cepted between two parallel planes; and when those planes are equally distant from the centre, it is called the middle zone of the sphere.

18. The height of a solid is a perpendicular, drawn from its vertex to the base or plane on which it is supposed to stand.

PROBLEM I.

To find the solidity of a cube, the height of one of its sides

being given.

RULE.*

Multiply the side of the cube by itself, and that product again by the side, and it will give the solidity required.

EXAMPLES.

1. The side AB, or BC, of the cube ABCDFGHE, is 25.5: what is the solidity?

* Demon. Conceive the base of the cube to be divided into a number of little squares, each equal to the superficial measuring unit. Then will those squares be the bases of a like number of small cubes, which are each equal to the solid measuring unit.

But the number of little squares contained in the base of the cube are equal to the square of the side of that base, as has been shown already.

And consequently, the number of small cubes contained in the whole figure, must be equal to the square of the side of the base multiplied by the height of that figure; or, which is the same thing,

L

Here AB3-22.5|3=25.5× 25.5 × 25.5=25.5 x 650.25 =16581.375 the answer, or content of the cube.

2. The side of a cube is 15 inches: what is the solidity? Ans. 1ft. 11in. 5pa.

3. What is the solidity of a cube whose side is 17.5 inches? Ans. 3.1015 feet.

PROBLEM II.

To find the solidity of a parallelopipedon.

RULE.*.

Multiply the length by the breadth, and that product again by the depth or altitude, and it will give the solidity required.

EXAMPLES.

1. Required the solidity of a parallelopipedon ABCD FEHG, whose length AB is 8 feet, its breadth FD 44 feet, and the depth or altitude AD 63 feet?

the square of the side of the base multiplied by the base, is equal to the solidity. Q. E. D.

Note. The surface of the cube is equal to six times the square of its side.

*The reason of this rule, as well as of the following ones for the prism and cylinder, is the same as that for the cube.

Note. The surface of the parallelopipedon is equal to the sum of the areas of each of its sides or ends.

Here ABX ADX FD 8x 6.75 x 4.554x 4.5 243 solid feet, the contents of the parallelopipedon required. 2. The length of a parallelopipedon is 15 feet, and each side of its square base 21 inches: what is the solidity? Ans 45.9375 feet.

of marble, whose

3. What is the solidity of a block length is 10 feet, its breadth 5 feet, and the depth 3

feet?

Ans. 201.25 feet.

PROBLEM III.

To find the solidity of a prism.

RULE.*

Multiply the area of the base into the perpendicular height of the prism, and the product will be the solidity.

EXAMPLES.

1. What is the solidity of the triangular prism ABCF ED, whose length AB is 10 feet, and either of the equal sides, BC, CD, or DB, of one of its equilateral ends BCD, 2 feet?

*The surface of a prism is equal to the sum of the areas of the two ends and each of its sides.

* Here × 2.5a × √3=‡× 6.25× √3=1.5625× √3= 1.5625 × 1.732=2.70625=area of the base BCD.

Or,

2.5+2.5+2.5
2

7.5
= = 3.75 =
2

sum of the sides,

BC, CD, DB, of the triangle CDB.

And 3.75—2.5=1.25, .. 1.25, 1.25 and 1.25=3 differ

ences.

Whence √3.75 × 1.25 × 1.25 × 1.25=√ 3.75 × 1.252= ✓7.32421875=2.7063=area of the base as before,

And 2.7063×10=27.063 solid feet, the content of the prism required.

2. What is the solidity of a triangular prism, whose length is 18 feet, and one side of the equilateral end 14 feet? Ans. 17.5370265 feet.

3. Required the solidity of a prism whose base is a hexagon, supposing each of the equal sides to be 1 foot 4 inches, and the length of the prism 15 feet. Ans. 69.282ft.

PROBLEM IV..

To find the convex surface of a cylinder.

RULE.†

Multiply the periphery or circumference of the base, by the height of the cylinder, and the product will be the convex surface required.

EXAMPLES.

1. What is the convex surface of the right cylinder ABCD, whose length BC is 20 feet, and the diameter of its base AB 2 feet?

* See Notes to Prob. III. Cor. 2. p. 56.

+Demon. If the periphery of the base be conceived to move in a direction parallel to itself, it will generate the convex superficies of the cylinder; and, therefore, the said periphery being multiplied by the length of the cylinder, will be equal to that superficies. Q. E. D.

Note. If twice the area of either of the ends be added to the convex surface, it will give the whole surface of the cylinder.

Here 3.1416x2=6.2832-periphery of the base AB. And 6.2832 x 20=125.6640 square feet, the convexity required.

2. What is the convex surface of a right cylinder, the diameter of whose base is 30 inches, and the length 60 inches? Ans. 5654.88 inches. 3. Required the convex superficies of a right cylinder, whose circumference is 8 feet 4 inches, and its length 14 feet. Ans. 116.666, &c. feet.

PROBLEM V.

To find the solidity of a cylinder.

RULE.*

Multiply the area of the base by the perpendicular height of the cylinder, and the product will be the solidity.

*The four following cases contain all the rules for finding the superficies and solidities of cylindric ungulas.

1. When the section is parallel to the axis of the cylinder.