dron. Now it is evident that all these pyramids will have for their common altitude the radius of the sphere, so that each pyramid will be equal to a face of the polyedron, which serves as a base, multiplied by a third of the radius; therefore the entire polyedron will be equal to its surface multiplied by a third of the radius of the inscribed sphere.

It will be perceived by this that the solidities of polyedrons circumscribed about a sphere are to each other as the surfaces of these same polyedrons. Thus the property which we have demonstrated for the circumscribed cylinder is common to an infinite number of other bodies.

We might have remarked also that the surfaces of polygons circumscribed about a circle are to each other as their perimeters.

PROBLEM.

551. The circular segment BMD (fig. 271) being supposed to Fig. 271. revolve about a diameter exterior to this segment, to find the value of the solid generated.

Solution. Let fall upon the axis the perpendiculars BE, DF, and upon the chord BD the perpendicular CI, and draw the radii CB, CD.

2

-2

The solid generated by the sector BCA = × CB × АE (546); the solid generated by the sector DCA = { π × CB3× AF ; consequently the difference of these two solids, or the solid generated by the sector DCB, will be equal to

¦ π × CB × (AF —AE)=¦ñ × CB × EF. But the solid generated by the isosceles triangles DCB has for

its measure

× CI × EF (543); consequently the solid generated by the segment BMD=37 × EF × (CB-CI). Now

--2 --2

CB-CI-BI=

in the right-angled triangle CBI we have CB — CI =BI = 1 BD; therefore the solid generated by the segment BMD has for its

measure л × EF × ¦ BD), or ¦ л BD × EF.

552. Scholium. The solid generated by the segment BMD is

--2

to the sphere whose diameter is BD, as ¦ л × BD × EF is to

-3

л × BD, or :: EF : BD.

Fig. 271.

THEOREM.

553. Every segment of a sphere, comprehended between two parallel planes, has for its measure the half sum of its bases multiplied by its altitude, plus the solidity of the sphere of which this same altitude is the diameter.

Demonstration. Let BE, DF (fig. 271), be the radii of the bases of the segment, EF its altitude, so that the segment may be formed by the revolution of the circular space BMDFE about the axis FE. The solid generated by the segment BMD

will be equal to ¦ л × BD × EF (552), the frustum of a cone
π BD3×
generated by the trapezoid BDFE will be equal to

× EF × (BE + DF + BE × DF) (527); consequently the segment of the sphere which is sum of these two solids = × EF × (2BE+2DF+2BEXDF+BD.) But, by drawing BO parallel to EF, we shall have DO=DF-BE,

-2

--2

-2

2DF × BE + BE (182), and consequently

-2

--2

BD = BO+ DO=EF2 + DF — 2DF × BE + BE. Putting this

-2

value in the place of BD in the expression for the segment, and reducing it, we shall have for the solidity of the segment

¿ ñ × EF × (3BE +3DF2+ EF3),

an expression which may be decomposed into two parts; the

one ↓ 7× EF × (3BE+3DF), or EF ×

-2

π π

(*X BE + 1 X DF").

2

is the half sum of the bases multiplied by the altitude; the other

× EF represents the sphere of which EF is the diameter (548); therefore the segment of the sphere &c.

554. Corollary. If one of the bases is nothing, the segment in question becomes a spherical segment having only one base; therefore every spherical segment having only one base is equivalent to half of the cylinder of the same base and same altitude, plus the sphere of which this altitude is the diameter.

General Scholium.

555. Let R be the radius of the base of a cylinder, H its altitude; the solidity of the cylinder will be л Ra × Н, or л R2 Н.

Let R be the radius of the base of a cone, H its altitude; the solidity of the cone will be л R2 × ¦ H, or л R2 H.

Let A, B, be the radii of the bases of the frustum of a cone, H its altitude, the solidity of the frustum will be

Let R be the radius of a sphere; its solidity will be л R3. Let R be the radius of a spherical sector, H the altitude of the zone, which answers as a base; the solidity of the sector will be π R3 H.

