sphere, and for its base one of the polyedron's faces. Now it is evident that all these pyramids will have the radius of the sphere for their common altitude: so that each pyramid will be equal to one face of the polyedron multiplied by a third of the radius: hence the whole polyedron will be equal to its surface multiplied by a third of the radius of the inscribed sphere.

It is therefore manifest, that, the solidities of polyedrons circumscribed about the sphere are to each other as the surfaces of those polyedrons. Thus the property, which we have shown to be true with regard to the circumscribed cylinder, is also true with regard to an infinite number of other bodies. We might likewise have observed that the surfaces of polygons, circumscribed about the circle, are to each other as their perimeters.

PROBLEM.

551. If a circular segment be supposed to make a revolution about a diameter exterior to it, required the value of the solid so produced.

Let the segment BMD revolve about AC.

On the axis, let fall the perpendiculars BE, DF; from the centre C, draw CI perpendicular to the chord BD; also draw the radii CB, CD.

D

M

B

E

F

The solid described by the sector BCA is equal to 3. CB2.AE (546.); the solid described by the sector DCA=3T. CB.2 AF; hence the difference of these two solids, or the solid described by the sector DCB=3. CB3. (AF-AE)=. CB.EF. But the solid described by the isosceles triangle DCB (543.) has for its measure 3T.CI2.EF; hence the solid described by the segment BMD=3«.EF. (CB-CI3). Now, in the right-angled triangle CBI, we have CB2-CI'=BI=1BD2; hence the solid described by the segment BMD will have for its measure 3.EF.4BD, or 7. BD2. EF.

552. Scholium. The solid described by the segment BMD is to the sphere, which has BD for its diameter, as 7.BD2.EF is to 7.BD3, or as EF to BD.

THEOREM.

553. Every segment of a sphere, included between two parallel planes, is measured by the half-sum of its bases multiplied by its altitude, plus the solidity of a sphere whose diameter is this same altitude.

Let BE, DF (see the preceding figure), be the radii of the segment's two bases, EF its altitude, the segment being produced by the revolution of the circular space BMDFE about the axis FE. The solid described by the segment BMD is equal to .BD2.EF (552.); and (527.) the truncated cone described by the trapezoid BDFE is equal to 7. EF (BE3+ DF2+BE.DF); hence the segment of the sphere, which is the sum of those two solids, must be equal to 17.EF. (2BE2+ 2DF2+2BE.DF+BD). But, drawing BO parallel to EF, we shall have DO=DF-BE, hence (182.) DO2=DF-2 DF.BE+BE2; and consequently BD2=BÓ2+DO2=EF2+ DF 2 DF.BE+BE. Put this value in place of BD3 in the expression for the value of the segment, omitting the parts which destroy each other; we shall obtain for the solidity of the segment,

EF. (3 BE2+3 DF2+EF2),

an expression whsch may be decomposed into two parts; the T.BE+.DF one, Jr.EF. (3BE2+3DF), or EF 2

F*) being

the half sum of the bases multiplied by the altitude; while the other. EF represents (548.) the sphere of which EF is the diameter: hence every segment of a sphere, &c.

554.

Cor. If either of the bases is nothing, the segment in question becomes a spherical segment with a single base; hence any spherical segment, with a single base, is equivalent to half the cylinder having the same base and the same altitude, plus the sphere of which this altitude is the diameter.

General Scholium.

555. Let R be the radius of a cylinder's base, H its altitude the solidity of the cylinder will be RXH, or RH.

Let R be the radius of a cone's base, H its altitude: the solidity of the cone will be TRXH, or R3H.

Let A and B be the radii of the bases of a truncated cone, H its altitude: the solidity of the truncated cone will be 7. H. (A+B+AB).

Let R be the radius of a sphere; its solidity will be 4R3: Let R be the radius of a spherical sector, H the altitude of the zone, which forms its base: the solidity of the sector will be R H.

Let P and Q be the two bases of a spherical segment, H its P+Q altitude: the solidity of the segment will be • H + j *. H2 >

2

If the spherical segment has but one base, the other being nothing its solidity will be PH+1⁄4 ̃H3.

APPENDIX TO BOOKS VI. AND, VII.

OF SPHERICAL ISOPERIMETRICAL POLYGONS.

THEOREM.

556. Let S be the number of solid angles in a polyedron, H the number of its faces, A the number of its edges; then in all cases we shall have S+H=A+2.

