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 : b2 : : a2 —x2 × (a2—x2)=y2 by the prop erty of the ellipsis. a2 And, since ACD is an ellipsis similar to EmF, it will be ry bryb =z; as is shown by the writers on Conics. But the fluxion of the solid AEFD is pyzx=pyx× ry pry2x prx b2× (a2—x2) == prbxx a2** And the fluent prbxx a2 Which, by substituting for a2 its And this again, by putting z for its equal, becomes 2. Multiply the sum thus found, by the distance of the ends or the height of the frustrum, and the product again by .2618, and it will give the solidity required. EXAMPLES. 1. In the middle frustrum 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 .2618 3960 x 18 x .261871280 × .2618 = 18661.104 solidity required. = 2. In the middle frustrum 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 frustrum 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 frustrum? 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. RULE.* 1. Divide the square of the revolving axis by the square of the fixed axe, and multiply the quotient by the difference between three times the fixed axe and twice the height of the segment. 2. Multiply the product, thus found, by the square of the height of the segment, and this product again by .5236, and it will give the solidity required. EXAMPLES. 1. In the prolate spheroid DEFD, the transverse axis 2 DO is 100, the conjugate AC 60, and the height Dn of the segment EDF 10; what is the solidity? D n 602 1002 Here × 300—20) × 101 × .5236=.36 × 280 × 102 X.5236 100.80 x 100 x .5236 = 10080 x .5236=5277 .888 solidity required. 2. The axes of a prolate spheroid are 50 and 30; what is the solidity of that segment whose height is 5, and its basc perpendicular to the fixed axe? Ans. 659.736. * This rule is formed from the theorem for the segment in the demonstration to problem XX. The content of the segment may also be found by the following theorem: (D2+4d2) ×nh-content of the segment; D being the diameter of the base, d=diameter in the middle, h=height, and n=.7854= area of a circle whose diameter is 1. 3. The diameters of an oblate spheroid are 100 and 60, what is the solidity of that segment whose height is 12, and its base perpendicular to the conjugate axe? Ans. 32672.64. CASE II. When the base is perpendicular to the revolving axis. RULE.* 1. Divide the fixed axe by the revolving axe, and multiply the quotient by the difference between three times the revolving axe and twice the height of the segment. 2. Multiply the product, thus found, by the square of the height of the segment, and this product again by .5236, and it will give the solidity required. EXAMPLES. 1. In the prolate spheroid aEbF, the transverse axe EF is 100, the conjugate ab 60, and the height an of the segment aAD 12; what is the solidity? * Demon. Put ao=a, Eo=b, om=r, an=x, an=y, nc and p=3.1416. Then will a2: b2:: a2—(a—x)2 or (2ax b2×2ax-z2) =%9 -x2) : a2 y' by the property of the ellipse. And since ACD is an ellipse similar to EmF, it will be ry br::y: b =z; as is shown by the writers on Conics. But the fluxion of the solid aACD = ry b pry'x prx b2× (2ax—x2) = a2 pyzx = whose fluent is рух х prb = -x2 a ; which, when x=h-the height of the segment, prb becomes (3ah2-h3) ×· Whence, since r=a, we shall have (3ab3-h3) × =solidity of the segment. 3a Q. E. D. Here 156 (=diff. of 3ab and 2an) 13 (=EF÷ab× 144 (=square of an)× .5236= 156 × 5 x 144× .5236 = 52 × 5 = X 144 x .5236=260 × 144 x .5236 = 37440 x .5236 = 19603.584 solidity required. = 2. Required the content of the segment of a prolate spheroid: its height being 6, and the axes 50 and 30. Ans. 2450.448. PROBLEM XIX. To find the solidity of a parabolic conoid. RULE.* Multiply the area of the base by half the altitude, and the product will be the content. Demon. Let Dm=α, вm= =b, Dn=x, En=y, and p= 3.1416. Then, by the nature of the parabola a: b2 :: x: y2, or b2x pb2x; =y; wherefore (=py2;) the fluxion of the a solid, and pb2x2 2a = a = ― a, its fluent;, which, when a becomes ispab2 for the whole solid, or for any segment whose height is a, and the radius of its base - b. Q. E. D. = Coroll. The parabolic conoid is its circumscribing cylinder. Note. The rule given above will hold for any segment of the paraboloid, whether the base be perpendicular or oblique to the axe of the solid. |