If a cone and a pyramid have equal bases and altitudes, they are equal in their solidity. Consequently, the rule already given for the frustrum of a pyramid is equally applicable to the frustrum of a cone. (Art. 41.) Ex. 1. How many cubic feet in a piece of round timber, the diameter of the greater end being 18 inches, and that of the smaller, 9 inches, and the length 14.25 feet ? Then, 254.4696x63.6174-127.23-square root of product of the two areas. And, 254.4696 63.6174 127.234 H 445.3210 sum of areas of two ends in inches, and square root of their product. Then, 14.25 feet-171 inches. Therefore, 445.3210×171÷1728-14.6894 solid feet. Ex. 2. How many gallons of ale are contained in a cistern in the form of a conic frustrum, if the larger diameter be 9 feet, and the smaller diameter 7 feet, and, the depth 9 feet? PROBLEM VII. To find the Surface of a Sphere or Globe. ART. 56. Rule.-I. Multiply the diameter of the sphere by its circumference, and the product will be the surface. Or, II. Multiply the square of the diameter by 3.14159. Ex. 1. What is the surface of a sphere whose diameter is 7 feet? OPERATION. First, 7x3.14159-21.99113= circumference. Then, 21.99113x7-153.93791 sq. ft.-surface. Ex. 2. The dome of the Capitol at Washington is hemispherical, and is 95 feet across the base; how many square feet in its convex surface? Ans. 28352.85 sq. ft. NOTE. By the same rule the mechanic may readily determine the number of square feet of sheathing, painting, wainscoting, or plastering, in the external or internal surface of any dome, however large or small. If the dome be not hemispherical, it is the segment or zone of a sphere, and its convex surface is found by Art. 57, following. Ex. 3. How many square feet of lead will it require to cover a hemispherical dome whose base is 13 feet across? Ans. 265.5. NOTE.-The surface of a sphere is equal to the convex surface of a cylinder whose diameter is equal to the diameter of the sphere, and whose height is also equal to the altitude of the sphere. Or, the surface of a sphere is equal to 4 times the area of a circle of the same diameter. Therefore, to find the surface of a sphere, we have the fol lowing concise Rule: Multiply the area of a circle of the same diameter of the sphere into 4, and the product will be the entire surface. Consequently, the area of a hemisphere is equal to twice the area of the base. For the area of a circle is equal to the product of half the diameter into half the circumference, or, what is the same in the result, of the product of the diameter and circumference. Therefore, if the circumference of a circle be multiplied by its diameter, the product will be 4 times the area of the circle. PROBLEM VIII. To find the Convex surface of a Spherical Zone or Segment. ART. 57. Rule.-Multiply the height of the zone or segment by the whole circumference of the sphere of which it is a part, and the product will be the convex surface. NOTE. The convex surface of any spherical segment or zone is equal to that of the circumscribed cylinder, by Art. 58, (Note.) Ex. 1. If the axis of a sphere, be 42 inches, what is the convex surface of a segment or zone ABD, A. whose height ED is 9 inches? OPERATION. First, 42x3.14159-131.9468-circumference. 9-height. D B E 1187.5212-surface in square inches. Ex. 2. If the diameter of the earth be 7930 miles, what is the convex surface of the torrid zone extending 23°20' on each side of the equator. Ans. 78.669.700.sq. miles. Ex. 3. The diameter of a sphere is 25 feet and the height of the zone is 4 feet, what is the convex surface of the zone? Ans. 314.159. PROBLEM IX. To find the Solidity of a Sphere or Globe. ART. 58. Rule I.-Multiply the surface by the diameter, and of the product will be the solidity. Or, II. Multiply the square of the diameter by the circumference, and of the product will be the contents. Or, III. Multiply the cube of the diameter by the decimal .5236. NOTE.-If we put D=the diameter, C-circumference, and S-the surface of the sphere, or its equal the circumscribing cylinder; also, A the number 3.14159; Then of S is the base of the cylinder, and D is the height of the cylinder; therefore, DXS is the solidity of the cylinder. But it is proved that a sphere is two thirds of its circumseribing cylinder (Sup. Euc. 21.3.) Therefore, of DX S, or its equal, DXS the solidity of the sphere according to the first rule. 2. But since the solidity of the cylinder is equal to its height multiplied into the area of its base, (Art. 51,) and since the height and diameter of the cylinder are each equal to the diameter of the sphere, if we put D again for the diameter, we shall have by the 3d rule: D'x .7854XD, or D3x.7854 solidity of circumscribing cylinder. A nd since the solidity of the sphere is of this, we have D3x.5236=the solidity of the sphere. 3. There are certain numbers frequently occurring in mathematical investigations, and particularly in determining the areas of circles and solidities of spheres, &c., which cannot be made too familiar to the mind. They are the following: 3,14159; .7854; and .5236. The first represents the ratio of the circumference of a circle to its diameter; the second expresses the ratio of the area of a circle to the square of the diameter; and the third, the ratio of the solidity of a sphere to the cube of its diameter. If we divide 3.14159 by 4, the quotient is .7854 very nearly, and again 3.14159 divided by 6=.5236. 12x3.14159-452.38996-surface of the sphere. -3 Or thus: 12-1728-cube of the diameter. And, 1728x.5236-904.78-solid contents. Ex. 2. What is the solidity of the earth, if it be a sphere, 7930 miles diameter ? Ans. 261.107.000.000. cubic miles. Ex. 3. The diameter of a sphere is 16; what is its solidity? Ans. 2144.6656. Ex. 4. Required the solidity of a globe whose diameter is 18 inches. Ans. 3315.23 cubic inches. ART. 59.-Knowing the solidity of a sphere, we may determine the diameter by reversing the third rule in this article; that is, by dividing the solidity by .5236 and extracting the cube root of the quotient. Ex. 1. What is the diameter of a sphere whose solidity is 4.188.8000. In this case we divide the solidity by .5236, and extract the cube root of the quotient for the diameter; and so the diameter of any solid OPERATION. .5236)4188.8000(8000 4188.8 0 may be found, whose capa- And, 8000-20=diameter. city is given. Ex. 2. Required the diameter of a sphere whose solidity is 113.0976. Ans. 6. Ex 3. What must be the diameter of a globe to contain 16755 pounds of water? Ans. 8 ft. PROBLEM X. To find the Solidity of a Spherical Segment. ART. 60. Rule.-To three times the square of the radius |