Ex. 1. Find the solid content of a cube whose side is 24 inches. Ans. 13824. Ex. 2. How many cubic feet are in a block of marble, its length being 3 feet 2 inches, breadth 2 feet 8 inches, and thickness 2 feet 6 inches? Ans. 21.

Ex. 3. How many gallons of water will the cistern contain whose dimensions are the same as in the last example, when 277.274 cubic inches are contained in one gallon? Ans. 131.566. Ex. 4. Required the solidity of a triangular prism whose length is 10 feet, and the three sides of its triangular end or base are 3, 4, 5 feet. Ans. 60. Ex. 5. Required the content of a round pillar, or cylinder, whose length is 20 feet, and circumference 5 feet 6 inches. Ans. 48.1459.

PROBLEM V.

To find the volume of any pyramid or cone.

Find the area of the base, and multiply that area by the perpendicular height; then take one third of the product for the volume. (See Prop. XI., Solid Geom., and corollaries.)

Ex. 1. Required the solidity of the square pyramid, each side of its base being 30, and its perpendicular height 25. Ans. 7500. Ex. 2. To find the content of a triangular pyramid whose perpendicular height is 30, and each side of the base 3. Ans. 38.97117. Ex. 3. To find the content of a triangular pyramid, its height being 14 feet 6 inches, and the three sides of its base 5, 6, 7. Ans. 71.0352. Ex. 4. What is the content of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet? Ans. 27.5276. Ex. 5. What is the content of the hexagonal pyramid whose height is 6·4, and each side of its base 6 inches? Ans. 1.38564 feet. Ex. 6. Required the content of a cone, its height being 10 feet, and the circumference of its base 9 feet. Ans. 22.56093.

PROBLEM VI.

To find the volume of the frustum of a cone or pyramid. RULE I. Add into one sum the areas of the two ends, and the mean proportional between them, or the square root of the product, and one third of that sum will be a mean area; which, being multiplied by the perpendicular height or length of the frustum, will give its content.

RULE II. For the cone. Add together the squares of the radii of the two bases and their product, and multiply the sum by 3·1416, and the product by one third of the altitude. (See Prop. XII., Solid Geom., and corol.)

Ex. 1. To find the number of solid feet in a piece of timber whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches; also, the length or perpendicular altitude 24 feet? Ans. 19.

Ex. 2. Required the content of a pentagonal frustum whose height is 5 feet, each side of the base 18 inches, and each side of the top or less end 6 inches. Ans. 9'31925 feet.

Ex. 3. To find the content of a conic frustum, the altitude being 18, the greatest diameter 8, and the least diameter 4. Ans. 527.7888.

Ex. 4. What is the volume of the frustum of a cone, the altitude being 25; also, the circumference at the greater end being 20, and at the less end 10? Ans. 464.216.

Ex. 5. If a cask, which is two equal conic frustums joined together at the bases, have its bung diameter 28 inches, the head diameter 20 inches, and length 40 inches, how many gallons of wine will it hold? Ans. 79.0613.

PROBLEM VII.

To find the surface of a sphere, or any segment. RULE I. Multiply the circumference of the sphere

by its diameter, and the product will be the whole. surface of it.*

RULE II. Multiply the square of the diameter by 3.1416, and the product will be the surface.

Note. For the surface of a segment, multiply the circumference of a great circle of the sphere by the altitude of the segment.

Ex. 1. Required the convex superficies of a sphere whose diameter is 7, and circumference 22.

*For if a regular semi-polygon be revolved about a diameter of the figure, each of the trapezoids, as BGHC, will describe the frustum of a cone, the convex surface of which will be measured by the circumference of MN, described by the middle point of its inclined side, multi- C plied by the slant height BC (Prob. 3, Rule 2). But by the similarity of the triangles IMN and BCO, whose sides are respectively perpendicular, BC: BO:: IM: MN: circum. IM: circum. MN (Geom., th. 71).

