EXAMPLES. 1. Required the solidity of the cone ACB, whose diameter AB is 20, and its perpendicular height CS 24. C Here .7854 20 = .7854 x 400=314.16=area of the base AB. 24 And 314.16 x =314.16 x 8=2513.28 = solidity re 3 quired. 2. Required the solidity of the hexagonal pyramid ECBD, each of the equal sides of its base being 40, and the perpendicular height CS 60. AX But one x x=fluxion of the pyramid Ceob, and its correct fluent=AX s the same as in the cone; and this rule is general, let the figure of the base be what it will. Here 2.598076 (multiplier when the side is 1). 402= 2.598076 x 1600=4156.9216=area of the base. 60 And 4156.9216 x 3 =4156.9216 X 20 =83138.432 solidity required. 3. Required the solidity of a triangular pyramid, whose height is 30, and each side of the base 3. Ans. 38.97117. 4. Required the solidity of a square pyramid, each side of whose base is 30, and the perpendicular height 20. Ans. 6000 5. What is the solidity of a cone, the diameter of whose base is 18 inches, and its altitude 15 feet? Ans. 8.83575 feet. 6. If the circumference of the base of a cone be 40 feet, and the height 50 feet; what is the solidity ? Ans. 2122.1333 feet. 7. What is the content of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet? Ans. 27.5276 PROBLEM IX. To find the solidity of a frustrum of a cone or pyramid. RULE.* 1. For the frustrum of a cone, the diameters, or circumferences of the two ends, and the height being given. * Demon. First let D= =diameter AB, d=ED, p=.7854, h=ss= the height of the frustrum ABDE of the cone. See the last figures. Then d:d :: cs : cs, and D-d :d :: cs- -Cs (h) : dh dh =cs=height of the cone EDC. But dh =solidity of the whole cone 3 dh solidity of the cone Therefore (h+ 3 D pd? pd = the pp? ECD. D Add together the square of the diameter of the greater end, the square of the diameter of the less end, and the product of the two diameters; multiply the sum by .7854, and the product by the height; } of the last product wilí be the solidity. Or, Add together the square of the circumference of greater end, the square of the circumference of the less end, and the product of the two circumferences ; multiply. he sum by .07958, and the product by the height; $ of the .ast product will be the solidity. II. For the frustrum of a pyramid whose sides are regular polygons. Add together the square of a side of the greater end, the square of a side of the less end, and the product of ( pd D dh dh dh р d? x d hp D? — dx Х (v? + ď® + d) 3 the solidity of the frustrum ABDE, which is the same as the rule. And, since the circumferences of circles have the same ratio that their diameters have, if C be put for the circumference of the greater end, c=that of the less end, and p=.07958, the demonstration of the rule, when the circumferences are given, will differ in nothing from the above. Again, for the polygon, let S=ED, s=ed, and m=proper multi. plier in the table of polygons; then S:8:: CS : Cs, and S-8:8:: hs CS--C8 (1): S 8 But ms? and ms? are the areas of polygons whose sides ms2 hs are s and 8 respectively. And therefore 3 +(h+ h -(ms+ ms2-ms3 x Х 3 =(ms? + msi 3 h mh +mss) * (83 +52 +ss) * =solidity of the frustrum 3 3 exdib which is the same as the rule. S -S S these two sides; multiply the sum by the proper number in the table, Prob. VIII. of Superficies, and the product by the height: $ of the last product will be the solidity. Note. When the ends of the pyramids are not regular polygons.' Add together the areas of the two ends and the square root of their product; multiply the sum by the height, and of the product will be the solidity. EXAMPLES. 1. What is the solidity of the frustrum of the cone EABD, the diameter of whose greater end AB is 5 feet, that of the less end ED, 3 feet, and the perpendicular height Ss, 9 feet? (52 + 32 +5 x 3).7854 x 9 346.3614 =115.4538 solid 3 3 feet, the content of the frustrum. 2. What is the solidity of the frustrum eEDBb of an hexagonal pyramid, the side ED of whose greater end is 4 feet, that eb of the less end 3 feet, and the height Ss, 9 feet? (4* + 32 +4*3)* 2.598076.9 865.159308 =288.386436 3 3 solid feet, the solidity required. 3. What is the solidity of the frustrum of a cone, the diameter of the greater end being 4 feet, that of the less end 2 feet, and the altitude 9 feet? Ans. 65.9736. The following cases contain all the rules for finding the superficies and solidities of conical ungulas. 1. When the section passes through the opposite extremities of the ends of the frustrum. С D А B Let D=AB the diameter of the greater end; d=CD, the diameter of the less end; h=perpendicular height of the frustrum, and n=.7854. d2—d dd ndh Х =solidity of the greater elliptic 3 ungula ADB. Dvod-da ndh Х =solidity of the less ungula ACD. da nh Х difference of these hoofs. -d 3 D+d And v 4h2 +(1—da) X (D2- V dd) =curve 2 surface of ADB. II. When the section cuts off part of the base, and makes the angle DrB less than the angle CAB. с Dd 3 D n D |