breadth of the zone, and this sum multiplied by the said breadth, and the product again by 1.5708, will give the solidity. EXAMPLES. 1. What is the solid content of the zone ABCD, whose greater diameter AB is 20 inches, the less diameter CD 15 inches, and the distance nm of the two ends 10 inches? 102 Here 10 +7.52+)x10x 1.5708 (100+56.25+ 33.33) x 10 x 1.5708: = 189.58 x 10 x 1.5708 = 1895.8 x 1.5708-2977.92264 solid inches, the answer. 2. What is the solid content of a zone, whose greater diameter is 24 inches, the less diameter 20 inches, and the distance of the ends 4 inches? Ans. 1566.6112 inches. 3. Required the solidity of the middle zone of a sphere, whose top and bottom diameters are each 3 feet, and the breadth of the zone 4 feet? Ans. 61.7848 feet. and exterminating H and h by the means of the two equations H a2 pa and н-ha,we shall have (R2+12+) X, which is the = If it be the middle zone of the sphere, the solidity will be d2+ h) x.7854h; where d-diameter of each end, and hits height. i 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 Then a2: b2 :: x × (a−x): 2× (ax—x2)=y2 by the property of the ellipsis. And therefore the fluxion of the solid (=py'x)= (axx-xx); and its fluent pb2 × (‡ax2 — }x3)=segment pb9 pab nam. Which, when x=a becomes × ({a3 —}a3)=* =content of the whole spheroid. Q. E. D. 6 If ƒ be put fixed axe, r-revolving axe, q=f2 r2)÷ f, and p-3.1416, &c. Then will prf√1+1g=surface of the oblate spheroid and prf1-19-that of the prolate spheroid. Here DB2x AC× .5236=702× 90x .5236=4900 × 90× .5236-441000x.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? PROBLEM XVII. Ans. 314160. To find the content of the middle frustrum 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 Aо=a, Do=b, En=h, no=c, ro=x, re= v, and p=3.14159, &c. Then a2: b2 :: a2—x2 a by the property of the × (a2 —x2)= b2 b2 x2 b2 : a2 And also a2: b2 : : a2 —c2 : b2c2 — (b2—h2.) multiplied by the length of the frustrum, and the product again by .2618, will give the solidity. Where note that .2618 of 3.1416. EXAMPLES. 1. In the middle frustrum 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? Ꭰ = Here (502×2+40) x 18 x .2618 (2500 × 2+1600) × 18.2618=(5000 + 1600) × 18×.2618 = 6600×18× 2618 118800 x .2613 = =31101.84 cubic inches, the answer. 2. What is the solidity of the middle frustrum 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 equation, we shall have y b2 — b4x2-b2h2x3 b2c2 = b2 And consequently the fluxion of the solid (py';)=pb3x pa× (b2—h2); the fluent of which is = pb*x — (b-h2); which, when x=c, becomes pb3c pcx (2b2+h2) pc + 3 12×86+462. Q. E. D. 3 3. What is the solidity of the middle frustrum of an oblate spheroid, the middle diameter being 100, that of either of the ends 80, and the distance of the ends 36? CASE II. Ans. 248814.72 When the ends are elliptical or perpendicular to the revolving axis. RULE.* 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 =z, and p=3.14159, &c. b2 Then a : ba : : aa—x2 : /= × (a®—x2)=ya by_the_property of the ellipsis. And, since ACD is an ellipsis similar to EmF, it will be z; as is shown by the writers on Conics. ry bir y T But the fluxion of the solid AEFD is pyzx=pyxx pryx prx b2x (a2—x2) ry pry2x = b b X b2x2 value22, a2x2 Which, by substituting for a2 its 2b2+y3· ry3 becomes prx×· 36 =px × & rb+36• And this again, by putting z for its equal, becomes px ne 12 px. ×2rb+yz frustrum EFDA. Or X(2EFX 20m+ ADX2ne)=middle frustrum ABCD. Q. E. D. 3 |