* Here .7854 x 30.7854 x 900=706.86-area of the base AB. 3. From of the cube of the right sine of half the arc of the base, subtract the product of the area of the base and the cosine of the said half arc. 4. Multiply the difference, thus found, by the quotient arising from the height divided by the versed sine, and the product will be the solidity. IV. When the section passes obliquely through both ends of the cylinder. Rule 1. Conceive the section to be continued, till it meets the side of the cylinder produced; then say, as the difference of the versed sines of half the arcs of the two ends of the ungula, is to the versed sine of half the arc of the less end, so is the height of the cylinder to the part of the side produced. 2. Find the surface of each of the ungulas, thus formed, by Case III. and their difference will be the surface required. 3. In like manner find the solidities of each of the ungulas, and their difference will be the solidity required. * In working the examples in this and the following rules, .7854 is used for the area, and 3.1416, the circumference of a circle whose diameter is 1; where greater accuracy is required, .7853981634 may be used for the area, and 3.14159265359 for the circumference. See Note to Prob. IX. Superficies. 35343 And 706.86x50=35343 cubic inches; or 1728 20.4531 solid feet, the answer required. 2. What is the solidity of a cylinder whose height is 5 feet, and the diameter of the end 2 feet? Ans. 15.708 feet. 3. What is the solidity of a cylinder whose height is 20 feet, and the circumference of its base 20 feet also? Ans. 636.64 feet. 4. The circumference of the base of an oblique cylinder is 20 feet, and the perpendicular height 19.318; what is the solidity? Ans. 614.93 feet. PROBLEM VI. To find the convex surface of a right cone. RULE.* Multiply the circumference of the base by the slant height, or the length of the side of the cone, and half the product will be the surface required. * Demon. Let AB=a, BC=b, 3.1416=p, and ED=y. fluent: phy pha convex sur 2a 2 which, when y=a, becomes face of the whole cone. Q. E. D. To find the surface of a right pyramid. Rule. Multiply the perimeter of the base by the length of the side, or slant height of the cone, and half the product will be the surface required. EXAMPLES. 1. The diameter of the base AB is 3 feet, and the slant height AC or BC 15 feet; required the convex surface of the cone ACB. Bl=b Here 3.1416x3=9.4248-circumference of the base AB, 9.4248×15 141.3720 And = =70.686 square feet, the convex surface required. 2. The diameter of a right cone is 4.5 feet, and the slant height 20 feet; required the convex surface. Ans. 141.372 feet. 3. The circumference of the base is 10.75, and the slant height 18.25; what is the convex surface? PROBLEM VII. Ans. 98.09375. To find the convex surface of the frustrum of a right cone. RULE.* Multiply the sum of the perimeters of the two ends, by the slant height of the frustrum, and half the product will be the surface required. * Demon. Let the perimeter of the circle AB=P, that of DE=p, BD=h, and the rest as in the last problem. EXAMPLES. 1. In the frustrum ABDE, the circumferences of the two ends AB and DE are 22.5 and 15.75 respectively, and the slant height BD is 26; what is the convex surface? Then P:p:: (BC): CD; and, by division, P=p: p:: ph PP ; but p × (h+: ph P-p twice the con b-CD(h): CD; vex surface of the whole cone, by the last rule; and also px ph P-P -=twice the convex surface of the part ECD. There ph ph ph P-p hp P+pxh=twice the convex surface of the frustrum -PX- -—hp+P—p× = hp + P-P P-P fore PX)h+ ABDE; and the half thereof is (P+p)xh which is the same 2 as the rule. Q. E. D. To find the surface of the frustrum of a right pyramid. Rule. Multiply the sum of the perimeters of the ends by the slant height, and half the product will be the surface required. 1. What is the convex surface of the frustrum of a right cone, the circumference of the greater end being 30 feet, that of the less end 10 feet, and the length of the slant side 20 feet? Ans. 400 feet. 2. What is the convex surface of the frustrum of a right cone, the diameters of the ends being 8 and 4 feet, and the length of the slant side 20 feet? Ans. 376.992 feet. 3. If a segment of 6 feet slant height be cut off a cone whose slant height is 30 feet, and circumference of its base 10 feet; what is the surface of the frustrum? Ans. 144 feet. PROBLEM VIII. To find the solidity of a cone or pyramid. RULE.* Multiply the area of the base by one third of the perpendicular height of the cone or pyramid, and the product will be the solidity. * Demon. Let sc=a, cs=x, and A=area of the base of the cone ACB. Then a2(cs2): x2(Cs2) :: AB2 : ED2 (by sim. As) :: A: Ax2 a (area of the circle ED) because all the circles are as the squares of their diameters. Ax2 But xx-fluxion of the cone ECD, and its fluent= a2 which, when 3a2 dity of the whole cone. Aa a 3 becomes =AX for the soli- In the pyramid CEDB it will be a2(cs2): x2(Cs2) :: CE3 Ce2:: ED2: eo2 (by sim. As) :: A (area of the base EB) Ax2 (area of the polygon eb) because all similar figures are as the squares of their like sides. |