PROBLEM III. To find the Surface of the Frustum of a Cone or RULE.-Multiply the sum of the perimeters of the two ends by the slant height, and half the product will be the slant surface; to which add the areas of the two ends, and the product will be the whole surface. EXAMPLE.-Required the convex surface of the frustum of a cone A B C D, whose base A B = 20 inches, the slant height B C = 19, and top end CD = 11. 3.1416 x 20 +3.1416 x 11 x 19 2 =925.2012 square inches, and B PROBLEM IV. To find the Solid Content of the Frustum of a Cone. RULE. To the product of the diameters of the two ends, add the sum of their squares; multiply this sum by the perpendicular height and by .2618, the product is the solid content. EXAMPLE 1.-Required the solid content of the frustum in problem 3, whose perpendicular E F = 18 inches. 20 × 11 220 and 220 + 202 + 112 × 18 x .2618 8491.8884 cubic inches, and divided by 1728 2.0208 cubic feet nearly. EXAMPLE 2.-Required the content in imperial gallons, of the inverted frustum of a cone A B CD, whose inner dimensions are 34 feet deep, 18 inches diameter at bottom, and 22 inches diameter at top. 22 × 18=396, and 396+222 +182 D PROBLEM V. To find the Solid Content of the Frustum of a Pyramid. RULE. To the sum of the areas of the two ends, add the square root of, their product, multiply this sum by the perpendicular height; and of the product is the solid content. EXAMPLE.-Required the solid content of the frustum of a pyramid A B C D, whose perpendicular height= =24 inches, the area of the base 144 inches, and area of the top end = 64. 14464208 and 144 × 64 =96 then 208 +96 × 24 3 =2432 cubic inches, and ÷ 1728=1.4074 D B PROBLEM VI. To find the Solidity of a Wedge. RULE. To the length of the edge add twice the length of the base; multiply that sum by the height, and by the breadth of the base, and of the product will be the solidity. EXAMPLE.-Required the content in cubic inches of the wedge A B C D E, whose base A B C = 12 inches long and 4 inches broad, the length of the To find the Convex Surface and Solid Content of a Sphere or Globe. RULE 1.-Multiply the square of the diameter by 3.1416, the product will be the convex superficies. RULE 2.-Multiply the cube of the diameter by .5236, and the product is the solid content. EXAMPLE 1.-Required the convex surface of a sphere, whose diameter A B = 25 inches. EXAMPLE 2.-Required the solid content of a sphere, whose diameter A B = 25 inches. 25.53 x .5236 8682.00795 cubic inches 17285.0243 cubic feet nearly. PROBLEM VIII. To find the Convex Surface and Solid Content of the Segment of a Sphere. RULE 1.-Multiply the height of the segment by the whole circumference of the sphere, and the product is the curve surface. RULE 2.-Add the square of the height to three times the square of the radius of the base; multiply that sum by the height, and by .5236, and the product is the solid content. EXAMPLE 1.-The diameter A B of the sphere ABCD 20 inches; what is the convex surface of that segment of it, whose height E D = 8 inches? F 3.1416 x 20 x 8502.656 square inches 144 3.49 A superficial feet nearly. E EXAMPLE 2.--The base F G of the segment FDG = 18 inches, and perpendicular ED 8, what is the solid content? 82 64., and 92 × 3 = 243, then 243 +64 x 8 x .5236 1728 EXAMPLE 3.-Suppose A B C D to be a sugar pan, and that the diameter of the mouth A B is 4 feet, the depth D C being 25 inches, how many imperial gallons will it contain ? RULE.-Multiply the square of the revolving axis by the fixed axis, and by 5236, and the product will be the solidity. EXAMPLE 1. Required the solid content of the prolate spheroid A B C D, whose fixed axis AC is 50, and revolving axis B D 30. 302 x 50 x .5236 = 23562 the solidity. A B EXAMPLE 2.-What is the solid content of an oblate spheroid, the fixed axis being 30, and revolving axis 50? 502 x 30 x .5236 = 39270 the solid content. PROBLEM X. To find the Solidity of the Segment of a Spheroid when the base is circular or parallel to the revolving axis. RULE. From triple the fixed axis, take double the height of the segment; multiply the difference by the square of the height, and by .5236; then say, as the square of the fixed axis is to the square of the revolving axis, so is the former product to the solidity. EXAMPLE 1.-Required the solid content of the segment A B C; whose height Br is 10; the revolving axis E F being 40, and fixed axis B D 25. EXAMPLE 2.-What is the solid content of the segment of a spheroid, whose height 20 inches, the revolving axis being 25, and fixed axis 50. 50 x 3 20 x 2 = 110, and 110 × 202 × .523623038.4; then, as 502 252 :: 23038.4 : 5759.6 inches, the solid content. |