.CA.

For this double reason, CP. EG is greater than AD. The first is the measure of the solid described by the polygonal sector; the second, by hypothesis, is that of the spherical sector described by the circular sector ECF : hence the solid described by the polygonal sector must be greater than the spherical sector; whereas, in reality, it is less, being contained in the latter : hence our hypothesis was false ; hence, in the first place, the zone or base of a spherical sector multiplied by a third of the radius, cannot measure a greater spherical sector.

We are next to shew that it cannot measure a less spherical sector. Let CEF be the circular sector, which, by its revoJution, generates the given spherical sector; and suppose, if possible, that .CE3.EG is the measure of some smaller spherical sector, say of that produced by the circular sector ACB.

The construction remaining as above, the solid described by the polygonal sector will still have for its measure 2.CI. EG. But CI is less than CE: hence the solid is less than 3.CE.EG, which, according to our supposition, is the measure of the spherical sector described by the circular sector ACB. Hence the solid described by the polygonal sector must be less than the spherical sector described by ACB; whereas, in reality, it is greater, the latter being contained in the former. Hence, in the second place, it is impossible that the zone of a spherical sector, multiplied by a third of the radius, can be the measure of a smaller spherical sector.

Hence, every spherical sector is measured by the zone which forms its base, multiplied by a third of the radius.

:

A circular sector ACB may increase till it becomes equal to a semicircle in which case, the spherical sector described by its revolution is the whole sphere. Hence the solidity of a sphere is equal to its surface multiplied by a third of the radius.

547. Cor. The surfaces of spheres being as the squares of their radii, these surfaces being multiplied by the radii, shews their solidities to be as the cubes of the radii. Hence the solidities of two spheres are as the cubes of their radii, or as the cubes of their diameters.

548. Scholium. Let R be the radius of a sphere, its surface will be 4R; its solidity 4 R XR, or .R. If the diameter is named D, we shall have RD, and R'D' ; hence the solidity may likewise be expressed by XD3, or fπD3.

THEOREM.

The so

549. The surface of a sphere is to the whole surface of the circumscribed cylinder (including its bases), as 2 is to 3. lidities of these two bodies are to each other in the same ratio.

Let MPNQ be a great circle of the sphere; ABCD the circumscribed D square if the semicircle PMQ and the half square PADQ are at the same time made to revolve about the diameter PQ, the semicircle will generate the sphere, while the half-square will generate the cylinder circumscribed about that sphere.

M

B

P

The altitude AD of that cylinder is equal to the diameter PQ; the base of the cylinder is equal to the great circle, its diameter AB being equal to MN; hence (523.), the convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter. This measure (535.) is the same as that of the surface of the sphere: hence the surface of the sphere is equal to the convex surface of the circumscribed cylinder.

But the surface of the sphere is equal to four great circles; hence the convex surface of the cylinder is also equal to four great circles: and adding the two bases, each equal to a great circle, the total surface of the circumscribed cylinder will be equal to six great circles; hence the surface of the sphere is to the total surface of the circumscribed cylinder as 4 is to 6, or as 2 is t. 3; which was the first branch of our Proposition.

In the next place, since the base of the circumscribed cylinder is equal to a great circle, and its altitude to the diameter, the solidity of the cylinder (516.) will be equal to a great circle multiplied by its diameter. But (546.), the solidity of the sphere is equal to four great circles multiplied by a third of the radius, in other terms, to one great circle multiplied by of the radius, or by of the diameter; hence the sphere is to the circumscribed cylinder as 2 to 3, and consequently the solidities of these two bodies are as their surfaces.

550. Scholium. Conceive a polyedron, all of whose faces touch the sphere; this polyedron may be considered as formed of pyramids, each having for its vertex the centre of the

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.

551.

PROBLEM.

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.

2

D

M

B

C

The solid described by the sector BCA is equal to . CB. AE (546.); the solid described by the sector DCA=3. CB.2 AF; hence the difference of these two solids, or the solid described by the sector DCB=3. CB2. (AF—AE)=&T. CB.EF. But the solid described by the isosceles triangle DCB (543.) has for its measure .CI.EF; hence the solid described by the segment BMD=17. EF. (CB-CI). Now, in the right-angled triangle CBI, we have CB2 CI=BI=BD2; hence the solid described by the segment BMD will have for its measure 3. EF. BD3, or . BD. EF.

552. Scholium. The solid described by the segment BMD is to the sphere, which has BD for its diameter, as 17. BD. EF is to . 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 .BD EF (552.); and (527.) the truncated cone described by the trapezoid BDFE is equal to .EF. (BE+DF+BE DF); hence the segment of the sphere, which is the sum of those two solids, must be equal to . EF. (2BE2+2DF2+2BE.DF+BD). But, drawing BO parallel to EF, we shall have DO=DF-BE, hence (182.) DO=DF —2 DF. BE+BE2; and consequently BD2=B02 +DO EF+DF2-2 DF.BE+BE2. Put this value in place of BD2 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 BE +3 DF2+EF2),

an expression which may be decomposed into two parts;

one, .EF. (3 BE +3 DF), or EF.

T.BE2+7.DF2,
2

the

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 Let R be the radius of a cone's base, solidity of the cone will be R2X÷H, or

R×H, or R3H. H its altitude: the 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 1.H. (A2+B2+AB).

Let R be the radius of a sphere; its solidity will be. R3. 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 RH.

Let P and Q be the two bases of a spherical segment, Hits

altitude: the solidity of the segment will be

P+Q
2

.H+¦ ̃.H3.

If the spherical segment has but one base, the other being nothing, its solidity will be PH+; ̃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 takeu