radius (P. XIII., C.). But this portion of the semicircle is a circular sector, the volume which it generates is a spherical sector, and the surface generated by the arc is a zone: hence, the volume of a spherical sector is equal to the zone which forms its base multiplied by one third of the radius.

Cor. 2. If we denote the volume of a sphere by V, and its radius by R, the area of the surface will be equal to 47R2 (P. X., C. 1), and the volume of the sphere will be equal to 47R2 × R; consequently, we have,

Again, if we denote the diameter of the sphere by D, we shall have R equal to 4D, and Rs equal to D3, and consequently,

hence, the volumes of spheres are to each other as the cubes of their radii, or as the cubes of their diameters.

Scholium. If the figure EBDF, formed by drawing lines from the extremities of the arc BD perpendicular to CA, be revolved about CA, as an axis, it will generate a segment of a sphere whose volume may be found by adding to the spherical sector generated by CDB, the cone generated by CBE, and subtracting from their sum the cone generated by

D

B

CDF. If the arc BD is so taken that the points E and F fall on opposite sides of the centre C, the latter cone must be added, instead of subtracted. The area of the zone BD is equal to 2-CD x EF (P. X., C. 2); hence,

segment EBDF = π (2CD × EF + BE2 x CE DF3× CF).

PROPOSITION XV. THEOREM.

The surface of a sphere is to the entire surface of the circumscribed cylinder, including its bases, as 2 is to 3; and the volumes are to each other in the same ratio.

Let PMQ be a semicircle, and PADQ a rectangle, whose sides PA and QD are tangent to the semicircle at P and Q, and whose side AD, is tangent to the semicircle at M. If the semicircle and the rectangle be revolved about PQ, as an axis, the former will generate a sphere, and the latter a circumscribed cylinder.

1°. The surface of the sphere is to the entire surface of the cylinder, as 2 is to 3.

M

For, the surface of the sphere is equal to four great circles (P. X., C. 1), the convex surface of the cylinder is equal to the circumference of its base multiplied by its altitude (P. I.); that is, it is equal to the circumference of a great circle multiplied by its diameter, or to four great circles (B. V., P. XV.); adding to this the two bases, each of which is equal to a great circle, we have the entire surface of the cylinder equal to six great circles : hence, the surface of the sphere is to the entire surface of the circumscribed cylinder, as 4 is to 6, or as 2 is to 3; which was to be proved.

P

2. The volume of the sphere is to the volume of the cylinder as 2 is to 3.

For, the volume of the sphere is equal to πR3 (P. XIV., C. 2); the volume of the cylinder is equal to its base multiplied by its altitude (P. II.); that is, it is equal to

R2 x 2R, or to

R: hence, the volume of the sphere is cylinder as 4 is to 6, or as 2 is to 3; which was to be proved.

to that of the

Cor. The surface of a sphere is to the entire surface of a circumscribed cylinder, as the volume of the sphere is to the volume of the cylinder.

Scholium. Any polyedron which is circumscribed about a sphere, that is, whose faces are all tangent to the sphere, may be regarded as made up of pyramids, whose bases are the faces of the polyedron, whose common vertex is at the centre of the sphere, and each of whose altitudes is equal to the radius of the sphere. But, the volume of any one of these pyramids is equal to its base multiplied by one third of its altitude: hence, the volume of a circumscribed polyedron is equal to its surface multiplied by one third of the radius of the inscribed sphere.

Now, because the volume of the sphere is also equal to its surface multiplied by one third of its radius, it follows that the volume of a sphere is to the volume of any circumscribed polyedron, as the surface of the sphere is to the surface of the polyedron.

Polyedrons circumscribed about the same, or about equal spheres, are proportional to their surfaces.

GENERAL FORMULAS.

If we denote the convex surface of a cylinder by S, its volume by V, the radius of its base by R, and its altitude by H, we have (P. I., II.),

If we denote the convex surface of a cone by S, its volume by V, the radius of its base by R, its altitude by H, and its slant height by H', we have (P. III., V.),

If we denote the convex surface of a frustum of a cone by S, its volume by V, the radius of its lower base by R, the radius of its upper base by R', its altitude by H, and its slant height by H', we have (P. IV., VI.),

If we denote the surface of a sphere by S, its volume. by V, its radius by R, and its diameter by D, we have (P. X., C. 1, XIV., C. 2, XIV., C. 1),

If we denote the radius of a sphere by R, the area of any zone of the sphere by S, its altitude by H, and the volume of the corresponding spherical sector by V, we shall have (P. X., C. 2, XIV., C. 1),

S = 2TR X H

V

= TR2 X H

(9.)

. (10.)

If we denote the volume of the corresponding spherical segment by V, its altitude by H, the radius of its upper base by R', the radius of its lower base by R", the distance of its upper base from the centre by H', and of its lower base from the centre by H", we shall have (P. XIV., S.):

V = (2R2XH + RH' FR2xH") (11.)

EXERCISES.

1. The radius of the base of a cylinder is 2 feet, and its altitude 6 feet; find its entire surface, including the bases.

2. The volume of a cylinder, of which the radius of the base is 10 feet, is 6283.2 cubic feet; find the volume of a similar cylinder of which the diameter of the base is 16 feet, and find also the altitude of each cylinder.

3. Two similar cones have the radii of the bases equal, respectively, to 4 and 6 feet, and the convex surface of the first is 667.59 square feet; find the convex surface of the second and the volume of both.

4. A line 12 feet long is revolved about another line as an axis; the distance of one extremity of the line from the axis is 4 feet and of the other extremity 6 feet; find the area of the surface generated.

5. Find the convex surface and the volume of the frustum of a cone the altitude of which is 6 feet, the radius of the lower base being 4 feet and that of the upper base 2 feet.

6. Find the surface and the volume of the cone of which the frustum in the preceding example is a frustum.

7. A small circle, the radius of which is 4 feet, is 3 feet from the centre of a sphere; find the circumference of a great circle of the same sphere.

8. The radius of a sphere is 10 feet; find the area of a small circle distant from the centre 6 feet.

9. Find the area of the surface generated by the semiperimeter of a regular semihexagon revolving about its axis, the radius of the inscribed circle being 5.2 feet and the axis 12 feet.

10. The area of the surface generated by the semi