647. DEF. A section of a cylinder is the figure formed if the cylinder is intersected by a plane; a right section is a section formed by a plane perpendicular to the elements. PROPOSITION XXIII. THEOREM 648. Every section of a cylinder made by a plane passing through an element is a parallelogram. Given ABCD, a section of cylinder AC, made by plane through element AB. To prove ABCD is a parallelogram. Proof. Through D, in plane AC, draw a line | AB. This line is an element of the cylinder (638). Since this line is in the plane AC and is an element of the cylindrical surface, it must be their intersection, and therefore coincides with DC. Also .. DC is a straight line | AB. AD is a straight line || BC. .. ABCD is a parallelogram. (Why?) Q. E. D. 649. COR. Every section of a right cylinder made by a plane passing through an element is a rectangle. PROPOSITION XXIV. THEOREM 650. The bases of a cylinder are congruent. B Given ABC and A'B'C', the bases of the cylinder BC. Proof. Let A, B, and C be any three points in the lower base, and AA', BB', CC' the corresponding elements. Place the upper base upon the lower base so that A'B' coincides with AB. Then 'must coincide with C. But as c' is any point in the perimeter of the upper base, all points in the upper base coincide with the corresponding points in the lower base. .. the bases are congruent. Q. E. D. 651. COR. 1. Any two parallel sections cutting all the elements of a cylinder are congruent. 652. COR. 2. Any section of a cylinder parallel to the base is congruent to the base. 653. DEF. The area of the lateral surface of a cylinder is the limit which the lateral area of an inscribed prism ap proaches, if the number of its lateral faces is indefinitely increased. The volume of a cylinder is the limit which the volume of the inscribed prism approaches if the number of its faces is increased indefinitely. PROPOSITION XXV. THEOREM 654. The lateral area of a cylinder is equal to the product of the perimeter of a right section of the cylinder by an element. Given L the lateral area, P the perimeter of a right section, and E an element of the cylinder AK; L' the lateral area, P' the perimeter of a section of a prism with a regular polygon as base, inscribed in cylinder AK. Proof. The edge of the inscribed prism coincides with an element of the cylinder. If the number of faces of the inscribed prism be increased, L' will approach L as a limit, and P' will approach P as a limit. 655. DEF. A cylinder of revolution is a right circular cylinder because it may be generated by a rectangle revolving about one of its sides as an axis. 656. DEF. Similar cylinders of revolution are cylinders generated by similar rectangles revolving about homologous sides as axes. 657. COR. 1. is the product of The lateral area of a cylinder of revolution the circumference of its base by its altitude. 658. COR. 2. If I denote the lateral area, T the total area, H the altitude, and R the radius of a cylinder of revolution, L=2 TRH, and Ex. 1593. Find the total area of a cylinder of revolution, if H = 6, and R = 2. PROPOSITION XXVI. THEOREM 659. The volume of a circular cylinder is equal to the product of its base by its altitude. K Given A the volume, B the base, and H the altitude, of cylinder AK; ' the volume, B' the regular polygon forming the base of a prism inscribed in AK. HINT. Use the Theorem of Limits as in preceding proposition.* * Assume that B is the limit which B' approaches, if the number of sides is indefinitely increased. R, 660. COR. For a cylinder of revolution with radius of base V=TR2 X H. 661. The lateral areas, or the total areas, of similar cylinders of revolution are to each other as the squares of their radii, or as the squares of their altitudes; and their volumes are to each other as the cubes of their radii, or as the cubes of their altitudes. Given L, L', the lateral areas; T, T', the total areas; V, V', the volumes; R, R', the radii; and H, H', the altitudes of two similar cylinders of revolution. L : L' = T : T' = R2 : R'2 = H2 : H12, V : V' = R3 : R′3 — H3 : H′3. To prove and = H+R L 2 TRH L' 2 πR'H' 2 TR (H+R) 2 TR' (H'+R') πR2H V TR'2H' = = = X R/2 |