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multiply the sum by the height, and the product again by 1.5708 for the folidity.

RULE 2. From the folidity of the whole sphere, fubtract the folidity of the segments ABC and DEF; the remainder is the folidity of the zone.

RULE 3. Add into one fum twice the square of the sphere's diameter, and the square of the diameter of the zone's base ; divide this sum by 3.8197, and multiply the quotient by the zone's height; the product is the soliditý.

EXAMPLE I.
Required the folidity of the middle zone of a sphere, whose
diameter is 80 inches; the diameter of the zone's base being
48, and height 64 inches.

By RULE I.
24
24

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Ex. 2. Required the solidity of a zone, whose greater diameter is 2 feet, the less 1 foot 4 inches, and the height į foot 8 inches.

inf. 10723.328 inches. Ex. 3. What is the solid content of a zone, whose height is 30, and end diameters 60 and 40 inches?

Anf. 75398.4 cubic inches. Ex. 4. What is the solidity of a zone, whose height is 8 inches, and diameter of the ends 12 inches ?

Anf. 1172 864 cubic inches.

PROBLEM XVI. Fig. 96.

To find the area of a circular Spindle.

RULE.

Multiply the length of the spindle by the radius of the revolving arch; again multiply the distance between the centre of the revolving arch and the centre of the spindle Iby the length

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of the revolving arch ; subtract this last product from the former, and multiply the remainder by 6.2832 for the superficies.

EXAMPLE. Required the area of a circular spindle, whose length is 40 and thickness 30 inches.

AD2+BD- = AB the chord of the arch ABC; that is,

400+225 = 25

AD

=DH and DH+BD=FB rad. also FB-BD=DF cent. dift.

BD

2

400

=26.6 and 26.6+16=20.83 rad. also 20.83–15=5.83=DF 15

2

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Then fay, As 360°: 147° 2 :: 3.1416 X 41.6:53.58 leng. of arch,

Or thus :

25

200
40

3 160

53.3 the arch nearly.

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Ex. 2. Required the number of square inches which will cover a circular spindle, whose length is 80 and thickness 16 inches ?

Anf. 2747.3166 336. Ex. 3. Required the area of a circular spindle, whose length is 12, and thicknefs y inches.

Anf. 294.3621 sq. inches.

PROBLEM XVII.

To find the solidity of a circular spindle.

Multiply the area of the revolving segment by the distance between the centres of the arch and spindle, subtract the prcduct from the cube of half the length of the spindle, then multiply the remainder by 4, and this product again by 3.1416 for the solidity. See the last figure.

EXAMPLE I.

Required the area of a circular spindle, whose length is 60 and diameter 4). Dd

AD

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