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2. Required the content of a zone, whose greater diameter is 12, less diameter 10, and height 2.

Ans. 195*8264. 3. What is the content of a middle zone, whose height is 8 feet, and the diameter of each end 6?

Ans. 494'2784 feet.

PROBLEM XV.

To find the surface of a circular spindle

RULE.

Multiply the length AB of the spindle by the radius OC of the revolving arc. Multiply also the said arc ACB by the central distance OE, or distance between the centre of the spindle and centre of the revolving arc. Subtract the latter product from the former, and multiply double the remainder by 3•1416, or the single remainder by 6*2832, for the surface.

Note. The same rule will serve for any segment or zone cut off perpendicular to the chord of the revolving arc, only using the particular length of the part, and the part of the arc, which describes it, instead of the whole length and whole arc.

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1. Required the surface of a circular spindle, whose length AB is 40, and its thickness CD 30 inches.

Here
The chord AC=VAE“ +CE`=V20° +15= 25,

25 And 2CE : AC :: AC : CO

2015

30 Hence OE=OC-CE=201-1535

2

Also,

Also, by Prob. X. Rule 2, MENSURATION OF SUPERFICIES.

25 AC

3)160

53 arc ACB.

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Then, by the rule,

53
55

B

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2. What is the surface of a circular spindle, whose length is 24, and thickness in the middle 18 ?

Ans. 1177*4485

PROBLEM PROBLEM XVI.

20

To find the solidity of a circular spindle.

RULE. Multiply the central distance OE by half the area of the revolving segment ACBEA. Subtract the product from

of the cube of AE, half the length of the spindle. Then multiply the remainder by 12.5664, or 4 times 3.1416, for the whole content.

EXAMPLES 1. Required the content of the circular spindle, whose length AB is 40, and middle diameter CD 30.

[See ļast Figure.] By the work of the last. Problem, We have OE-6

20 half length. And are AC26 And rad. OC=20$

400
533
22

3)8000
Sector OACB 555
AEXQE=OAB116

2666

12805 2)438

1386 seg. ACB 2197 OE 5

20

or 1386-44

4665'21 mult. invert. 1097 183 nearly. 138644

27739 1280

6932
832
83
5

174235 answer.

2. What

2. What is the solidity of a circular spindle, whose length is 24, and middle diameter 18 ?

Ans. 3739*93.

PROBLEM XVII.

To find the solidity of the middle frustum, or zone, of a cir

cilar spiridle.

RULE,

From the square of half the length of the whole spindie take of the square of half the length of the middle frustum, and multiply the remainder by the said half length of the frustum. Multiply the central distance by the revolving area, which generates the middle frustum. Subtract this latter product from the former ; and the remainder, multiplied by 6'2832, or twice 3-1416, will give the content.

EXAMPLES.

1. Required the solidity of a frustum, whose length mn is 40 inches, its greatest diameter ET 32, and its least diameter AD or BC 24.

I

Draw

2

6

Draw DG parallel to mn, then we have
DG=mn=20,
And EG=EF- AD=4,
Chord DE'=DG:+GE=416,

And DE EG=*** = 104 the diameter of the generating circle, or the radius OE=52 ;

Hence OI=52–16=36 the central distance, And HI' -OH-OI=52-36*=1408, {DGʻ=of 400 = 133

1274

2

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20

25493 ist prod.

4 104

2 6

GE-=-20E=1=id = '03846 a ver, sine.

. Its tabular segment

'00994 But 1942 is

10816 43264 97344 97344

Area of seg. DECGD 107'51104

mDXmn=12X40 480
Gener. area mDECn 587 51104

ΟΙ

36

352506624
176253312
21150*39744 2d product,

. 25493-33333 ist product. 434293589 23826 mult. invert.

260576

8686
3474
130

9
27287*5 answer.

2. What

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