ticular length of the part, and the part of the arc which describes it, instead of the whole length and whole arc.

Ex. Required the surface of a circular spindle, whose length AB is 40, and its thickness CD 30 inches.

Here, by the remarks at pa. 688.

The chord AC = √ AE2 + CE2 = √ 202 + 152 = 25,

Prob. 16. To find the Solidity of a Circular Spindle.

Multiply the central distance OE by half the area of the revoling segment ACBEA. Subtract the product from of the cube of EA,

half the length of the spindle. Then multiply the remainder by 12.5664, or 4 times 3.1416, for the whole contents

Ex. Required the contents of the circular spindle, whose length AB is 40, and middle diameter CD 30.

Prob. 17. To find the Solidity of the Middle Frustum or Zone of a Circular Spindle.

3

From the square of half the length of the whole spindle, take of the square of half the length of the middle frustum, and multiply the remainder by the said half length of the frustum.-Multiply the central distance by the revolving area, which generates the middle frustum.-Subtract this latter product from the former; and the remainder multiplied by 6.2832, or twice 3'1416, will give the

contents.

Ex. Required the solidity of the frustum, whose length m n is 40 inches, also its greatest diameter EF is 32, and least diameter AD or BC 24.

Draw DG parallel to m n, then we
mn = 20,

have DG =

Prob. 18. To find the Superficies or Solidity of any Regular Body. 1. Multiply the tabular area (taken from the following table) by the square of the linear edge of the body for the superficies.

2. Multiply the tabular solidity by the cube of the linear edge, for the solid contents.

Er. If the linear edge or side of a tetraedron be 3, required its surface and solidity.

The square of 3 is 9, and the cube 27,. Then,

solidity 3-18195

Prob. 19. To find the Surface of a Cylindrical Ring.

This figure being only a cylinder bent round into a ring, its surface and solidity may be found as in the cylinder, namely, by multiplying the axis, or length of the cylinder, by the circumference of the ring or section, for the surface; and by the area of a section, for the solidity. Or use the following rules:

For the surface.———To the thickness of the ring add the inner diameter; multiply this sum by the thickness, and the product again by 98696, or the square of 31416.

Er. Required the superficies of a ring, whose thickness AB is 2 inches, and inner diameter BC is 12 inches.

Prob. 20. To find the Solidity of a Cylindrical Ring.

To the thickness of the ring add the inner diameter; then multiply the sum by the square of the thickness, and the product again by 2:4674, or of the square of 3·1416, for the solidity.

Er. Required the solidity of the ring whose thickness is 2 inches, and its inner diameter 12.