EXAMPLES. XXXII. 1. If a cannon ball, 3 inches in diameter, weigh 6 lbs., find the weight of a ball of the same metal 5 inches in diameter. 2. If the model of a steam engine weigh 80 lbs., find the weight of the engine itself, supposing it made of the same substance as the model and of nine times its lineal dimensions. 3. The heights of two similar right circular cylinders are 7 inches and 10 inches respectively: shew that a similar cylinder 11'03 inches high is less than the sum of the two, and a similar cylinder 11'04 inches is greater than the sum of the two. 4. The height of a pyramid is 16 inches, and its volume is 400 cubic inches: the pyramid is divided into two parts by a plane parallel to the base and distant 4 inches from it: find the volumes of the parts. The following examples involve the extraction of the cube root: 5. If a cannon ball, 3 inches in diameter, weigh 6 lbs., find the diameter of a ball of the same metal which weighs 20 lbs. 6. The height of a right circular cylinder is 4 feet: find the height of a similar cylinder of nine times the volume. 7. The diameters of the ends of the frustum of a cone are respectively 20 feet and 16 feet, and the height of the frustum is 5 feet; the frustum is divided into two equal parts by a plane parallel to the ends: find the distance of the plane from the smaller end. 8. If the frustum in the preceding Example is divided into three equal parts by planes parallel to the ends, find the distance of the planes from the smaller end. 9. A frustum of a pyramid on a square base is trimmed just enough to reduce it to a frustum of a cone: shew that rather more than of the original volume is removed. 10. Every edge of a pyramid on a square base is 1 foot: shew that the volume of the pyramid is of a cubic foot; No2 and that the volume of any pyramid on a square base which has all its edges equal may be obtained by multiplying the cube of an edge by 2 6 185 FIFTH SECTION AREAS OF THE SURFACES OF SOLIDS. XXXIII. PLANE SURFACES. 314. WE now proceed to the measurement of the surfaces of solids; this subject is properly connected with the second Section of our work, but we have thought it more convenient for beginners to treat of the volumes of solids before treating of the areas of the surfaces of solids. 315. The area of any plane surface of a solid must be found by the Rules given in the third Section of the work. We will mention the various cases that can arise. The faces of a rectangular parallelepiped are all rectangles; the faces of any other parallelepiped are parallelograms, two or four of which may be rectangles. The ends of a prism are triangles or other rectilineal figures; the other faces are rectangles or parallelograms according as the prism is right or oblique. The base of a pyramid is a triangle or other rectilineal figure, and the other faces are triangles. The ends of a prismoid, or of a frustum of a pyramid, are triangles or other rectilineal figures, and the other faces are trapezoids. Two of the faces of a wedge are triangles; each of the other three faces is a trapezoid, or a parallelogram, or a rectangle. In all these cases the surfaces are plane rectilineal figures, and their areas can be found by Rules already given. A Rule has also been given for finding the area of a circle; and the following cases will occur for the application of the Rule: the ends of a circular cylinder, the base of a circular cone, the ends of a frustum of a circular cone, the base of a segment of a sphere, and the ends of a zone of a sphere. In fact any Rule given in the third Section of the book might find an application in the present Section. Thus, for example, the Rule for finding the area of a segment of a circle will enable us to find the areas of the ends of the segments of cylinders which are considered in Art. 255. 316. Examples: (1) Find the area of the whole surface of a cube which is 8 inches long. Each face of the cube is a square containing 64 square inches; and there are six faces: thus the area of the whole surface in square inches is 6 × 64, that is 384. (2) A pyramid stands on a square base which is 10 inches long, and each of the four faces which meet at the vertex is an equilateral triangle: find the area of the whole surface of the pyramid. The area of the base is 100 square inches. The area of each of the triangular faces is about 43.3 square inches, by Art. 206; therefore the area of the four triangular faces is about 173.2 square inches. Thus the area of the whole surface of the pyramid is about 273-2 square inches. (3) A vessel is to be made in the form of a rectangular parallelepiped without a lid; externally the length is 4 feet, the breadth 3 feet, and the height 2 feet: find the area of the whole external surface. The surface consists of two rectangles each measuring 4 feet by 2 feet, two rectangles each measuring 3 feet by 2 feet, and one rectangle measuring 4 feet by 3 feet: the total area is 40 square feet. Suppose that the vessel is to be formed of metal half an inch thick; and that we have to find how much metal is required. The answer for practical purposes is this: we shall require 40 square feet of the assigned thickness. And in this manner any such question is usually solved when the thickness of the metal is small compared with the dimensions of the vessel. The exact method of solution has been given in Art. 245, from which it appears that the exact result is 2814 cubic inches. Now a sheet of metal of 40 square feet in area and half an inch thick will be found to contain 2880 cubic inches. Thus we see that the approximate result is slightly in excess of the exact result. The thinner the material of which the vessel is composed, the smaller will be the difference between the approximate result and the true result. (4) A vessel is to be made in the form of a rectangular parallelepiped on a square base without a lid, to hold a cubic foot; the height is to be half the length: find the area of the whole internal surface. A vessel of the same base but of twice the height would be in the shape of a cube, and would hold 2 cubic feet, that is 3456 cubic inches. Hence the length of a side of the base in inches will be the cube root of 3456: thus we shall obtain for a side of the base 15'119... inches. Since the height is half the length, the area of the base is double that of any of the other four faces; and therefore the area of the whole internal surface is three times that of the base; so that in square inches it is three times the square of 15'119...: it will be found that this is 685.75... Hence, as in the preceding Example, if the vessel is to be made of metal of an assigned small thickness we shall require about 686 square inches of metal of that thickness. 317. The principle illustrated in the last two Examples may be thus stated: in order to construct a vessel of material having an assigned small thickness we require a sheet of the material equivalent to the external surface of the vessel. Admitting this principle we can give an interesting practical form to some results of calculation. Thus, for instance, by comparing the results of Examples 37... 41 at the end of the present Chapter, |