PROBLEM XVI. To find the solidity of a spheroid. RULE.* Multiply the square of the revolving axe by the fixed axe, and this product again by .5236, and it will give the solidity required. Where note that .5236 is=1 of 3.1416. EXAMPLES. 1. In the prolate spheroid ABCD, the transverse, or fixed axe AC is 90, and the conjugate, or revolving axe DB is 70: what is the solidity? * Demon. Let Ac=a, DB=b, ar=x, rn=y, and p= 3.14159, &c. b2 a2 × (ax—x2) = y2 by the of the solid (=py2x) pb2 = pab2 6 =content of the whole spheroid. Q. E. D. If f be put-fixed axe, r-revolving axe, q=(fur2)÷ f2, and p=3.1416, &c. Then will prf✓1+q=surface of the oblate spheroid, and prf√1-14 that of the prolate spheroid, Here DB2 X ACX.5236-702 X 90 X.5236=4900× 90 X.5236=441000 X.5236=230907.6- solidity required. 2. What is the solidity of a prolate spheroid, whose fixed axe is 100, and its revolving axe 60? Ans. 188496. 3. What is the solidity of an oblate spheroid, whose fixed axe is 60, and its revolving axe is 100? To find the content of the middle frustum of a spheroid, its length, the middle diameter, and that of either of the ends being given. CASE I. When the ends are circular, or parallel to the revolving axis. RULE.* To twice the square of the middle diameter add the square of the diameter of either of the ends, and this sum *Demon. Let AO-a, Dob, En-h, no- c, ro=x, re =y, and p=3.14159, &c. Then a2 b2 :: a2x2: : by the property of the ellipsis. b2 b2x2 a2 ·× (a2 — x2) = b2. a2 -b2 × (a2—c2), h2; or a2 = a2 And also a2 b2 :: a2-c2: : bac2 ÷ (b2-h2.) multiplied by the length of the frustum, and the product again by .2618, will give the solidity. Where note that .2618 of 3.1416. EXAMPLES. 1. In the middle frustum of a spheroid EFGH, the middle diameter DB is 50 inches, and that of either of the ends EF or GH is 40 inches, and its length nm 18 inches what is its solidity? Here (502 x2+402) x 18 x.2618 (2500x2+1600) × 18 × .2618 (5000 + 1600) × 18 × .26186600 × 18× .2618 118800x.2613=31101.84 cubic inches, the an swer. = 2. What is the solidity of the middle frustum of a prolate spheroid, the middle diameter being 60, that of either of the two ends 36, and the distance of the ends 80 ? Ans. 177940.224. Whence, by substituting this value of a2 in the former px2x And consequently the fluxion of the solid (py2)=pb2x ca pcb2-pch 2 (b2—h2); which, when x⇒c, becomes pb2 c _pc> (2b2+h2) __pc 3 pc x 8 b2+4b2. Q. E. D. 12 3 3. What is the solidity of the middle frustum of an oblate spheroid, the middle diameter being 100, that of either of the ends 80, and the distance of the ends 36? Ans. 248814.72. CASE II. When the ends are elliptical or perpendicular to the revolv ing axis. 1. Multiply twice the transverse diameter of the middle section by its conjugate diameter, and to this product add the product of the transverse and conjugate diameters of either of the ends. * Demon. Put so=a, Eo=b, om=r, on=x, an=y, xc= p=3 3.14159, &. and a2 ·× (a2—x2)=ya by the pro perty of the ellipsis. And, since ACD is an ellipsis similar to EmF, it will be b:r::y: ry Ъ =z; as is shown by the writers on Conics. But the fluxion of the solid AEFD is pyzx=pyx× ry pry2x prx b3× (a2 —x2) a2. X =prbx x And And this again, by putting z for its equal =2rb+yz=frustum EFDA. Orex (2EF× 20m+ AD × 2ne)=middle frustum ABCD. Q. E. D. 2. Multiply the sum thus found, by the distance of the ends or the height of the frustum, and the product again by .2618, and it will give the solidity required. EXAMPLES. 1. In the middle frustum ABCD of an oblate spheroid, the diameters of the middle section EF are 50 and 30 ; those of the end AD 40 and 24; and its height ne 18; what is the solidity? Here (50 × 2 × 30+40 × 24) × 18.2618 (3000+960) x 18 x .26183960 × 18 x .261871280 × .2618 : 18661.104 solidity required. 2. In the middle frustum of a prolate spheroid, the diameters of the middle section are 100 and 60; those of the end 80 and 48; and the length 36: what is the solidity ? Ans. 149288.832. 3. In the middle frustum of an oblate spheroid, the diameters of the middle section are 100 and 60; those of the end 60 and 36; and the length 80: what is the solidity of the frustum ? Ans. 296567.04. PROBLEM XVIII. To find the solidity of the segment of a spheroid. CASE I. When the base is parallel to the revolving axis. |