3. If the diameter of the sphere is given, the foregoing rule may be conveniently stated in another form, thus:

Rule III.-The volume of a sphere is equal to the cube of the diameter, multiplied by 52359.

4. In rules 2 and 3 the volume of the sphere is compared with that of the circumscribing cube, to which it bears a constant ratio; and for rough calculations it is a useful rule that it is equal to a little more than half the volume of the circumscribing cube. We will now compare it to the circumscribing cylinder, and we shall find that its volume is always equal to two-thirds of the circumscribing cylinder.

For the volume of the circumscribing cylinder is the area of the circular base, multiplied by the height (ie., by the diameter); or

If we take two-thirds of this we have r3, which is the expression denoting the volume of the sphere.

Rule IV.-The volume of a sphere is equal to two-thirds of the volume of the circumscribing cylinder.

EXERCISES (P).

1. A globe, whose radius is 28 in., has to be gilded at a cost of 2s. 4d. per square foot; find the cost.

2. The radius of a sphere is 43 in.; find its solidity.

3. What is the solidity of a spherical shell, whose inner and outer radii are 6 in. and 7 in.

4. A sphere contains 5946 cubic inches; find its radius.

5. What must be the radius of a sphere which contains exactly 1 cubic yard?

6. The surface of a sphere contains 6734 square ft.; find its radius. 7. The radius of a sphere is 182 ft.; find its surface.

8. If the cubical content of the earth is taken as 236,855,164,967

cubic ft., what is its circumference, and the area of its surface? 9. A sphere contains 904 708 cubic ft.; find its radius.

10. The circumference of a spherical ball is 4 yds.; what is the area of its surface?

11. The radius of a sphere is 41 in.; find its solidity.

12. The diameter of a balloon is 6 ft. 2 in.; how many cubic feet of gas will it hold?

13. The radius of a sphere is 1 ft. 4 in.; what is its surface.

14. The surface of a globe contains 6 sq. yds. 2.7449 ft.; find its diameter.

15. A spherical body contains 2.80957 cub. ft.; what is its circumference?

16. The diameter of the sun is 112 times greater than that of the earth; how many times greater is its volume?

LESSON XX.-FRUSTRUM OF PYRAMID OR CONE.

A frustrum is a solid figure formed by cutting off the top of a pyramid or cone by a plane parallel to the base.

In most instances the solidity of a frustrum may be conveniently found by first finding that of the whole pyramid or cone of which it is a part, and then subtracting the solidity of the piece which is cut off. But in cases where this not convenient, the calculation may be made as follows:Let us begin with the frustrum of a triangular pyramid, DCBFE.

Now, if a plane be made to pass through ABD, and another through ABE, the frustrum will be divided into three pyramids, viz., ABCD, whose base is the base of the frustrum, and height the

height of the frustrum; ABFE, whose base is the top of the

frustrum AFE, and height the height of the frustrum; and AEDB, whose base is the triangle BDE, and whose vertex is at A. But this last pyramid can be proved to be equal to a pyramid whose height is that of the frustrum, and base a mean proportional between the top and base of the frustrum.

Now these three pyramids taken together make up the whole frustrum, and if A represent the area of the base, a, that of the top, and h the height, then

Now it is evident that a frustrum with a polygonal base may be treated in the same way, and that since the circle is merely a regular polygon with innumerable sides, the frustrum of a cone may also be similarly treated, and the same rule made to hold good for all.

This rule is most convenient for use in the form of the formula given above, but, as in previous cases it may be expressed in words :—

To find the volume of a frustrum of a pyramid or cone, multiply the sum of the areas of the two ends, and of the square root of their product, by one-third the height of the frustrum.

LESSON XXI.-THE WEDGE.

A wedge is a solid five-sided figure, having a rectangular base, two rectangular or trapezoidal sides meeting in an edge, and two triangular ends.

If the upper edge, or ridge, of the wedge be of the same length as the base (fig. I.), it is evident that the wedge is

only one-half of a rectangular prism, and its volume may be calculated very easily. But if, as in most cases, the edge of the wedge is less than the length (fig. II.), the calculation must be made thus:

Let us suppose that through B a plane BDG is passed parallel to the end AEF. The whole wedge consists of the solid ABDEFG, and the pyramid BGDCH.

Let us for convenience write the initial letters for the length, EC, of the base; b for the breadth FE; e for the length of the edge AB; and h for the altitude or perpendicular height.

Then the volume of the solid ABDEFG, which is half the rectangular prism

This rule may be expressed in words thus :—

Multiply twice the sum of the length of the base, and

the length of the edge by the product of the breadth of the base, and the height of the wedge; and one-sixth of the last product will be the solidity.

The surface of a wedge may be obtained by finding the area of each surface separately, and then adding them together into one sum. The frustrum of a wedge is a prismoid, i.e., a solid whose sides are four plane trapezoids, and its ends two right-angled parallelograms which are parallel, but not similar to each other.

It is evident that the frustrum of a wedge is equal to two wedges, whose bases are the thick and thin ends, and common height the distance between them.

By bearing this in mind, you will be able to calculate the volume of haystacks, stone-heaps, colliery waggons, etc.

EXERCISES (Q).

1. A block of marble is in the form of a frustrum of a square pyramid; the side of the smaller end is 1 ft., and that of the larger end 2 ft. 6 in., and the length of the block 16 yds.; find its volume.

2. The sides of the ends of the frustrum of a square pyramid are 2 ft. 6 in. and 3 ft., and their distance apart 5 ft.; find the solidity of the frustrum.

3. Find the solidity of a frustrum of a pyramid, the sides of the octagonal ends being 1 ft. and 8. respectively, and their distance apart 3 yds.

4. The girth of a log of timber is 2 ft. 6 in. at the larger end, and 1 ft. 3 in. at the smaller; and its length 25 ft. 9 in.; what is its value at one shilling per cubic foot?

5. The base of a wedge is 5 ft. 9 in. long, and 3 ft. 6 in. broad, the length of its edge 4 ft. 3 in., and the perpendicular height 9 ft.; find its volume.

6. A frustrum of a cone is 25 ft. high, and the circumference of its greater and smaller ends 20 ft. and 10 ft. respectively; find its volume.

7. The sides of the ends of a frustrum of a hexagonal pyramid are 1 ft. and 1 ft. 6 in., and the height of the frustrum 6 ft.; find its solidity.

8. A pentagonal frustrum is 5 ft. high, and the sides of its ends 18 in. and 6 in.; find its volume.

LESSON XXII.-SPHERICAL SEGMENT AND ZONE.

In this lesson we shall treat, in a simple manner, of two solids which have as yet been passed over; giving short