by 6; the height of the prismoid is 4 feet: the prismoid is divided by a plane parallel to the ends and midway between them: find the volume of each part. 7. The ends of a prismoid are rectangles, the corresponding dimensions of which are 16 feet by 11, and 10 feet by 8; the height of the prismoid is 9 feet: the prismoid is divided into three parts, each 3 feet high, by planes parallel to the ends: find the volume of each of the parts. 8. The ends of a prismoid are rectangles, the corresponding dimensions of which are 20 feet by 16, and 14 feet by 12; the height of the prismoid is 5 feet: the prismoid is cut into two wedges by a plane which passes through one of the longer sides of one end, and the opposite longer side of the other end: find the volume of each part. 9. The edge of a wedge is 24 inches, the length of the base 8 inches, and the breadth 7 inches; the height of the wedge is 16 inches; the wedge is divided into two parts by a plane parallel to the base midway between the edge and the base: find the volume of each part. 10. The edge of a wedge is 27 inches; the length of the base is 18 inches, and its breadth 15 inches; the height of the wedge is 12 inches; the wedge is divided into three parts of equal height by two planes parallel to the base: find the volume of each part. 11. The ends of a prismoid are rectangles, the corresponding dimensions of which are 18 feet by 10, and 12 feet by 16; the height of the prismoid is 9 feet; a section is made by a plane parallel to the ends at the distance of 3 feet from the larger end: shew that the section is a square. 12. Find the volumes of the two parts in the preceding Example. XXIX. SPHERE. 291. To find the volume of a Sphere. RULE. Multiply the cube of the diameter by one-sixth of 31416, that is, by 5236. 292. Examples. (1) The diameter of a sphere is 10 inches. The cube of 10 is 1000; 5236 × 1000=523'6. Thus the volume of the sphere is about 523'6 cubic inches. (2) The diameter of a sphere is 3 feet. The cube of 3·5 is 42·875; 42·875 × ·5236=22.44935. Thus the volume of the sphere is very nearly 22:45 cubic feet. 293. The volume of a spherical shell will of course be obtained by subtracting the volume of a sphere, having its diameter equal to the inner diameter of the shell from the volume of a sphere having its diameter equal to the outer diameter of the shell. Thus we obtain the Rule which will now be given. 294. To find the volume of a spherical shell. RULE. Subtract the cube of the inner diameter from the cube of the outer diameter, and multiply the remainder by 5236. 295. Examples. (1) The outer diameter of a spherical shell is 9 inches, and the thickness of the shell is 1 inch. Here the inner diameter will be 7 inches. The cube of 9 is 729; the cube of 7 is 343; 729-343=386; •5236 x 386=202.1096. Thus the volume of the shell is very nearly 202.11 cubic inches. (2) The inner diameter of a spherical shell is 10 inches, and the thickness of the shell is 131⁄2 inches. Here the outer diameter will be 13 inches. The cube of 13 is 2197; the cube of 10 is 1000; 2197-1000=1197; 5236 × 1197 = 626°7492. Thus the volume of the shell is very nearly 626·75 cubic inches. 296. If one sphere fall entirely within the other, it is obvious that the Rule of Art. 294 will give the volume of the space between the surfaces of the two spheres, even when the spheres are not concentric. 297. We will now solve some exercises. (1) The circumference of a great circle of a sphere is 28 inches: find the volume of the sphere. We first determine the diameter of the sphere; by Art. 111 this will be about 8.9 inches: then by Art. 291 we shall obtain for the volume of the sphere about 369.12 cubic inches. (2) Find the weight of a leaden ball 5 inches in diameter, supposing a cubic inch of lead to weigh 6·6 ounces. The cube of 5 is 125; 125 × 5236 = 65'45. Thus the volume of the ball is 65 45 cubic inches; and therefore its weight in ounces = = 65'45 × 6'6=431.97. (3) If a cubic inch of gold weighs 11.194 ounces, find the diameter of a ball of gold which weighs 1000 ounces. The number of cubic inches in the ball will be 1000 11.194' that is, about 89.334; this number then is equal to the product of the cube of the diameter into 5236. Thus to obtain the cube of the diameter we must divide 89.334 by *5236; the quotient will be found to be 170 615. The cube root of this number will be the diameter; we shall find that this cube root is 5'546. Thus the diameter of the ball is about 5'55 inches. EXAMPLES. XXIX. Find the volumes of spheres having the following dia meters: 1. 11 inches. 2. 8 feet. 3. 24 feet. 4. 32.5 feet. Find to the nearest hundredth of a cubic foot the volumes of spheres having the following circumferences of great circles: 5. 6 feet. 6. 8 feet. 7. 10 feet. 8. 12 feet. Find in cubic inches the volumes of spherical shells having the following dimensions: 9. External diameter 5 inches, internal 4. 10. External diameter 8 inches, internal 6. 11. External diameter 10 inches, internal 7. 12. External diameter 16 inches, internal 12. 13. Find how many gallons a hemispherical bowl, 2 feet 4 inches in diameter, will hold. 14. Find how long it will take to fill a hemispherical tank of 10 feet diameter, from a cistern which supplies by a pipe 6 gallons of water per minute. 15. A solid is in the form of a right circular cylinder with hemispherical ends; the extreme length is 29 feet and the diameter 3 feet: find the volume. 16. A solid is in the form of a right circular cylinder with hemispherical ends; the extreme length is 22 feet and the diameter 2 feet 6 inches: find what will be the weight of water equal in bulk to this solid. 17. A sphere, 4 inches in diameter, is cut out of a cube of wood, the edge of which is 4 inches: find the quantity of wood which is cut away. 18. Find the weight of a spherical shot of iron, 6 inches in diameter, supposing a cubic inch of iron to weigh 4.2 ounces. 19. If a sphere of lead, 4 inches in diameter, weighs 2213 ounces, find the weight of a sphere of lead 5 inches in diameter. 20. Find the weight of gunpowder required to fill a hollow sphere of 7 inches diameter, supposing that 30 cubic inches of gunpowder weigh one lb. 21. Find the weight of gunpowder required to fill a hollow sphere of 9 inches diameter. 22. Find the weight of a spherical shell one inch thick, the external diameter of which is 10 inches, composed of a substance a cubic foot of which weighs 216 lbs. 23. Find the weight of a spherical shell 1 inches thick, the external diameter of which is 11 inches, composed of iron weighing 4 cwt. to the cubic foot. 24. The external diameter of a shell is 8-4 inches and the internal 7.2 inches: find the weight of the shell if it is composed of a substance of which a cubic foot weighs 7860 ounces. 25. Find the weight of a shell 1g inches thick, the external diameter of which is 13 inches, composed of metal a cubic inch of which weighs 44 ounces. 26. Find the weight of a shell 3 inches thick, the external diameter of which is 1 foot 5 inches, composed of metal a cubic foot of which weighs 480 lbs. 27. If an iron ball, 4 inches in diameter, weigh 9 lbs., find the weight of an iron shell 2 inches thick, whose external diameter is 20 inches. 28. If a shell, the external and internal diameters of which are 5 inches and 3 inches, weighs 8 lbs., find the weight of a shell composed of the same substance, the external and internal diameters of which are 7 inches and 4 inches. 29. Shew that the weight of a cone, 7 inches high on a circular base, of which the radius is 2 inches, is equal to that of a spherical shell of the same material, of which the external diameter is 4 inches and the thickness 1 inch. 30. Find the weight of a pyramid of iron, such that its height is 8 inches and its base is an equilateral triangle, each side being 2 inches, supposing a ball of iron 4 inches in diameter to weigh 9 lbs. |