BOOK X. MEASUREMENT OF THE THREE ROUND BODIES. CYLINDERS. 666. A prism is inscribed in a cylinder when its bases are inscribed in the bases of the cylinder, and its lateral edges are elements of the lateral surface of the cylinder. 667. A prism is circumscribed about a cylinder when its bases are circumscribed about the bases of the cylinder. 668. The lateral area of a cylinder is the area of its lateral surface. PROPOSITION I. THEOREM. 669. The lateral area of a cylinder of revolution is measured by the product of the circumference of its base by its altitude. B B Given C, the circumference of the base of a cylinder of revolution ABCD-C'; H, its altitude; and S, its lateral surface; Inscribe in the cylinder a regular prism ABCD-C', whose bases are regular polygons inscribed in the bases of the cylinder. If we denote by s the lateral surface of the prism, and by p the perimeter of each of the base polygons; then, H being also its altitude, s=px H, whatever be the number of lateral faces of the prism. (514) Let the number of lateral faces be indefinitely increased by continually doubling the number of sides of the base polygons. Then as p has for limit c (392), and the lateral edges of the prism are, if indefinitely increased in number, the elements of the surface s, 670. COR. 1. If the cylinder be generated by a rectangle whose sides are H and R, revolving about H, then H is the altitude of the cylinder and R the radius of the base. Hence the perimeter of the base is 2R (396), and for the lateral area S we obtain the expression, S = 2π · R. H. 671. COR. 2. Since the area of each base is π· R2 (398), we obtain for T, the total area of a cylinder of revolution, the expression, T 2π · R(H+ R). = · 672. DEFINITION. Similar cylinders of revolution are generated by similar rectangles revolving about homologous sides. 673. COR. 3. The lateral areas, or the total areas, of similar cylinders of revolution, are as the squares of their altitudes or radii. For let s and s' denote the lateral areas of two similar cylinders of revolution; R and R', the radii of their bases; H and H', their altitudes; T and T', their total areas; then, since the generating rectangles are similar, (Hyp.) H: H' = R : R' = H + H' : R + R'; .. S: S' and T: T' 2 2. RH: 2π R'. H' = H2: H'2 = R2: R2, (246) · R(H+R): 2π• R' (H'+R') = H2 : H'2 = R2 : R'2. 674. SCHOLIUM. The lateral area of any cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of its surface. This may be proved by a method similar to that employed in the proof of Prop. I., assuming that the bases of the cylinders, when not circular, are still the limits of inscribed polygons, the number of whose sides is indefinitely great. 675. The volume of a cylinder of revolution is measured by the product of its base by its altitude. B Given: V, the volume of a cylinder of revolution AB, whose base is B and altitude H; Let ' denote the volume of a regular inscribed prism of any number of faces; B', its base; H will also be its altitude. Now whatever be the number of sides of the prism, But when the number of faces is indefinitely increased, B' has for limit B, and v' has for limit v; 676. COR. 1. Let v be the volume of the cylinder, R the radius of its base, and H its altitude; then, since B = π · R2, • R2. H. 677. COR. 2. The volumes of similar cylinders of revolution are to each other as the cubes of their altitudes or radii. For if v and v' be the volumes of two similar cylinders of revolution, R and R' the radii of their bases, H and H' their altitudes; since the generating rectangles are similar, (Hyp.) .'. V : V' = π · R2 · H : π R12. H = H3 : H13 = R3 : R13. 678. SCHOLIUM. The volume of any cylinder is measured by the product of its base by its altitude. This may be proved by the same method as that employed in Prop. II., making the assumption before referred to (674). EXERCISE 836. Show that, by the following construction, two lines, M and N, can be found such that M: N = 355: 113. Take AB=10 units of any convenient length, and on it lay off AC= 5, and AD = 3. 1. Join AE, and draw CF, DG, each 1 to F, G resp. Take M3 AB+ AC + CF, Draw BE1 to AB and = CONES. 679. A pyramid is inscribed in a cone when its base is inscribed in the base of the cone, and its vertex is that of the cone, the lateral edges of the pyramid thus being elements of the surface of the cone. 680. A pyramid is circumscribed about a cone when its base is circumscribed about the base of the cone, and its vertex is that of the cone. 681. The altitude of a cone is the perpendicular distance from its vertex to its base. 682. The slant height of a cone of revolution is equal to the hypotenuse of the generating triangle. 683. The lateral area of a cone is the area of its lateral surface. 684. The lateral area of a cone of revolution is measured by the product of the circumference of its base by one half its slant height. S A Given S, the lateral area; C, the circumference of the base; and I, the slant height of a cone of revolution S-ADF; Inscribe in the base any regular polygon ADF; and upon this polygon as base construct the regular inscribed pyra |