EXAMPLE. Taking the fame example, we fhall have I. For the Oblong Spheroid. And X5203 x 6283.1853 4177775203. 5884.2029 = the furface of the oblong fpheroid very nearly. 2. For the Oblate Spheroid. And × 6036 × 6283·1853 = 6826·6806 = the furface of the oblate fpheroid nearly. RULE VII. For Both Spheroids. From the fum or difference of 1 and of q, according as it is for the oblate or oblong fpheroid, take of the fquare of q, and call of the re's mainder c.* 8 27 Then Farther, by tranfpofing the first three terms of the original feries, and reducing the refult, to make the first term of it equal to that of the last remainder, we shall have 2o1⁄2 × [3—4abp(1±¿d2 —‚'¿d*)]=4abp × (± 5d9 &c), 2.7.27 6.8.81 Then from A fubtract the fum of в and c, multiply the remainder by P, and 27 of the product will be the furface of the fpheroid, ftill nearer. That is, 27 PX (ABC) the furface nearly; ×(1±19-1899), 27 where c = 8 I and the value of the other letters as before. EXAMPLE. Taking still the fame example, in which the axes are 40 and 50; 1. For the Oblong Spheroid. 8 27 Here again q=36, P = 6283.1853, A = •93808, B=417777 &c, and c, × (1 — 9 — — 099) - ' qq) =277558. - Then 27 PX (A-B-C) = 6283.1853 × 93628 5882-8206 the furface of the oblong fpheroid very near. 2. For the Oblate Spheroid. Here q == •5625, P = 6283·1853, A = 1.089725, B=486111, and c× (1+ 19 - 699) =32173. = 27 - + Then 27 X (A-B-C) XP = 1.0872 X 6283.1853 =6831079 = the furface of the oblate fpheroid very nearly. 8 ~ 4abp × [√I± }dd—†(1 ±}dd) — 2 (1 ± {d2 — od+)] = 5ds $. 12.81 &c, and confequently s = 27 x 4abp × PROBLEM X. To find the Curve Surface of the Fruftum of a Spheroid, contained between two Planes cutting the Spheroid Perpendicular to the Fixed Axe, and one of them paffing through the Center. Let f denote the fixed, and r the revolving axe; put p = 3.14159, and q = ffcorr ff as in the last problem; alfo the height CA of the fruftum, and z = ff be expreffed by ; then will the value of the furface Where A, B, C, &c, are the feveral preceding terms, and the upper or under figns to be ufed, according as it is the oblate or oblong fpheroid.* * See the investigation of rule 1 of the laft problem, I E X EXAMPLE I. What is the furface of the fruftum of a fpheroid, whofe height is 15; the diameter of the greater end being 40, and that of the lefs 32? Here co= 40 and the revolving axe, 2 AB = 32, and HB = CA = 15; hence CH = CC— AB = 4, HOCC + AB = 36; and confequently, by cafe 3 prob. 2, we have as ✓CH X HO= 12; HB = 15 :: co= 40: TR = 50 the fixed axe; which fhews that the fruftum is that of an oblong fpheroid. Then, as in the example to rule 1 of the last problem, ƒ= 50, r = 40, 9=23=36. But h=15, 4 × 9 × 15 × 15 = 1296. and z = 4gb b 25 X 50 X 50 Hence the ift term A = =-0.0216 + I' = -0.0004199+ = 0.0000195 the fum of the negative terms is -0*0220407, which taken from the firft term 1*0000000, leaves O'9779593 for the value of the infinite feries; which being drawn into prb 3.14159 X 40 X 15 1884 955592, produces 1843 4098512 for the furface required. EXAMPLE II. It is required to find the curve furface of the fruftum 2CABC of a fpheroid, whofe height is 16; the diameter of the greater end being 50, and that of the lefs 30. CA 10, Here the revolving axe TR 50, CH AB 16, and HB CA 15; hence AR = CR and TA = TC +CA 40; then, by case 3 prob. 2, we shall have as AR X AT 20: AB = 16 :: TR≈ 50: co = 40 the fixed axe; which indicates the fruftum to be that of an oblate fpheroid. Wherefore ƒ 40, r = 50, 9=,9%, b = 16, and 2= = ff 16 X 40 X 40 = 36 100 36, which be ing the fame as the converging quantity q in the ex ample to rule 1 of the laft prob. the feveral terms of the feries must be the fame as there found, viz. the ift term A = +I 2d their difference = 10571159 the value of the feries; which being drawn into prb = 3.14159 X 50 X 16 = 2513.27412287, will produce 2656.822036 for the furface of the oblate fruftum required. RULE |