter of a regular inscribed polygon; this half perimeter is called a regular semi-perimeter. The figure bounded by the regular semiperimeter and the diameter of the semi-circumference is called a regular semi-polygon. The diameter itself is called the axis of the semi-polygon.

2o. If lines be drawn from the extremities of any side, and perpendicular to the axis, the intercepted portion of the axis is called the projection of that side.

B

H

K

D

E

G

The broken line ABCDGP is a regular semi-perimeter; the figure bounded by it and the diameter AP, is a regular semi-polygon, AP is its axis, HK is the projection of the side BC, and the axis, AP, is the projection of the entire semi-perimeter.

PROPOSITION IX. LEMMA.

If a regular semi-polygon be revolved about its axis, the

surface generated by the semi-perimeter will be equal to the axis multiplied by the circumference of the inscribed circle.

Let ABCDEF be a regular semi-polygon, AF its axis, and ON its apothem then will the surface generated by the regular semi-perimeter be equal to AF x circ. ON.

From the extremities of any side, as DE, draw DI and EH perpendicular to AF; draw also NM perpendicular to AF, and EK perpendicular to DI. Now, the surface generated by ED is equal to DE × circ. NM

(P. IV., S.). But, because the triangles EDK and ONM are similar (B. IV., P. XXI.), we have,

DE: EK or II :: ON : NM :: circ.ON : circ.NM

whence,

DE x circ. NM = IH × circ.ON ;

that is, the surface generated by any side is equal to the projection of that side multiplied by the circumference of the inscribed circle: hence, the surface generated by the entire semi-perimeter is equal to the sum of the projections of its sides, or the axis, multiplied by the circumfer

ence of the inscribed circle; which was to be proved.

Cor. The surface generated by any portion of the perimeter, as CDE, is equal to its projection PH, multiplied by the circumference of the inscribed circle.

The surface of a sphere is equal to its diameter multiplied by the circumference of a great circle.

Let ABCDE be a semi-circumference, O its centre, and AE its diameter: then will the surface of the sphere generated by revolving the semi-circumference about AE, be equal to AEX circ. OE.

E

D

For, the semi-circumference may be regarded as a regular semi-perimeter with an infinite number of sides, whose axis is AE, and the radius of whose inscribed

B

circle

A

is OE hence (P. IX.), the surface generated by it is equal

to AE × circ. OE; which was to be proved.

Cor. 1. The circumference of a great circle is equal to

2 OE (B. V., P. XVI.): hence, the area of the surface

of the sphere is equal to 20E × 20E,

;

or to 40E2 that is, the area of the surface of a sphere is equal to four great circles.

Cor. 2. The surface generated by any arc of the semicircle, as BC, will be a zone, whose altitude is equal to the projection of that arc on the diameter. But, the arc BC is a portion of a semiperimeter having an infinite number of

sides, and the radius of whose inscribed circle is equal to that of the sphere: hence (P. IX., C.), the surface of a zone

is equal to its altitude multiplied by the circumference of a great circle of the sphere.

Cor. 3. Zones, on the same sphere, or on equal spheres, are to each other as their altitudes.

If a triangle and a rectangle having the same same base and equal altitudes, be revolved about the common base, the volume generated by the triangle will be one-third of that generated by the rectangle.

Let ABC be a triangle, and EFBC a rectangle, having the same base BC, and an equal altitude AD, and let them both be revolved about BC: then will the volume generated by ABC be one-third of that generated by EFB C.

For, the cone generated by the right-angled triangle ADB, is equal to one-third of the cylinder generated by

F

B

D

E

the rectangle ADBF (P. V., C. 1); and the cone generated by the triangle ADC, is equal to one-third of the cylinder generated by the rectangle ADCE. But, when AD falls within the triangle, the sum of the cones generated by ADB and ADC, is equal to the volume generated by the triangle ABC; and the sum of the cylinders generated by ADBF and ADCE, is equal to the volume generated by the rectangle EFB C. When AD falls without the triangle, the difference of the cones generated by ADB and ADC, is equal to the volume generated by ABC; and the difference of the cylinders generated by ADBF and ADCE, is equal to the volume generated by EFBC: hence, in either case, the volume generated by the triangle ABC, is equal to one-third of the volume generated by the rectangle EFBC; which was to be proved.

Cor. The volume of the cylinder generated by EFBC, is equal to the product of its base and altitude, or to

T

2

«ÃÐ2 × BC: hence, the volume generated by the triangle ABC, is equal to AD x BC.

PROPOSITION XII. LEMMA.

If an isosceles triangle be revolved about a straight line passing through its vertex, the volume generated will be equal to the surface generated by the base multiplied by one-third of the altitude.

Let CAB be an isosceles triangle, C its vertex, AB its base, CI its altitude, and let it be revolved about the line CD, as an axis: then will the volume generated be equal to surf. AB × CI.

There may be two cases: the base, or base produced,

may meet the axis; or, the base may be parallel to the axis.

1°. Suppose the base, when produced, to meet the axis at D; draw AM, IK, and BN, perpendicular to CD, and BO parallel to DC. Now, the volume generated by CAB is equal to the difference of the

A

I

B

MK N

volumes generated by CAD and CBD; hence (P. XI., C.),

volCAB=}«AM2 × CD—\πBN2 × CD=}*(AM2 — BÑ3) × CD.

But, AM2

BN is equal to (AM+ BN) (AM — BN), (B. IV., P. X.); and because AM+ BN is equal to 2IK (P. IV., S.), and AM - BN to A0, we have,

But, the right-angled triangles AOB and CDI lar (B. IV., P. XXI.); hence,

are sim

A0: AB :: CI : CD; or, A0 × CD – AB × CI.

=

Substituting, and changing the order of the factors, we have,

vol. CAB = AB × 2′′IK × 1CI.

But, AB × 2 IK is equal to the surface generated by AB; hence,

vol. CAB = surf. AB CI

This demonstration holds good when the axis CD coincides with one side of the triangle CAB.

2o. Suppose the base of the triangle to be parallel to the axis.