EXAMPLES. Ex. 1. Required the volume of a cylinder whose diameter of base is 30 inches, and the length 25 inches. Area of base = 30a X •7854706·86; .'. volume = 706·86 × 25 = 176715 cubic in. Ex. 2. Find the volume and weight of a brass cylinder, the diameter of which is 12 inches, and height 9 inches; a cubic foot of brass weighing 525 lbs. Volume Hence weight: = 7854 X 12a × 9 = 1017.8784 cubic inches. =30925125 lbs. 309 lbs. 4 oz. cubic inches. lbs. 1017.8784 X 525 233. To find the volume of a cone. RULE LXXXV. Proceed exactly as in the foregoing Rule, and divide the result by 3. EXAMPLE. Required the volume of a cone, the diameter of the base being 10 inches, and height 15 inches. Volume area of base of height= *7854 X 102 X 15 3 392'70 cubic inches. 234. Def.-The frustum of a cone is the part that is left after the top or end is cut off by a plane parallel to the base. To find the volume of the frustum of a cone. RULE LXXXVI. Add the squares of the diameters of the two ends to their product; this sum multiplied by one-third the perpendicular height, and again by 7854 will give the solidity of the frustum. 235. To find the volume of a sphere. Solidity of sphere whose diameter is D. Cube the diameter and multiply by 5236 (= 31416); the product will be the solidity. Ex. 1. Ex. 2. EXAMPLES. What is the solidity of a sphere whose diameter is 24 inches ? What is the cubical content of a sphere whose diameter is 7 inches ? Cubical content diam.3 X 5236 =73 X 5236 = 343 X *5236 EXAMPLES FOR PRACTICE. 1. The radius of a sphere is 44 inches: find its solidity. 2. 3. If the diameter of a balloon is 6 ft. 2 in.: how many cubic feet of gas will it contain ? A spherical body contains 2.80957 cubic feet: what is its surface? 4. The dimeter of the sun is 112 times greater than that of the earth: how many times greater is its volume? 5. A sphere contains 5946 cubic inches: find its radius. 6. If the cubical content of the earth is taken as 236855164967 cubic feet: what is its circumference. To find the volume of the segment of a sphere. RULE LXXXVIII. To three times the square of the radius of its base add the square of its height; then multiply the sum by the height, and the product by ·5236. To find the volume of a cylindrical ring. RULE LXXXIX. 3'14162 Add the inner diameter to the thickness of the ring, multiply the sum by the square of the thickness, and the product by 2'4674 (= ), and it will give the solidity. EXAMPLE. Ex. The thickness of a cylindrical ring is 2 inches, and the inner diameter 5 inches: required the solidity? (2 + 5) × 22 = 7 X 4 = 28; then 28 X 2°4674 69'0872 cubic inches, Ans. EXAMPLES FOR PRACTICE. I. Required the solidity of an iron ring whose axis forms the circumference of a circle; the diameter of a section of the ring 2 inches, and the inner diameter, from side to side, 18 inches. 2. The thickness of a cylindrical ring is 7 inches, and the inner diameter 20 inches: required its solidity. 3. What is the solidity of a circular ring whose thickness is 2 inches, and its diameter 12 inches ? SUPERFICIAL AREAS OF SOLID BODIES. 236. To find the superficies of a sphere. RULE XC. Multiply the diameter by the circumference, or multiply 3'1416 by the square of the diameter, and the product will be the surface. Or, multiply the square of the diameter by 3·1416. EXAMPLE. Ex. What is the surface of a globe whose diameter is 24 inches? Circumference 24 X 3'1416 = 75*3984 inches. Ans. Surface = 75°3984 X 24 = 1809°5616 square inches. I. 2. EXAMPLES FOR PRACTICE. What is the surface of the earth, its diameter being 7957 miles, and circumference 25000 miles? also find it from each of these data taken separately. The surface of a sphere contains 6734 square feet: find its radius. 3. The radius of a sphere is 1 ft. 4 in.: what is its surface? I 4. Find the surface of a sphere whose diameter is 7. 237. To find the convex surface of a cylinder, having given the diameter and length. RULE XCI. 1°. Multiply the circumference by the length of the cylinder, the product is the convex surface. NOTE. If the surface of the entire solid be required, the area of the two ends must be found and added to the convex surface, the sum is the entire surface required. I. 20 62.8320 EXAMPLES FOR PRACTICE. What is the number of square inches in the outside surface of a cylinder whose outer diameter is 15 inches, and height 20 inches? 2. What is the convex surface of a cylinder whose circumference is 76 inches and length 5 ft. 4 in. ? What is the surface of a cylinder, including the ends, its girth being 12 ft. and length 15 ft. ? 4. Required the interior surface of a cylinder the length being 12 ft, and the radius of the bore 23 inches. 5. Required the superficial content of a cylinder whose diameter is 21.5 in. and height 16 ft. ? 