PROBLEM XII. To find the convex surface of a sphere. RULE.* Multiply the diameter of the sphere by its circumference and the product will be the convex superficies required. Note.—The curve surface of any zone or segment will also be found by multiplying its height by the whole cir. cumference of the sphere. * Demon. Put the diameter BG=d, BA=X, AC=y, BC=%; and 3.1416=p. Then, since the triangles AOC and CED are similar, we shall d dx have ca (y): Co 2 :: CE (*): CD(2)=% But 2pyz is the general expression for the fluxion of any surface; and therefore, by dx substituting 2y for its equal 2, the fluxion will become pdx; and consequently pds=surface of any segment of a sphere whose heiglit is X, and pdd=that of the whole sphere. Q. E. D. Cor. 1. The surface of a sphere is also equal to the curve surface of its circumscribing cylinder. Cor. 2. The surface of a sphere is also equal to four times the area of a great circle of it. 1. To find the lunar surface included between two great circles of the sphere. Rule. Multiply the diameter into the breadth of the surface in the middle, and the product will be the superficies required. Or, As one right angle is to a great circle of the sphere; To the surface included by them. 2. To find the area of a spherical triangle, or the surface included by the intersecting arcs of three great circles of the sphere. Rule. As two right angles, or 180°, Is to a great circle of the sphere; EXAMPLES. 1. What is the convex superficies of a globe BCG whose diameter BG is 17 inches ? D Here 3.1416 x 17 x 17=53.4072 x 17=907.9224 square inches. And 907.9224--144=6.305 square feet, the answer. 2. What is the convex superficies of a sphere whose diameter is 14 feet, and the circumference 4.1888 feet? Ans. 5.58506 feet. 3. If the diameter, or axis of the earth be 79574 miles, what is the whole surface, supposing it to be a perfect sphere? Ans. 198944286.35235 sq. miles. 4. The diameter of a sphere is 21 inches; what is the convex superficies of that segment of it whose height is 41 inches? Ans. 296.8812 inches. 5. What is the convex surface of a spherical zone, whose breadth is 4 inches, and the diameter of the sphere, from which it was cut, 25 inches ? Ans. 314.16 inches. PROBLEM XIII. RULE.* Multiply the cube of the diameter by .5236, and the product will be the solidity. * Demón. Put ad=&, Dc=y, the diameter AB=d, and p=3.146. EXAMPLES. 1. What is the solidity of the sphere AEBC, whose di. ameter AB is 17 inches ? Here 173 x 5236 = 17 x 17 x 17 x.5236 = 289 x 17 x 6236=4913 X.5236=2572.4468 solid inches. And 2572.4468--1728=1.48868 solid feet, the answer. 2. What is the solidity of a sphere whose diameter is 15 feet? Ans. 1.2411 feet. 3. What is the solidity of the earth, supposing it to be perfectly spherical, and its diameter 7957% miles ? Ans. 263858149120 miles. = Then, by the property of the circle, dx-x=y. But the general expression for the fluxion of any solid is pyʻz ; and therefore by writing dx—æ2 for its equal y, we shall have på x dx-x?=pdxc-pa*. The fluent of which is pdz pix_3pdx'--2px®x 2 3 =content of the segment CAE. 6 3pd_2pd And if d be substituted for x, it will become 6 pa -=dx.5236, or .5236d; which is the same as the rule. 6 Coroll. A sphere, or globe, is equal to two-thirds of its circum scribing cylinder. i PROBLEM XIV. To find the solidity of the segment of a sphere. RULE.* To three times the square of the radius of its base add he square of its height; and this sum multiplied by the height, and the product again by .5236, will give the solidity. Or, From three times the diameter of the sphere subtract twice the height of the segment, multiply by the square of the height, and that product by .5236; the last product will be the solidity. EXAMPLES. 1. The radius Cn of the base of the segment CAD is 7 inches, and the height An 4 inches; what is the solidity ? * Demon. Let r=radius of the base of the segment, h =height of the segment, and the other letters as before. Then will (3dk?—2h)=solidity of the segment, as 6 is shown in the last problem. g2 +h2 But since h =d, by the property of the circle we shall N Here (72x3+42) x 4x.5236=(49x3+4) x 4x.5236 .=(147 +42) * 4x.5236=(147 +16)X 4x.5236 = 163 X 4 x.5236=652X.5236=341.3872 solid inches, the answer. 2. What is the solidity of the segment of a sphere, the diameter of whose base is 20, and its height 9? Ans. 1795.4244 3. What is the content of a spherical segment, whose height is 4 inches, and the radius of its base 8? Ans. 435.6352. 4. What is the solidity of a spherical segment, the di. ameter of its base being 17.23368, and its height 4.5? Ans. 572.5566. 5. The diameter of a sphere being six inches, required the solidity of the segment whose altitude is two inches. Ans. 29.3216 cubic inches. 6. Required the solidity of a spherical seginent, the height of which is 15, the diameter of the sphere being 18. Ans. 2827.44. PROBLEM XV. To find the solidity of a frustrum or zone of a sphere. RULE.* To the sum of the squares of the radii of the two ends, add one third of the square of their distance, or of the 3r%hx 3h4 р ph have х = solidity 6 6 of the segment, which is the same as the rule. Or, if d=diameter of the sphere, and h=height of the segment; then will .52361* x (3d-2h)=solidity. * Demon The difference between two segments of a sphere whose heights are H and h, and the radii of whose bases are Ñ and will P by the last problem=ã (3r2u+H3-3,2h-13)=zone whose height is H-h. And therefore hy putting a for the altitude of the frustrum, |