Let P, Q, be the two bases of a spherical segment, H its alti

tude, the solidity of this segment will be

P+
2

If the spherical segment have only one base P, its solidity

will be PH + ¦ π Н3.

Appendix to the Third Section of the Second Part.

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.

Demonstration. Within the polyedron, take a point, from which let straight lines be drawn 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 (fig. 240), and n the number of its Fig. 240 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 be estimated in a similar manner, and afterwards the whole 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 of 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 meet at any one point A are equal to four right angles, the sum of all the angles of 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; since the same edge is in every case a side to two faces. Hence we have 8 = 4S — 4+4H; or, dividing the whole by 4, 2=S-A+H; therefore S+ H= A + 2.

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

For, if we consider a face, the number of whose sides is n, we shall find that the sum of the angles of this face is equal to 2n-4 right angles (79). But the sum of all these 2n's, or twice the number of sides in all the faces, will be 4A; and 4, taken as many times as there are faces, will be 4H; hence the sum of the angles in all the faces is 4A-4H. Now, by the theorem just demonstrated, we have A-HS-2, and consequently 4.A — 4H = 4 (S -2). Therefore the sum of all the plane angles, &c.

THEOREM.

558. Of all the spherical triangles formed with two given sides Fig. 276. CA, CB (fig. 276), and a third assumed at pleasure, the greatest, ABC, is that in which the angle C, contained by the given sides, is equal to the sum of the two other angles, A and B.

Demonstration. Produce the two sides AC, AB, till they meet in D; we shall have a spherical triangle BCD, in which the angle DBC is also equal to the sum of the two other angles BDC, BCD. For, BCD + BCA, being equal to two right angles, and likewise CBA + CBD, we have

BCD+BCA = CBA + CBD;

and adding BDC = BAC, we shall have

BCD + BCA + BDC = CBA + CBD + BAC.
Now, by hypothesis, BCA CBA + BAC; hence

Draw BI making the angle CBI = BCD, and consequently IBD BDC; the two triangles IBD, IBC, will be isosceles, and we shall have IC= IB = ID. Hence the point I, the middle point of DC, is at equal distances from the three points B, C, D. For a similar reason, the point O, the middle of BA, is equally distant from the points A, B, C.

Now, suppose CA CA and the angle BCA > BCA; if A'B be joined, and the arcs A'C, A'B, produced till they meet in D', the arc D'CA' will be a semicircumference, as well as DCA; therefore, since we have CA' CA, we shall also have CD CD. But in the triangle CID', we have CI+ID> CD; hence ID > CD — CI, or ID>ID.

=

In the isosceles triangle CIB, bisect the angle I 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 shewn as in art. 41, that BL + LC < BI+IC; and, taking the half of each, that BL<BI. But in the triangle D'LC, we have D'L> DC— CL, and still more D'L> DC- CI, or D'L> DI, or D'L> BI; cnosequently D'L> BL. Hence, if in the arc EIF, we seek for a point equally distant from the three points B, C, D', it can be found only in the prolongation of El towards F. Let I' be the point required; we shall have D'I′ = BI' = CI′; the triangles I'CB, I'CD', I'BD', being isosceles, we shall have the equal angles IBC= I'CB, YBD' = I'D'B, I'CD' = I'D C. But the angles D'BC + CBA' are equal to two right angles, and

D'CB + BCA'

are likewise equal to two right angles; therefore

D'BI+IBC + CBA' = 2,
BCI-ICD+BCA' = 2.

Add together the two equations observing that I'BC='BCI, and DBI-ICD = BD'I' — F'D'C= CD'B= CA'B; and we shall 2 IBC+CA'B+ CBA' + BCA' = 4.

have

Hence CA'B+CBA' + BCA' - 2 (which measures the area of the triangle ABC (501) = 2—2 I'BC; so that we have area A'BC=2—2 angle I'BC;

likewise, in the triangle ABC, we should have

Now the angle I'BC has already been proved to be greater than IBC; hence the area A'BC is less than ABC.