238. To find the convex surface of a right cone. RULE XCII. Multiply the circumference of the base by the slant height or side of the cone, and take half the product for the area. 239. To find the convex surface of a cylindrical ring. RULE XCIII. Multiply the extreme diameter by the thickness of the ring, and the product by 9.8696 (= 3*14162) the result is the area. со SIMPSON'S RULE FOR AREAS. WHEN more than ordinary accuracy is required in finding the area of a space lying between a curve and a straight line. Divide the base of the figure (suppose it to be the following) into any number of equal parts, say eight (the more we take the greater accuracy we attain), by drawing lines at right-angles to the base; this will give an odd number of points of division. The measurements of the length of these perpendiculars, or ordinates, must be taken from a scale of equal parts, and used as in the following, which is called Simpson's Rule : = - Let S the sum of the lengths of the first and last ordinates. d Then the area = Whence the following 3 RULE XCIV. x d. To the sum of the lengths of the first and last ordinates add four times the sum of the lengths of all the other even ordinates, and twice the sum of the remaining, or odd ordinates; then this sum multiplied by one-third the common distance between the ordinates is equal to the area. For example: given the following figure, ABC: required the area. Area 7.37 This rule is much used by shipbuilders in calculating displacements, water lines, &c., and is on the general assumption that the curve is parabolic; it may also be applied to the "working off" the indicator diagram, I. 2. 3. EXAMPLES FOR PRACTICE. The length of a rectangle is three times its breadth, and its area is 5808 yards: what is the length in feet? The largest possible circle is cut out of a piece of paper in the form of a square whose side is 7 inches: what is the area of the remnants ? A circle of 2 inches diameter is cut out from one whose diameter is 3 inches: what is the area of the remainder? 4. Show that a circle of 5 inches diameter is as large as two others of 4 inches and 3 inches diameter together. 5. The area of the coal field of South Wales is 100 square miles, and the average thickness of the coal is 60 feet. If a cubic yard of coal weighs 1 ton, and the annual consumption of coal in Great Britain be 70,000,000 tons: find the number of years for which this coal field alone would supply Great Britain with coal at the present rate of consumption. If the coal annually consumed in this country were piled up into a pyramid, having for its base the great court of Trinity College, the dimensions of which are 110 by 90 yards: find the height of the pyramid. N.B.-The volume of pyramid is equal to the area of the base multiplied into one-third of the height. 6. The sides of a rectangle contain 12:19 feet and 375 feet: find its area. If the length of the sides were only known to be correct within 06 of an inch, how much might the true area exceed the approximate one? 7. What length must be cut off a board 6 inches wide, that the area may be a square foot? 8. What fraction of a square foot is a rectangle whose sides are of a foot and 18 of a foot ? 9. A ship's hold is a 100 feet long, 50 feet broad, and 4 feet deep: how many bales of goods, each 2 feet 4 inches long, 2 feet 1 inch broad, and 2 feet 1 inch deep, can be stowed into it, leaving a gangway of 3 feet broad, and of the same length and depth as the hold ? 10. II. 12. What is the area contained between two concentric circles whose radii are 2 feet 2 inches and 4 feet? Find the area of an annular valve of which the greater and less diameters are 9 and 5 inches respectively. The diameters of the circles are 10 and 6: required the area of the circular ring. 13. The area of an oval varies as its length x breadth. When the length and breadth are each inch, the area is square inch: what is the area of an oval whose length is 7 inches and breadth 5 inches? 14. The volume of a circular cylinder varies as the height × square of diameter. When the diameter and height are each 1 inch, the volume is cubic inch: what is the volume of a cylinder whose diameter is 4 inches and height 21 inches ? 15. Find the surface and volume of a sphere whose radius is 6 feet 9 inches. 16. Of two cylinders, one contains 154 lbs. of water, and its dimensions are half those of the other: how many cubic feet of water will the latter cylinder hold? A cubic foot of water weighing 1000 ounces nearly. 17. If the diameter of the earth is 8000 miles, and the interior to the depth of 5 miles is known: how much of the earth's interior would still be unknown? 18. Admitting the height of the atmosphere to be 45 miles, what would be its solid contents ? 19. If 405 million square inches of gilding can be made of a cubic foot of gold: find its thickness. 20. A valve is 6 inches in diameter: show that the pressure on it is 141 372 lbs., whether it be considered to be loaded to 5 lbs. per square inch or 3'927 per circular inch, |