sphere, as the altitude of the zone is to the diameter of the sphere.

Cor. 5. Let R denote the radius of a sphere, D its diameter, C the circumference of a great circle, and S the surface of the sphere, then we shall have

C=2πR, or πD (Prop. XIII., Cor. 2, B. VI.).

Also, S=2TR X2R=4πR2, or πD2.

If A represents the altitude of a zone, its area will be

2πRA.

PROPOSITION VIII. THEOREM.

The solidity of a sphere is equal to one third the product of its surface by the radius.

Let ACEG be the semicircle by the revolution of which the sphere is described. Inscribe in the semicircle a regular semi-polygon ABCDEFG, and draw the radii BO, ČO, DO, &c.

The solid described by the revolution of the polygon ABCDEFG about AG, is composed of the solids formed by the revolution of the triangles ABO, BCO, CDO, &c., about AG.

First. To find the value of the solid formed by the revolution of the triangle ABO.

G·

From O draw OH perpendicular to AB, and from B draw BK perpendicular to AO. The two triangles ABK, BKO, in their revolution about AO, will describe two cones having a common base, viz., the circle whose radius is BK. Let area BK represent the area of the circle described by the revolution of BK. Then the solid described by the triangle ABO will be represented by Area BK× AO (Prop. V.).

Now the convex surface of a cone is expressed by TRS (Prop. III., Cor.); and the base of the cone by R. Hence the convex surface: base : πRS: TR2,

::S:R (Prop. VIII., B. II.).

But AB describes the convex surface of a cone, of which BK describes the base; hence

the surface described by AB : area BK :: AB BK
:: AO: OH,
because the triangles ABK, AHO are similar. Hence
Area BK AO= OH x surface described by AB,
Area BK × AO=}OH × surface described by AB.

or

But we have proved that the solid described by the triangle ABO, is equal to area BKXAO; it is, therefore, equal to OHX surface described by AB.

Secondly. To find the value of the solid formed by the revolution of the triangle C BCO.

Produce BC until it meets AG produced in L. It is evident, from the preceding demonstration, that the solid described by the triangle LCO is equal to

OMX surface described by LC; and the solid described by the triangle LBO is equal to

N

D

OMX surface described by LB;

M

B

T

hence the solid described by the triangle BCO is equal to OMX surface described by BC.

AK

In the same manner, it may be proved that the solid described by the triangle CDO is equal to

1ONX surface described by CD;

and so on for the other triangles. But the perpendiculars OH, OM, ON, &c., are all equal; hence the solid described by the polygon ABCDEFG, is equal to the surface described by the perimeter of the polygon, multiplied by OH.

Let, now, the number of sides of the polygon be indefinitely increased, the perpendicular OH will become the radius ÓA, the perimeter ACEG will become the semi-circumference ADG, and the solid described by the polygon becomes a sphere; hence the solidity of a sphere is equal to one third of the product of its surface by the radius.

Cor. 1. The solidity of a spherical sector is equal to the prod uct of the zone which forms its base, by one third of its radius. For the solid described by the revolution of BCDO is equal to the surface described by BC+CD, multiplied by OM. But when the number of sides of the polygon is in definitely increased, the perpendicular OM becomes the radius OB, the quadrilateral BCDO becomes the sector BDO, and the solid described by the revolution of BCDO becomes a spherical sector. Hence the solidity of a spherical sector is equal to the product of the zone which forms its base, by one third of its radius.

Cor. 2. Let R represent the radius of a sphere, D its diameter, S its surface, and V its solidity, then we shall have S=4πR1 or πD' (Prop. VII., Cor. 5).

Also, V=RXS=4πR or πD;

hence the solidities of spheres are to each other as the cubes of their radii

If we put A to represent the altitude of the zone which forms the base of a sector, then the solidity of the sector will be represented by

2πRA× RπR'A.

Cor. 3. Every sphere is two thirds of the circumscribed cylinder.

For, since the base of the circumscribed cylinder is equal to a great circle, and its altitude to a diameter, the solidity of the cylinder is equal to a great circle, multiplied by the diameter (Prop. II.). But the solidity of a sphere is equal to four great circles, multiplied by one third of the radius; or one great circle, multiplied by of the radius, or of the diameter. Hence a sphere is two thirds of the circumscribed cylinder.

A spherical segment with one base, is equivalent to half of a cylinder having the same base and altitude, plus a sphere whose diameter is the altitude of the segment.

Let BD be the radius of the base of the segment, AD its altitude, and let the segment be generated by the revolution of the circular half segment AEBD about the axis AC. B Join CB, and from the center C draw CF perpendicular to AB.

The solid generated by the revolution of

E

the segment AEB, is equal to the difference of the solids generated by the sector ACBE, and the triangle ACB. Now, the solid generated by the sector ACBE is equal to

3

TCB XAD (Prop. VIII., Cor. 2).

And the solid generated by the triangle ACB, by Prop. VIII., is equal to CF, multiplied by the convex surface described by AB, which is 2πCF× AD (Prop. VII.), making for the solid generated by the triangle ACB,

2

TCFX AD.

3

Therefore the solid generated by the segment AEB, is equal

to

or

that is,

πAD×(CB2-CF2),
TADXBF2;
πADXAB2,
6

9

because CB2-CF is equal to BF2, and BF" is equal to one fourth of AB2.

Now the cone generated by the triangle ABD is equal to πAD×BD2 (Prop. V., Cor. 2).

Therefore the spherical segment in question, which is the sum of the solids described by AEB and ABD, is equal to

that is,

πAD(2BD2+AB2);
πAD(зBD+ AD),

because AB2 is equal to BD2+AD2.

This expression may be separated into the two parts
¿πAD × BD3,
¿πAD3.

and

The first part represents the solidity of a cylinder having the same base with the segment and half its altitude (Prop. II.); the other part part represents a sphere, of which AD is the diameter (Prop. VIII., Cor. 2). segment, &c.

Therefore, a spherical

Cor. The solidity of the spherical segment of two bases, generated by the revolution of BCDE about the axis AD, may be found by subtracting that of the segment of one base generated by ABE, from that of the segment of one base generated by ACD.

CONIC SECTIONS.

THERE are three curves whose properties are extensively applied in Astronomy, and many other branches of science, which, being the sections of a cone made by a plane in dif ferent positions, are called the conic sections. These are The Parabola, The Ellipse, and The Hyperbola.

PARABOLA.

Definitions.

1. A parabola is a plane curve, every point of which is equally distant from a fixed point, and a given straight line. 2. The fixed point is called the focus of the parabola and the given straight line is called the directrix.

Thus, if F be a fixed point, and BC a B given line, and the point A move about F in such a manner, that its distance from F D is always equal to the perpendicular distance from BC, the point A will describe a parabola, of which F is the focus, and BC the directrix.

3. A diameter is a straight line drawn through any point of the curve perpendicular to the directrix. The vertex of the diameter is the point in which it cuts the curve.

D

A

Thus, through any point of the curve, as A, draw a line DE perpendicular to the directrix BC; DE is a diameter of the parabola, and the point A is the vertex of this diameter.

4. The axis of the parabola is the diameter which passes through the focus; and the point in which it cuts the curve is called the principal vertex.

Thus, draw a diameter of the parabola, GH, through the