EXAMPLES. 1. What is the solidity of the cylinder ABCD, the diameter 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. 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. D B 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 'by the versed sine, and the product will be the curve surface. dividea.. * Here .7854 × 302.7854× 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 706.86 x 50=35343 cubic inches; or 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 Then a : b circle ED. by ED=y. -= ::y: =DC; and py=circumference of the by a a -fluxion of the surface of CED, and its But pyx fluent_phy? which, when pha 2a becomes y=a, 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. Here 3.1416×3=9.4248-circumference of the base AB. And 9.4248 X 15 141.3720 2 convex surface required. =70.686 square feet, the 2 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? Ans. 98.09375. PROBLEM VII. 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=4, and the rest as in the last problem. EXAMPLES. 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 Pp: b(BC): CD; and, by division, P-p:p :: ph P-P 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 fore PX (h+ P-P ph P-P P-P hp=P+pxh=twice the convex surface of the frustrum ABDE; and the half thereof is (P+p)×h which is the same 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. |