Here AB=22.5=25.5 x 25.5 x 25.5=25.5 x 650.25 =16581 375 the answer, or content of the tube. 2. The side of a cube is 15 inches : what is the solidity ? Aps. 1ft. 1lin. 5pa. 3. What is the solidity of a cube whose side is 17.5 inches ? Ans. 3.1015 feet. PROBLEM II. To find the solidity of a parallelopipedon. RULE.* Multiply the length by the breadth, and that product again by the depth or altitude, and it will give the solidity required. EXAMPLES. 1. Required the solidity of a parallelopipedon ABCD FEHG, whose length AB is 8 feet, its breadth FD 41 feet, and the depth or altitude AD 6 feet? the square of the side of the base multiplied by the base, is equal to the solidity. Q. E. D. Note.—The surface of the cube is equal to six times the square of its side. * The reason of this rule, as well as of the following ones for the prism and cylinder, is the same as that for the cube. Note.-The surface of the parallelopipedon is equal to the sum of the areas of each of its sides or ends. Here ABX ADXFD=8X6.75X4.5 = 54x4.5 = 243 solid feet, the contents of the parallelopipedon required. 2 The length of a parallelopipedon is 15 feet, and each side of its square base 21 inches : what is the solidity ? Ans. 45.9375 feet. 3. What is the solidity of a block of marble, whose length is 10 feet, its breadth 54 feet, and the depth 3} feet? Ans. 201.25 feet. PROBLEM III. To find the solidity of a prism. RULE.* Multiply the area of the base into the perpendicular height of the prism, and the product will be the solidity. EXAMPLES. 1. What is the solidity of the triangular prism ABCF ED, whose length AB is 10 feet, and either of the equal sides, BC, CD, or DB, of one of its equilateral ends BCD, 21 feet? E F * The surface of a prism is equal to the sum of the areas of the two ends and each of its sides. * Here x 2.5x 3=1x6.25 X V3=1.5625 x 3= 1.5625 1.732=2.70625=area of the base BCD. 2.5+2.5 +2.5 7.5 Or, 2 = 3.75 = f sum of the sides, 2 BC, CD, DB, of the triangle CDB. And 3.75—2.5=1.25.1.25, 1.25 and 1.25=3 differ ences. Whence 3.75 X 1.25 x 1.25 x 1.25=73.75x1.25= ✓ 7.32421875=2.7063=a =area of the base as before, And 2.7063 x 10=27.063 solid feet, the content of the prism required. 2. What is the solidity of a triangular prism, whose length is 18 feet, and one side of the equilateral end 1} feet? Ans. 17.5370265 feet. 3. Required the solidity of a prism whose base is a hexagon, supposing each of the equal sides to be 1 foot 4 inches, and the length of the prism 15 feet. Aps. 69.282ft. PROBLEM IV. To find the convex surface of a cylinder. RULE. Multiply the periphery or circumference of the base, by the height of the cylinder, and the product will be the convex surface required. EXAMPLES 1. What is the convex surface of the right cylinder ABCD, whose length BC is 20 feet, and the diameter of its base AB 2 feet? * See Notes to Prob. III. Cor. 2. p. 56. † Demon. If the periphery of the base be conceived to move in a direction parallel to itself, it will generate the convex superficies of the cylinder; and, therefore, the said periphery being multiplied by the length of the cylinder, will be equal to that superficies. Q. E. D. Note. If twice the area of either of the ends be added to the convex surface, it will give the whole surface of the cylinder. Here 3.1416 X 2=6.2832=periphery of the base AB. And 6.2832 X 20=125.6640 square feet, the convexity required. 2. What is the convex surface of a right cylinder, the diameter of whose base is 30 inches, and the length 60 inches ? Ans. 5654.88 inches. 3. Required the convex superficies of a right cylinder, whose circumference is 8 feet 4 inches, and its length 14 feet. Ans. 116.666, &c. feet. PROBLEM V. To find the solidity of a cylinder. RULE.* Multiply the area of the base by the perpendicular height of the cylinder, and the product will be the solidity. * The four following cases contain all the rules for finding the superficies and solidities of cylindric ungulas. I. When the section is parallel to the axis of the cylinder. F EXAMPLES. 1. What is the solidity of the cylinder ABCD, the dia meter of whose base AB is 30 inches, and the height BC 50 inches? Rule 1. Multiply the length of the arc line of the base by the height of the cylinder, and the product will be the curve surface. 2. Multiply the area of the base by the height of the cylinder, and the product will be the solidity. II. When the section passes obliquely through the opposite sides of the cylinder. H Rule 1. Multiply the circumference of the base of the cylinder by half the sum of the greatest and least lengths of the ungula, and the product will be the curve surface. 2. Multiply the area of the base of the cylinder by half the sum of the greatest and least lengths of the ungula, and the product will be the solidity. III. When the section passes through the base of the cylinder, and one of its sides. BE A G Rule 1. Multiply the sine of half the arc of the base by the diameter of the cylinder, and from this product subtract the product of the arc and cosine. 2. Multiply the difference thus found, by the quotient of the height divided by the versed sine, and the product will be the curve surface. |