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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 ?

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Here (50 x 2x 30+40 x 24)18.2618=(3000+960) * 18 X 2618= 3960 X 18 X .2618=71280 X .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 di. ámeters 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

ES

A

602 Here

1002X 300-20) * 10$x.5236=.36 x 280 x 102 X.5236 = 100.80 X 100 X.5236 = 10080x.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 base 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 heorem:

(D2 +422) x šnh=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 66, what is the solidity of that segment whose height is 12, and ts 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 requiredo

EXAMPLES.

1. In the prolate spheroid aEbF, the transverse axe EF is 100, the conjugats ab 60, and the height an of the seg. ment aAD 12; what is the solidity ?

:

* Demon. Put ao=A, E0=

=b, om=r, an=x, An=y, no =%, and p=3.1416. Then will a' :bo::a—a—) or (2ax

62 x 2ax-x)
a?

=yo by the property of the ellipse. And since Act is an ellipse similar to EMF, it will be

ry b :::

y

7

-=%; as is shown by the writers on Conics. But the fluxion of the solid AACD = = pyzz pyä x ry pry's prie box (2ax—?) 풍

x
a

whose fluent is 32*}= ; which, when x=h=the height of the segment,

prb becomes (3ah?— *)* 3a2. Whence, since r=a, we shall

pb have (3ab" ho) Sa=solidity of the segment.

Q. E. D

prb

a

2 =ъх prb

a

:n

E;

m

Here 156 (=diff. of 3ab and 2an) x 13 (=EF-abx 114

156 x 5 (=square of an) x.5236= x 144 X.5236 = 52 x 5

3 X 144 X .5236=260 x 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.

a

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* Demor Let om=0, bm=b, on=X, EN=y, and p= 3.1416.

Then, by the nature of the parabola a : b2 :: X : ys, or béx =y; wherefore

pb x;

(=py?:) = the fluxion of the pback solid, and : its fluent; which, when a becomes = a,

2a is į pab? for the whole solid, or for any segment whose height is= , 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 para · boloid, whether the base be perpendicular or oblique to the axe of the ·

EXAMPLES.

1. What is the solidity of the paraboloid ADB, whose height Dm is 84, and the diameter BA of its circular base 48?

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Here 482 x 17854 42 =} Dm) = 2304x .7854x42 = 1809.5616 x 42=76001.5872=solidity required.

2. What is the solidity of a paraboloid, whose height is 60, and the diameter of its circular base 100? Ans. 235620

3. Required the solidity of a paraboloid conoid, whose height is 30, and the diameter of its base 40 ? Ans. 18849.6

4. Required the solidity of a paraboloid conoid, whose height is 50, and the diameter of its base 100. ?

Ans. 196350.

PROBLEM XX. To find the solidity of the frustrum of a paraboloid when

its ends are perpendicular to the are of the solid

RULE. * Multiply the sum of the squares of the diameters of the wo ends by the height of the frustrum, and the product gain by .3927, and it will give the solidity.

* Demon. The segment whose base is B, and altitude A, is = AB, and that whose base is b and altitude a, is= } ab, by the est problem: wherefore the frustrum, or the difference of the reg

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