PROBLEM XIV. To find the Surface AEBF included between Two Interfeiting Little Circles of a Sphere. Through the poles D, C, and interfections A, B, of the two little circles AFB, AEB, draw the great circles A D, BD, AC, BC; and also the great circle DEF C. By prob. 12 find the area of the triangle BCD, having firft found the angles at B, C, and D, from the D E B F given fides, by the principles of fpherical trigonometry. By prob. 10 find the furface of the segment cut off by the circle of which AFB is a part, thus, viz. As the diameter is to the verfed fine of the arc D F, or DB, fo is the furface of the fphere to that of the fegment. Then as 4 right angles, to the BDC, In the fame manner find the furface CBE. Then from the fum of DBF and CBE take the triangle DBC, and the remainder will be the part BEF; the double of which will be the whole AEBFA. PROBLEM XV. To find the Solidity of a Sphere or Globe. RULE I. I Multiply the furface by of the radius, or by of the diameter; and the product will be the folidity. RULE * DEMONSTRATION. The fphere may be confidered as conftituted of an infinite number of pyramids, whofe bafes compofe the fpheric furface, and RULE II. Multiply the cube of the diameter by 5236, and the product will be the folidity. That is (of 31416 or) •5236d3 = the folidity. E X and all the vertices meeting in the center, their common height being equal to the radius of the fphere. And confequently the fphere, or any fpherical pyramid, being a part contained within right lines drawn from the furface to the center, is equal to a pyramid whofe bafe is equal to the fpherical furface, and height equal to the radius. And therefore the furface of the whole, or of any such part, being drawn into of the radius, will give the 13 folidity, as in rule 1. = Corol. 1. Since the furface of the fphere is pdd, we fhall have pdd (of the radius or) d = pd3 = •523598775 &c. x x d3 (which is rule 2) = of a cylinder of the fame diameter and height. = Corol. 2. If be the height, or verfed fine, of any fegment; then, fince pdb its furface, pdb = 523598775 &c. x d2b will be the folidity of the fpheric pyramid, or cone, whose bafe is the furface of the fegment." Corol. 3. Hence fpheres and their fimilar pyramids, and alfo any other fimilar parts of thein, are as the cubes of the diameters. Corol. 4. If to or from pd, the fpheric cone, be added or subtracted! px (p±b) xbx (d—b) == ! pd2b± 1⁄2 pd b2 = 1 ph3, the cone whofe bafe is the fame with the bafe of the fegment, and whofe vertex is in the center, the fum or difference 1⁄2 pd b2 = { ph3 = {ph2 × ({ d − b), will be the fpheric fegment whofe height is b, either greater or lefs than the hemifphere, or of whatever magnitude is, not exceeding d. Or if r be the radius of the fegment's bafe, fince db = r2 + b2, the fegment will be pr+ ¿pb3 = ¿pb × (3r2 + b2). = Corol. 5. Hence the difference between two fegments whofe heights are н, b, and the radii of their bafes R, r, will give for the fruftum or zone p× (3R2H + H3 − 3 r2 b − b3); which, putting a for the altitude of the fruftum, and exterminating H and by means of the two equations (R2 + H^) b = (p2 + b2) H, and a = H b, will become ap × (R2 + r2 + 1⁄2 a2). I Corol. EXAMPLE. Suppofing the earth to be fpherical, and its diameter 7957 miles, what is its folidity? 1. By rule 1. By exam. 1 of prob. 10 the furface is 198943750. Then 198943750 × 79574 × miles is the folidity. 2. By the 2d rule. 263857437760 Here 5236d=5236×(79574)3=263858149120 miles the folidity by this rule. The difference arifing by taking the number 5236 rather too great. PROBLEM XVI. To find the Solidity of the Segment of a Sphere. RULE I. To three times the fquare of the radius of its base, add the fquare of its height; multiply the fum by the height, and the product by 5236, for the folidity. That is, if r = DE the radius of its base, bGE the height; Then 5236 × (3rr + bb) = the folidity of the fegment DGF. By cor. 4 to the laft problem. [See the figure in page 198.] RULE Corol. 6. then R2 = ap × (r2 + If one end of the fruftum pafs through the center, d2 = r2 + a2, and the laft theorem will become a2) = ap × (jd1 — ja3). Corol. 7. Hence the middle zone, or the double of the last expreffion, will be 2ap x (r2 + } a2) = 2αp × ( d2 - } a2); where a is half its altitude, and half the diameter of each end. But if a be its whole altitude, and D the diameter of each end, thofe theorems will become Aр × (D2 + a2) = 4^p × (d2 — } A2) = } Ap × (d2 + 2r2). From three times the diameter of the fphere, fubtract twice the height of the fruftum; multiply the difference by the fquare of the height, and the product by 5236, for the folidity. That is, If dGH the diam. of the fph. b=GE the height of the fruft. [See fig. page 198.] Then 52366 × (3d2b) the folidity of DG F. By cor. 4 to the last problem. EXAMPLE. What is the folidity of each of the frigid zones of the earth; the axe being 79573 miles, and half the breadth, or arc DG, of the zone being 231 degrees? By rule 2. As I tabular radius: 3978% radius of the earth: 0829399 tab. verfed fine of 23 degrees: 3300074946, the verfed fine or height of the feg ment. Then 5236b2×(3d—2b) =•5236×330·0074946* X 232132350108 1323679710, the content. By rule 1. As I 39783987491 tabular fine of 231 degrees: 1586-57282526, the radius of the bafe. Then 5236 × (3r2 + b2) = •5236 × 3300074946 × 7660544'9361323680299 69, the folidity. PRO PROBLEM XVII. To find the Solidity of a Fruftum or Zone of a Sphere. Add together the fquares of the radii of the ends, and of the fquare of their distance, or of the height; multiply the fum by the faid height, and the product again by 1.5708 for the content. That is, (R2 + r2 +÷b2) × pb the folidity of the fruftum whofe height is b, and the radii of its ends R and r, p being 31416. By cor. 5 to prob. 15. A EXAMPLE I. I Ꮐ D F E K B C k H What is the folidity of the fruftum of a fphere, the diameter of whofe great end is 4 feet, the diameter of the lefs 3 feet, and the height 2 feet? Here (R+r+b2)×1•5708h=(22+1·53+÷×2·52) X 1.5708 X 2 = 8 × 3·927 = 32.725, the folidity of the fruftum required. EXAMPLE II. What is the folidity of each temperate zone of the earth, extending from 23 degrees to 664 degrees of latitude, and the diameter of the earth being 7957 miles? By the example to the laft problem, the radius of the top is 1586.57282526. And as 1 : 3978; :: ·9170601 = tabular fine of 66 degrees: 3648.86750538, the radius of the bafe. |