Then the surfaces of C and P are as their bases (Art. 47. and 88.); which are as the bases of C' and P', (Sup. Euc. 7, 1.); so that,

surfC:surfP::baseC; baseP::baseC': baseP':: surfC':surfP'

But the surface of C is, by supposition, equal to the surface of C'. Therefore, (Alg. 395.) the surface of P is equal to the surface of P'. And by the preceding article,

solidP: solidC:: surfP: surfC:: surfP': surfC' :: solidP': solidC

But the solidity of P is greater than that of P'. (Art. 92. Cor.) Therefore, the solidity of C is greater than that of C'.

Schol. A right cylinder whose height is equal to the diameter of its base, is that which circumscribes a sphere. It is also called Archimedes' cylinder; as he discovered the ratio of a sphere to its circumscribing cylinder; and these are the figures which were put upon his tomb.

Cor. Archimedes' cylinder has a less surface, than any other right cylinder of the same capacity.

PROPOSITION XV.

96. If a SPHERE BE CIRCUMSCRIBED by a solid bounded by plane surfaces; the capacities of the two solids are as their surfaces.

If planes be supposed to be drawn from the center of the sphere, to each of the edges of the circumscribing solid, they will divide it into as many pyramids as the solid has faces. The base of each pyramid will be one of the faces; and the height will be the radius of the sphere. The capacity of the pyramid will be equal, therefore, to its base multiplied into of the radius (Art. 48.); and the capacity of the whole circumscribing solid, must be equal to its whole surface multiplied into of the radius. But the capacity of the sphere is also equal to its surface multiplied into of its radius. (Art. 70.)

Cor. The capacities of different solids circumscribing the same sphere, are as their surfaces.

PROPOSITION XVI.

97. A SPHERE has a greater solidity than any regular polyedron of equal surface.

If a sphere and a regular polyedron have the same center, and equal surfaces; each of the faces of the polyedron must fall partly within the sphere. For the solidity of a circumscribing solid is greater than the solidity of the sphere, as the one includes the other: and therefore, by the preceding article, the surface of the former is greater than that of the latter.

But if the faces of the polyedron fall partly within the sphere, their perpendicular distance from the center must be less than the radius. And therefore, if the surface of the polyedron be only equal to that of the sphere, its solidity must be less. For the solidity of the polyedron is equal to its surface multiplied into of the distance from the center. (Art. 59.) And the solidity of the sphere is equal to its surface multiplied into of the radius.

Cor. A sphere has a less surface than any regular polyedron of the same capacity.

For other cases of Isoperimetry, see Fluxions.

72

APPENDIX.-PART I.

CONTAINING RULES, WITHOUT DEMONSTRATIONS, FOR THE MENSURATION OF THE CONIC SECTIONS, AND OTHER FIGURES NOT TREATED OF IN THE FLEMENTS OF EUCLID.*

PROBLEM I.

To find the area of an ELLIPSE.

101. Multiply the product of the transverse and conjugate axes into .7854.

Ex. What is the area of an ellipse whose transverse axis is 36 feet, and conjugate 28? Ans. 791.68 feet.

PROBLEM II.

To find the area of a SEGMENT of an ellipse, cut off by a line perpendicular to either axis.

102. If either axis of an ellipse be made the diameter of a circle; and if a line perpendicular to this axis cut off a segment from the ellipse, and from the circle;

The diameter of the circle, is to the other axis of the ellipse; As the circular segment, to the elliptic segment.

For demonstrations of these rules, see Conic Sections, Spherical Trigonometry, and Fluxions, or Hutton's Mensuration.

Ex. What is the area of a segment cut off from an ellipse whose transverse axis is 415 feet, and conjugate 332; if the height of the segment is 96 feet, and its base is perpendicular to the transverse axis?

The circular segment is 23680 feet.
And the elliptic segment

18944

PROBLEM III.

To find the area of a conic PARABOLA.

103. Multiply the base by of the height.

23

Ex. If the base of a parabola is 26 inches, and the height 9 feet; what is the area? Ans. 13 feet.

PROBLEM IV.

To find the area of a FRUSTUM of a parabola, cut off by a line parallel to the base.

104. Divide the difference of the cubes of the diameters of the two ends, by the difference of their squares; and multiply the quotient by of the perpendicular height.

Ex. What is the area of a parabolic frustum, whose height is 12 feet, and the diameters of its ends 20 and 12 feet? Ans. 196 feet.

PROBLEM V.

To find the area of a conic HYPERBOLA.

៖

105. Multiply the base by 3 of the height; and correct the from it the series

oduct by subtractiuble ordinate,

2bh (1.3.5

+ &c.)

+ 5.7.9 +7.9.11+ &c.

b=the base or

In which h-the height or abscissa,

z=the height divided by the sum of the height

and transverse axis.

The series converges so rapidly, that a few of the first terms will generally give the correction with sufficient exactness. This correction is the difference between the hyperbola, and a parabola of the same base and height.

Ex. If the base of a hyperbola be 24 feet, the height 10 and the transverse axis 30; what is the area?

To find the area of a SPHERICAL TRIANGLE formed by three arcs of great circles of a sphere.

106. As 8 right angles or 720°,

To the excess of the 3 given angles above 180°;
So is the whole surface of the sphere,

To the area of the spherical triangle.

Ex. What is the area of a spherical triangle, on a sphere whose diameter is 30 feet, if the angles are 130°, 1020, and 68°? Ans. 471.24 feet.

PROBLEM VII.

To find the area of a SPHERICAL POLYGON formed by arcs of great circles.

107. As 8 right angles, or 7200,

To the excess of all the given angles above the product of the number of angles-2 into 180°;

So is the whole surface of the sphere,

To the area of the spherical polygon.