OF FINDING THE CONVEX SURFACES OF SOLIDS. PROBLEM I. To find the convex surface of a right cylinder, the circumference and length of the cylinder being given. Multiply the circumference by the length of the cylinder, and the product will be the area. EXAMPLE. What is the convex superficies of a right cylinder, whose circumference is 2ft. 6in. and the length 5ft. 3in.? Then 5f. 31. × 2f. 6i. = 13f. 11. 6ii. the answer. Note. If the diameter is given, find the circumference, and proceed as before. VOL. I. те PROBLEM PROBLEM II. To find the convex superficies of a right cone, the circumference and slant side being given. Multiply the circumference by the slant side of the cone, and half the product will be the area, EXAMPLE. What is the convex superficies of a right cone, the circumference of the base being 4.5 ft. and the slant side 6·25 ft.? Then 6.25 x 4.5 14-0625 the superficies required. 2 Note. If the diameter is given, find the circumference, and proceed as before. PROBLEM III. To find the convex surface of the frustum of a right cone, the circumferences of both ends being given, and the slant side of the cone. Multiply the sum of the circumferences by the slant side of the cone, and half the product will be the area. EXAMPLE. What is the convex superficies of a frustum of a right cone, the circumference of the base being 46 ft. the circumference of the top being 3.25 ft. and the altitude 5'75 feet? Then 463·25 × 575 = 22.56875 the answer. 2 PROBLEM IV. To find the superficies of a sphere or globe, the greatest circumference being given. Multiply the square of the circumference by 3183, and the product will be the superficies. EXAMPLE. What is the superficies of a globe, the greatest circumference being 10.6 ft.? Then 106 × 106 × 3183 35 764188 the superficies required. PROBLEM V. To find the convex superficies of the segment of a sphere or globe, the diameter of the base of the segment, and its height, being given. To the square of the diameter of the base, add the square of twice the height; and the sum multiplied by 7854, will give the superficies. EXAMPLE. What is the convex surface of the segment of a globe, the diameter of the base being 17.25 ft. and the height 4'5 ft.? 2 × 459 twice the height. 9 × 9 = 81 square of twice the height. 2 17:25 = 297:5625 square of the diameter of the base. Then 297-562581 × 7854 perficies required. 297 3229875 the su PROBLEM VI. To find the convex surfaces of a spherical zone, the diameters of the ends and their distance being given. Find the diameter of the sphere by Prob. XIII. and Prob. XIV. in Mensuration of Superficies; then multiply the diameter of the sphere, and the distance of the parallel ends of the zone together, and the product by 3'1416, will give the superficies required. EXAMPLE. In a spherical zone, the distance of the parallel ends being 4in. the diameter of the greater end 24in. and that of the lesser end 20in. what is the convex superficies, when the centre of the sphere is without the zone? The distance of the greater chord from the centre, will be found to be 3.5 inches, by Prob. XIII. The radius will be found to be 25 inches, by Prob. XIV. or the diameter 50 inches. Then 50 x 4 × 31416 = 628 32 the answer. Note. If the diameter is given, find the circumference, and proceed as before. 1 |