Within the polyedron, take a point, from which draw straight lines to the vertices of all its angles; conceive next, that from the same point as a centre, a spherical surface is described, meeting all these straight lines in as many points; join these points by arcs of great circles, so as to form on the surface of the sphere polygons corresponding in position and number with the faces of the polyedron. Let ABCDE be one of these polygons, n the number of its sides; its surface will be s-2n+4, s being the sum of the angles A, B, C, D, E. (506.) If the surface of each polygon is estimated in a similar manner, and afterwards the whole are added together, we shall find their sum, or the surface of the sphere, represented by 8, to be equal to the sum of all the angles in the polygons minus twice the number of their sides, plus 4, taken as many times as there are faces. Now, since all the angles which lie round any one point A are equal to four right angles, the sum of all the angles in the polygons must be equal to 4 taken

as many times as there are solid angles; it is therefore equal to 4S. Also, twice the number of sides AB, BC, CD, &c. is equal to four times the number of edges, or to 4A; because the same edge is always a side in two faces. Hence we have 8=4S-4A+4H; or dividing all by 4, we have 2=S-A+H; hence S+H=A+2.

557. Cor. From this it follows, that the sum of all the plane angles, which form the solid angles of a polyedrons, is equal to as many times four right angles as there are units in S-2, S being the number of solid angles in the polyedron.

For, examining a face the number of whose sides is n, the sum of the angles in this face (79.) will be 2n-4 right angles. But the sum of all these 2n's, or twice the number of sides of all the faces will be 4A; and 4 taken as many times as there are faces, will be 4 H: hence the sum of the angles in all the faces is 4A-4H. Now by the Theorem just demonstrated, we have A-H-S-2, and consequently 4A-4H=4 (S-2). Hence the sum of all the plane angles, &c.

THEOREM.

558. Of all the spherical triangles formed with two given sides, and a third assumed at pleasure, the greatest is the one in which the angle contained by the given sides, is equal to the sum of the two other angles of the triangle.

Let A'C, CB, be the given sides, and C the contained angle. Produce the

two sides A'C,

AB till they

meet in D'; you F

will have a sphe

rical triangle,

BCD', in which

A

A

E

the angle DBC D

will also be equal

to the sum of

D

B

the two other angles BD'C, BCD'. For, BCD'+BCA' being equal to two right angles, and likewise CBA'+CBD', we have BCD'+BCA=CBA'+CBD'; and adding on both sides BD'C BA'C, we shall have BCD+BCA + BDC=CBA+CBD' +BAC. Now, by hypothesis, BCA'=CBA'+BA ̊C; hence CBD'=BCD+BD'C.

Draw BI, making the angle CBI=BCD', and consequently IBD'=BD'C; the two triangles IBD', IBC will be isosceles,

and we shall have IC=IB=ID'. Hence the point I, is at equal distances from the three points B, C, D'.

well as DCA'; therefore, since we have CA=CA', we shall also have CD CD. But in the triangle CID, we have CI +ID7CD; hence ID 7CD-CI, or ID 7ID'.

In the isosceles triangle CIB, bisect the angle I at the vertex, by the arc EIF, which will also bisect BC at right angles. If a point L is assumed between I and E, the distance BL, equal to LC, will be less than BI; for it might be shown as in Art 41. that BL+LC BI+IC; and taking the halves of each, that BLZ BI. But in the triangle DLC, we have DL7DC-CL, and still more DL7D ̊C-CI, or DL7D'I, or DL>BI; hence DL>BL. Hence if in the arc EIF, we seek for a point equally distant from the three points B, C, D, it can only be found in the prolongation of El towards F. Let I' be the point required, such that we have DI'=BI' =CI'; the triangles ICB, ICD, I'BD being isosceles, we shall have the equal angles IBC=ICB, IBD=IDB, ICD= IDC. But the angles DBC+CBA are equal to two right angles, and likewise DCB+CBA are equal to two right angles; hence

DBI'+I'BC+CBA=2
BCI-ICD+BCA=2.

Add the two sums, observing that IBC=BCI', and DBII'CD=BDI-I'DC-CDB=CAB; we shall have

21 BC+CAB+CBA+BCA=4.

Hence CAB+CBA+BCA-2 (which measures the area of the triangle ABC (501.)=2—21 BC; so that we have area A'BC-2-2 angle 1 BC; likewise, in the triangle ABC, we should have area ABC-2-2 angle IBC. Now the angle IBC has already been proved greater than IBC; hence the area ABC is less than A'BC.

The same demonstration would lead to the same conclusion, if taking always the arc CA=CA, the angle BCA (see the figure of the last page) were made greater than BCA'; hence A BC is the greatest of all those triangles, having two sides given, and the third to be assumed at will.