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described by the arcs DH and DF; these have for their measure respectively DQ× circ. CD and DO × circ. CD; therefore the zone described by FH has for its measure

(DQ-DO) x circ. CD or OQ x circ. CD.

We conclude, then, that every spherical zone with one or two bases has for its measure the altitude of this zone multiplied by the circumference of a great circle.

539. Corollary. Two zones are to each other as their altitudes, and any zone whatever is to the surface of the sphere as the altitude of this zone is to the diameter.

THEOREM.

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540. If the triangle BAC (fig. 264, 265) and the rectangle Fig. 94. BCEF of the same base and same altitude turn simultaneously about the common base BC, the solid generated by the revolution of the triangle will be a third of the cylinder generated by the revolution of the rectangle.

Demonstration. Let fall upon the axis the perpendicular AD (fig. 264); the coné generated by the triangle ABD is a third Fig. 264. of the cylinder generated by the rectangle AFBD (524); also the cone generated by the triangle ADC is a third of the cylinder generated by the rectangle ADCE; therefore the sum of the two cones, or the solid generated by ABC, is a third of the sum of the two cylinders, or of the cylinder generated by the rectangle BCEF.

If the perpendicular AD (fig. 265) fall without the triangle, Fig. 265. the solid generated by ABC will be the difference of the cones generated by ABD and ACD; but, at the same time, the cylinder generated by BCEF will be the difference of the cylinders generated by AFBD, AECD. Therefore the solid generated by the revolution of the triangle will be always the third of the cylinder generated by the revolution of the rectangle of the same base and same altitude.

541. Scholium. The circle of which AD is the radius has for its surface « × AD; consequently × AĎ× BC is the measure

X

of the cylinder generated by BCEF, and ¦ « × ADX BC is the measure of the solid generated by the triangle ABC.

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Fig 266.

Fig. 267.

PROBLEM.

542. The triangle CAB (fig. 266) being supposed to make a revolution about the line CD, drawn at pleasure without the triangle through the vertex C, to find the measure of the solid thus generated.

Solution. Produce the side AB until it meet the axis CD in D, and from the points A, B, let fall upon the axis the perpendiculars AM, BN.

The solid generated by the triangle CAD has for its measure †¤× AM × CD (540); the solid generated by the triangle CBD has for its measure - BÑ3× CD; therefore the difference of these solids, or the solid generated by ABC, will have for its measure « × (AM — BÑ3) × CD.

This expression will admit of another form. From the point I, the middle of AB, draw IK perpendicular to CD, and through the point B draw BO parallel to CD, we shall have

AM+BN=2IK (178),

and AM-BN=AO; consequently (AM+BN) × (AM—BN),
or
AM-BN (184), is equal to 2IK × AO. Accordingly the
measure of the solid under consideration will also be expressed
by × IK × AO × CD. But, if the perpendicular CP be let
fall upon AB, the triangles ABO, DCP will be similar, and will
give the proportion AO: CP :: AB: CD; whence

40 × CD = CP x AB;

moreover CP × AB is double of the area of the triangle ABC; thus we have AO × CD = 2ABC; consequently the solid generated by the triangle ABC has also for its measure

or, since circ. KI is

ABC circ. KI.

× ABC × KI,

equal to 2′′ × KI, this same measure will be Therefore, the solid generated by the revolution of the triangle ABC has for its measure the area of this triangle multiplied by two thirds of the circumference described by the point I the middle of the base.

543. Corollary. If the side AC CB (fig. 267), the line CI will be perpendicular to AB, the area ABC will be equal to AB × CI, and the solidity ¦ ′′ × ABC × IK will become

× AB × IK X CI. But the triangles ABO, CIK, are similar, and give the proportion AB : BO or MN :: CI: IK; consequently

ABX IX= MN × CI;

therefore the solid generated by the isosceles triangle ABC will have for its measure & « × MN × CỈ.

that

544. Scholium. The general solution seems to suppose the line AB produced would meet the axis, but the results would not be the less true, if the line AB were parallel to the axis.

Indeed the cylinder generated by AMNB (fig. 268) has for its Fig. 268 measure « × AM × MN, the cone generated by ACM is equal to ↓*× AM3× CM,

and the cone generated by BCN = 4« × AM × CN. Adding the two first solids together, and subtracting the third from the sum, we have for the solid generated by ABC

TM × AM × (MN + ¦ CM — ¦ CN);

and, since CM— } CN — — (} CN— } CM) = — MN, the

}

above expression reduces itself to ≈ × ÂM × 3 MN, or

* × CP3× MN,

which agrees with the results already found.

