Proposition 7. Theorem. 778. The volume of any cone is equal to one-third the product of its base and altitude. Hyp. Let V, B, H denote the volume of the cone, the area of its base, and its altitude, respectively. Proof. Inscribe in the cone a pyramid, and let V' and B' denote its volume and the area of its base. Now let the number of lateral faces of the pyramid be indefinitely increased. B' will approach B as its limit. ... V' will approach V as its limit. ... VB × H. (430) Q.E.D. 779. COR. 1. For the volume of a cone of revolution, whose altitude is H and radius of the base is R, we have 780. COR. 2. The volumes of similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. (759) EXERCISES. 1. Required the lateral area and volume of a right circular cone whose altitude is 24 inches and radius of the base 10 inches. Ans. 816.82 sq. ins.; 2513.3 cu. ins. 2. Required the entire surface of a right circular cone whose altitude is 16 inches and radius of the base 12 inches. Ans. 1206.37 sq. ins. Proposition 8. Theorem. 781. The volume of a frustum of any cone is equal to the sum of the volumes of three cones, whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum. Hyp. Let V, B, b, H denote the volume of the frustum, its bases, and its altitude, respectively. Proof. Inscribe in the frustum of the cone the frustum of a pyramid, and let V', B', b' denote its volume and the areas of its bases. Then, H V′ = ¦(B′ + b′ + √B′b′). V' (635) Now let the number of lateral faces of the inscribed frustum be indefinitely increased. B' and 'will approach B and b respectively, as their limits. ... V' will approach V as its limit. (430) Q. E. D. 782. COR. 1. If the frustum is that of a right circular cone, and the radii of its bases are R and r, we have B = πR2, b = πr2. · · . V = }πH(R2 + 22 + Rr). 783. COR. 2. This formula may be put into the form [(+)+1/(B)] 3 2 NOTE.-The volume of a cask may be found approximately by this formula, in which H = total height of cask, R end. = radius of mid-section, and r = radius of The nearest approximation in the case of most casks is given by the formula THE SPHERE. DEFINITIONS. 784. A zone is a portion of the surface of a sphere included between two parallel planes. The altitude of the zone is the perpendicular distance between the parallel planes. The bases of the zone are the circumferences of the circles which bound the zone. If one of the parallel planes touches the sphere, the zone is called a zone of one base. 785. A spherical segment is a portion of the volume of a sphere included between two parallel planes. The altitude of the segment is the perpendicular distance between the parallel planes. The bases of the segment are the sections of the sphere made by the parallel planes. A segment of one base is a segment one of whose bounding planes touches the sphere. 786. A spherical sector is a portion of the volume of a sphere generated by the revolution of a circular sector about a diameter of the circle. A G H 787. Let the sphere be generated by the revolution of the semicircle ACDEFB about its diameter AB as an axis; and let CG and DH be drawn perpendicular to the axis. The arc CD generates a zone whose altitude is GH, and the figure CDHG generates a spherical segment whose altitude is GH. The circumferences generated by the points C and E D are the bases of the zone, and the circles generated by CG and DH are the bases of the segment, The arc AC generates a zone of one base, and the figure ACG a spherical segment of one base. The circular sector OEF generates a spherical sector whose base is the zone generated by the arc EF; the other bounding surfaces are the conical surfaces generated by the radii OE and OF. If OF coincides with OB, the spherical sector is bounded by a conical surface and a zone of one base. If OE is perpendicular to OB, the spherical sector is bounded by a plane surface, a conical surface, and a zone. Proposition 9. Theorem. 788. The area generated by a straight line revolving about an axis in its plane, is equal to the product of the projection of the line on the axis by the circumference whose radius is the perpendicular erected at the middle point of the line and terminated by the axis. Hyp. Let AB be the revolving line, CD its projection on the axis GO, and EO the at the mid. pt. of AB and termi nating in the axis. To prove area AB = CD × 2πEO. Proof. Draw EFL, and AH || to GO. The area generated by AB is the lateral B F area of a frustum of a cone of revolution, whose slant height is AB and axis CD. The ... area AB = AВ × 2πEF. AS ABH and EOF are similar. ... AB X EF = AH × EO = CD × EO. ... area AB = CD × 2πЕО. (777) (316) If AB meets the axis, or is || to it, thus generating a conical, or a cylindrical surface, the result is the same from (773) and (753). Q.E.D. Proposition 10. Theorem. 789. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Hyp. Let the sphere be generated by the revolution of the semicircle ABDF about the diameter AF, let O be the centre, R the radius, and denote the surface of the sphere by S. Proof. Inscribe in the semicircle a regular semi-polygon ABCDEF, of any number of sides. C Draw Bb, Cc, etc., I to AF and OHL to AB. area AB Аb × 2πОН. B TH A C e F (201) (788) Then, Similarly, area BC = bc × 27OH, and so on. In equal os equal chords are equally distant from the centre (206). Now the sum of the projections Ab, bc, etc., of all the sides of the semi-polygon make up the diameter AF. .. the entire surface generated by the revolving semipolygon AF X 27ΟΗ. = Now let the number of sides of the inscribed semi-polygon be indefinitely increased. The semi-perimeter will approach the semi-circumference as its limit, and OH will approach the radius R as its limit. (430) ... the surface of revolution will approach the surface of the sphere as its limit. ... S AF × 2πR. Q. E.D, |