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3485.08.6 folid inches as before. Ex. 2. Required the solidity of a segment, whose base dia meter is 100, and its height 13.5 inches.

Anf. 54302.75235 cubic inches. Ex. 3. How many folid miles are in either frigid zone, the height being 329 miles, and diameter of its base 3168 miles ?

Anf. 1315766512 folid miles.

PROBLEM XV.

To find the folidity of the middle zone of a sphere.

RULE I.

When the ends are unequal, add into one fum the squares of the radii of both ends, and the square of the zone's height;

multiply

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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 1.
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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 , 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.

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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

of

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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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