THEOREM.

545. Let AB, BC, CD (fig. 262), be several successive sides of a 1īg. 262. regular polygon, O its centre, OI the radius of the inscribed circle; if we suppose the polygonal sector AOD, situated on the same side of the diameter FG, to make a revolution about this diameter, the

solid generated will have for its measure & « × OỈ× MQ, MQ
ΟΙ
being the portion of the axis terminated by the extreme perpen-
diculars AM, DQ.

Demonstration. Since the polygon is regular, all the triangles AOB, BOC, &c., are equal and isosceles. Now, by the corollary of the preceding proposition, the solid generated by the isosceles triangle AOB has for its measure & « × OỈ× MN, the solid generated by the triangle BOC has for its measure « × OỈ× NP, and the solid generated by the triangle COD has for its measure « × OỈ3× PQ; therefore the sum of these solids, or the entire solid generated by the polygonal sector AOD, has for its measure & « × TỈ (MN+NP+PQ), or 3 « × TỈ × MQ.

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Fig. 269.

THEOREM.

546. Every spherical sector has for its measure the zone which serves as a base multiplied by a third of the radius, and the entire sphere has for its measure its surface multiplied by a third of the radius.

Demonstration. Let ABC (fig. 269) be the circular sector, which, by its revolution about AC, generates the spherical sector; the zone described by AB being AD × circ. AC, or

2X ACX AD (538),

we say that the spherical sector will have for its measure this AC, or ‡ « × AС × AD.

zone multiplied by

1. Let us suppose,

if it be possible, that this quantity

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is the measure of a greater spherical sector, of the spherical sector, for example, generated by the circular sector ECF similar to ACB.

Inscribe in the arc EF a portion of a regular polygon EMNF the sides of which shall not meet the arc AB, then suppose the polygonal sector ENFC to turn about EC at the same time with the circular sector ECF. Let CI be the radius of a circle inscribed in the polygon, and let FG be drawn perpendicular to EC. The solid generated by the polygonal sector will have for its measure - « × CỈ3× EG (545); now CI is greater than AC, by construction, and EG is greater than AD; for, if we join AB, EF, the triangles EFG, ABD, which are similar, give the proportion EG: AD:: FG: BD:: CF: CB; therefore EG>AD.

For this double reason 3 « × CỈ× EG greater than

3 « × CÂ3× AD;

the first expression is the measure of the solid generated by the polygonal sector, the second is, by hypothesis, that of the spherical sector generated by the. circular sector ECF; consequently the solid generated by the polygonal sector would be greater than the spherical sector generated by the circular sector. But, on the contrary, the solid in question is less than the spherical sector, since it is contained in it; accordingly the hypothesis

with which we set out cannot be maintained; therefore the zone or base of a spherical sector multiplied by a third of the radius cannot be the measure of a greater spherical sector.

2. We say that this same product cannot be the measure of a less spherical sector. For, let CEF be the circular sector which by its revolution generates the given spherical sector, and let us suppose, if it be possible, that × CEX EG is the measure of a less spherical sector, of that, for example, generated by the circular sector ACB.

The preceding construction remaining the same, the solid generated by the polygonal sector will always have for its measure 3 « × CỈ × EG. But CI is less than CE; consequently the solid is less than «× CE3× EG, which, by hypothesis, is the measure of the spherical sector generated by the circular sector ACB. Therefore the solid generated by the polygonal sector would be less than the solid generated by the spherical sector; but, on the contrary, it is greater, since it contains it. Therefore it is impossible that the zone of a spherical sector multiplied by a third of the radius should be the measure of a less spherical

sector.

We conclude, then, that every spherical sector has for its measure the zone which answers as a base multiplied by a third of the radius.

A circular sector ACB may be increased till it becomes equal to a semicircle; then the spherical sector generated by its revolution is an entire sphere. Therefore the solidity of a sphere is equal to its surface multiplied by a third of the radius.

547. Corollary. The surfaces of spheres being as the squares of their radii, these surfaces multiplied by the radii are as the cubes of the radii. Therefore the solidities of two spheres are as the cubes of their radii, or as the cubes of their diameters.

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548. Scholium. Let R be the radius of a sphere, its surface will be 4 R2, and its solidity 4 R XR, or R3. If we call D the diameter, we shall have RD, and R' = D3 ; } therefore the solidity will also be expressed by ¦ « × ÷ D3, or jπ D3.

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