12. What is the side of the greatest cube that can be cut from a sphere 12 inches in diameter ? Ans. 6.9282 inches.

13. How many square inches of gold-leaf will gild a sphere one foot in diameter ? Ans. 452.39 inches.

14. What is the surface of a sphere whose diameter is 2 feet 8 inches? Ans. 22.34 feet.

To find the spherical surface of any segment or zone of a sphere :

Multiply the circumference of the sphere (of which it is a part) by the height of the segment or zone, and the product will be the convex surface. Or,

Multiply the diameter of the sphere by 3.1416, and the product by the height of the segment or zone, and the product will be the area required.

To find the solidity of the spherical sector:

Multiply the area of the convex surface by one-third of the radius of the sphere, and the product will be the solidity.

To find the solidity of a spherical segment:

1. Find the solidity of the whole sector, bacs, from which subtract the solidity of the cone bcs, and the remainder will be the solidity of the segment bac.

3. Or, From three times the diameter of the sphere, subtract twice the height of the segment; multiply the remainder by the square of the height, and that product by .5236, and the result will be the solidity.

EXAMPLES.

1. What is the area of the convex surface of a spherical sector, the diameter of the sphere being 10 feet, and the height of the segment ar 2 feet?

Ans. 62.832 feet.

spherical segment, whose

2. Find the convex surface of a height is 3 feet 6 inches, and the diameter of the sphere 10 feet. Ans. 109.96 feet.

3. Find the convex surface of a spherical zone, whose height is 4 inches, and the diameter of the sphere 1 foot.

Ans. 150.7968 inches.

4. Find the convex surface of a spherical zone, the height of which is 5 inches, and the diameter of the sphere 25 inches. Ans. 392.7 inches.

5. Find the convex surface of a spherical segment, whose . height is 9 inches, and the diameter of the sphere 3 feet 6 inches. Ans. 1187.525 inches.

6. What is the solidity of a spherical sector, the height of the convex part being 2 inches, and the diameter of the sphere 9 inches? Ans. 84.8232 inches.

7. What is the solidity of a spherical sector, whose height is 4 feet, and the diameter of the sphere 70 feet?

Ans. 1026.256 feet.

8. Find the solidity of a spherical segment, whose height is 4 inches, and the radius of the base 8 inches.

Ans. 435.6352 inches.

9. What is the solidity of a spherical segment, the radius of whose base is 25 inches, and its height 6.75 inches?

Ans. 6787.844 inches.

10. Find the solidity of a spherical segment, the height of which is 2 feet, and the diameter of the sphere 10 feet.

Ans. 54.4544 feet.

11. Find the contents of a spherical segment in ale and wine gallons, the diameter of its base being 83.32 inches, and its height 28 inches.

Ans. 311.51 ale, and 380.29 wine gallons.

When the diameter of the segment's base and its height are given, to find the diameter of the sphere :-Divide the square of the radius of the base by the height, and to the quotient add the height. See 24, page 82.

The solidity of a spherical segment is equal to that of half of its least circumscribing cylinder, plus the solidity of a sphere whose diameter is equal to the height of the segment. Therefore,

To gauge a spherical segment:

Find the contents of one-half of its least circumscribing cylinder, and the contents of a sphere whose diameter is equal to the height of the segment, and the sum of their contents will be the solidity of the spherical segment.

12. What is the solidity of a spherical segment, whose height is 4 feet, and the diameter of its base 16 feet?

Ans. 435.635 feet.

Place 2 feet over 13.54 on D, the gauge point for feet in a cylinder, and over 192, the diameter of the segment's base in inches, you will find 302.3 feet, the solidity of one-half the

least circumscribing cylinder; then place 4 feet over 16.58, the gauge point for feet in a sphere, and over 48, the height of the segment in inches, you will find 33.4 feet, which, being added to 302.3, equals 435.7 feet for the solidity of the segment.

T 47. SPHERICAL ZONE.

A spherical zone is the frustum of a sphere formed by cutting off two segments by planes parallel to each other.

1. To find the solidity of a spherical zone :-Subtract the solidity of the two segments from the solidity of the whole sphere, and the remainder will be the solidity of the zone. Or,

2. Add together the squares of the radii of the two ends, and one-third of the square of the height, which sum multiply by the height, and this product by 1.5708, and the result will be the solidity.

EXAMPLES.

1. Find the solidity of a spherical zone, the diameter of the ends being 4 and 3 inches, and its height 2 inches.

Ans. 23.824 cubic inches.

2. Find the solidity of the middle zone of a sphere, the diameters of the ends being 4 feet each, and its height 6 feet. Ans. 188.496 inches.

3. The diameters of the ends of a spherical zone are 8 and 12 inches, and its height 10 inches; what is its solidity?

Ans. 1340.416 inches.

T48. SOLIDS GENERATED BY THE REVOLUTION OF THE CONIC SECTIONS.

The solids generated by the revolution of the conic sections are the ellipsoids, (called the oblate and prolate spheroids,) the paraboloid, or parabolic conoid, and the hyperboloid.

An oblate spheroid is a solid generated by the revolution of an ellipse about its minor axis.

An oblate spheroid is equal to two-thirds of a cylinder, whose diameter is equal to the greater or equatorial diameter of the spheroid, and its length equal to the polar diameter, or polar axis. Therefore,

To find the solidity of an oblate spheroid :

Multiply the square of the equatorial diameter by the minor or polar diameter, and the product by .5236, (that is, by twothirds of .7854,) and the result will be the solidity.

EXAMPLES.

1. What is the solidity of an oblate spheroid, whose polar axis is 15, and equatorial axis 25? Ans. 4908.75.

2. What is the content of an oblate spheroid in feet and bushels, its greater axis being 4 feet, and its less axis 3 feet? Answers, 25.1328 feet; and 20.2 bushels, nearly.