M

N

H

D

K

L

F

... BC X circum. MN BO X circum. IM. In the same manner, the convex surface of the frustum described by the revolution of the trapezoid HCDK may be shown to be measured by HK × circum. IM. Of that described by the revolution of DKLE by KL X circum. IM. And, by addition, the surface described by the portion of the perimeter BCDE is measured by GLX circum. IM. The same result will be obtained when the number of sides of the semi-polygon is infinite and it becomes a semicircle, generating a sphere by its revolution; and the portion BCDE generating a zone, of which GL is the altitude. The circum. IM in this case becomes the circumference of a great circle of the sphere. When the whole semi-polygon or semicircle revolves, the altitude becomes the diameter AF, and the surface is measured by the circumference of a great circle multiplied by its diameter. This is equal to four times the area of a great circle (see th. 73, Geom.).

Corol. The convex surface of a cylinder circumscribing a sphere is measured by the rectangle of the circumference of the base by the altitude, which, being equal to the diameter of the sphere, and the base of the cylinder equal a great circle, it follows that the measure of the surface of the sphere is equal to that of the convex surface of the cylinder. If now we add the two bases of the cylinder, since the surface of the sphere is equal to four great circles, we shall have the surface of the cylinder equal six great circles, so that the surfaces of the sphere and circumscribed cylinder are as 4 to 6, or as 2 to 3. Rule 2 follows obviously from Rule 1.

Ex. 2. Required the superficies of a globe whose diameter is 24 inches. Ans. 1809.5616. Ex. 3. Required the area of the whole surface of the earth, its diameter being 79573 miles, and its circumference 25000 miles.

Ans. 198943750 sq. miles. Ex. 4. The axis of a sphere being 42 inches, what is the convex superficies of the segment whose height is 9 inches? Ans. 1187.5248 inches. Ex. 5. Required the convex surface of a spherical zone whose breadth or height is 2 feet, and cut from a sphere of 121 feet diameter. Ans. 78.54 feet.

PROBLEM VIII.

To find the surface of a lune.

Multiply the arc which measures the angle of the lune by the diameter of the sphere. For the lune is to the whole surface of the sphere as its arc is to a circumference.

Cor. 1. The measure of a spherical wedge, or ungula, is for a similar reason the product of the lune which serves for its base, multiplied by one third the radius of the sphere (see next Prob.).

Cor. 2. The measure of a spherical triangle is the arc of a great circle subtending half the excess of the sum of its angles over two right angles, multiplied by the diameter of the sphere. This depends on the above and Prop. XVII., cor. 1, Spher. Geom.

The measure of the surface of a spherical polygon is the arc of a great circle subtending half the excess of the sum of its angles over as many times two right angles as the figure has sides, wanting two, multiplied by the diameter of the sphere.*

Or in symbols, s denoting the sum of the angles of the polygon in fractions of a right angle, n the number of its sides,

[s—2 (n-2)]÷8,

*This may be easily proved by dividing the polygon into triangles.

will express the fraction which the polygon is of the whole surface of the sphere.

is 8 and diameter 10.

Ex. 1. Required the surface of the lune whose arc Ans. 8 x 10 = 80. Ex. 2. Required the measure of the lune whose angle is 30° 20', and diameter 12.

Ans. 12 X 3·1416 × 12 X

30° 21'

3600

=38.139.* *

Ex. 3. Required the area of a spherical triangle, of which the three angles are 30°, 100°, and 80°, the diameter of the sphere being 40.

150 Ans.

3600

X 3.1416 x 40 × 40.

Ex. 4. Required the area of a spherical pentagon, the angles of which are 60°, 110°, 150°, 160°, and 100°, the diameter of the sphere being 50.

Ans. {[60+110+150+160+100-(5-2)

180°] 360×3.1416 × 50 × 50. Ex. 5. What fraction of the whole surface of a sphere is a spherical heptagon, the angles of which are in fractions of a right angle 11, 17, 12, 11, 12, 12, 14.

Ans. 11-10-13÷8=
8

PROBLEM IX.

To find the volume of a sphere or globe.

11

RULE I. Multiply the surface by the diameter, and take one sixth of the product for the content.

RULE II. Multiply the cube of the diameter by the decimal 5236 for the content.

Ex. 1. To find the content of a sphere whose axis is 12. Ans. 904-7808. Ex. 2. To find the solid content of the globe of the earth, supposing its circumference to be 25,000 miles. Ans. 263,857,437,760 miles.

* This answer is of the same denomination as the diameter, except that it is square units instead of linear. Logarithms may here be conveniently applied, using the arithmetical complement of the logarithm of the divisor 360° reduced to minutes.