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 31416, 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.

EXAMPLES.

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

Here

The chord AC√√√AE1+CE2=✔✅✔✅/202 + 152=25,

252
30

And 2CE: AC :: AC: CO= =20

Hence OE-OC-CE=205—15—5%.

Also,

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

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

'Ans. 1177 4485.

PROBLEM

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 31416, for the whole content.

EXAMPLES.

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

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 circular spindle..

RULE.

From the square of half the length of the whole spindle 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 EF 32, and its least diameter AD or BC 24.

I

Draw

ating circle, or the radius OE52;

Hence OI52-16-36 the central distance,
And HI1OH'—OI' — 52 —36—1408,