and the sphere, are called the three solids of revolution, or the three round bodies. 2. MENSURATION OF THE SURFACES. The Prism and Cylinder. 188. The convex surface of a prism is composed of parallelograms having an equal altitude. The bases of the prism are equal polygons. Find the area of each face, and the sum of the whole will be the superficial contents of the prism. 189. Since the cube is bounded by 6 squares, and consequently by equal faces, its superficial contents may be found by taking the area of one face 6 times. For example, if the side of a cube 2 feet, then the area of one face 2×2= 4 square feet, and the superficial contents of the whole cube will be equal to 4 X 624 square feet. = 190. The surface of a perpendicular cylinder consists of two equal circles, and of a convex surface which is equivalent to a parallelogram, whose base is equal to the circumference of one of the bases of the cylinder, and whose altitude is equal to the altitude of the cylinder, (fig. 6.) Therefore the area of the convex surface of the cylinder is found by multiplying the circumference of one of the bases by the altitude of the cylinder. To this product add the areas of the bases, and we shall have the superficial contents of the whole cylinder. The Pyramid and Cone. 191. The base of a pyramid is a polygon, and its convex surface is composed of triangles. Find the area of each separately, and the sum of the whole will be the superficial contents of the pyramid. If the base is a regular figure, and the pyramid is perpendicular, that is, if a perpendicular let fall from the vertex (or point where all the triangles meet) to the base, passes through the centre of the base, then all the triangles of the convex surface will be equal, and its area may be found by multiplying the sum of the sides of the base by the altitude of one of the triangles. 192. The Cone may be considered as a perpendicular pyramid whose convex surface is composed of an infinite number of equal triangles. Consequently the convex surface of the cone is equal to the product of the circumference of its base multiplied by half its side, as the straight line AB, (fig. 11,) drawn from the vertex of the cone to the circumference of the base, is called. By examining the skeleton of the cone, (fig. 11,) you will find that its convex surface is equal to a sector of a circle, whose radius is equal to the side of the cone, and whose arc is equal to the circumference of the base of the cone. The area of the sector is found by multiplying its arc by half its radius, consequently the area of the convex surface of the cone is found by multiplying the circumference of its base by half its side. To this product add the area of the base, and we have the superficial contents of the cone. Example. What are the superficial contents of a cone whose side is 12 feet, and the radius of whose base is 5 feet? Answer. The circumference of the base is 314 X 10 100 313 feet; therefore the area of the base is 312 × 78 square feet. The convex surface is 12 313=1883 square feet. The superficial content of the cone is 781 × 1882=266% square feet. Regular Solids. 193. All the faces of the regular solids are regular, and therefore equal figures. Consequently the superficial contents of a regular solid are equal to the area of one of its faces multiplied by the number of faces. The Sphere. 194. As a regular polygon is inscribed in a circle, and the circumference of the circle compared with the perimeter of the polygon, in order to get the approximate ratio of the circumference to the diameter of a circle; so we can suppose a regular polyedron, as a solid bounded by plane faces is called, to be enclosed within the sphere, and the superficies of the sphere to be compared with that of the polyedron. In this manner the method of calculating the superficial contents of the sphere has been learnt. It is found that it is equal to the product of the diameter of the generating circle multiplied by its circumference. This circle, and indeed. all circles, whose centres coincide with that of the sphere, are called great circles of the sphere. The diameter of a sphere is a line passing through the centre and terminating both ways in the surface. The area of a circle is equal to the product of its circumference by half its radius, or one fourth of its diameter, (182;) consequently the surface of a sphere is equal to four times the area of a great circle. Example. If the diameter of the sphere is 10 inches, then the circumference of a great circle of that sphere 10 sphere will be 314 31 × 10 square inches; that is, the surface of the sphere is equal to the circumference of a great circle multiplied by the diameter of the sphere. 3. MENSURATION OF THE SOLIDITY. 195. The Cube is the most simple of solids, since its three dimensions are the same. Therefore, a cube whose side is a linear unit has been adopted as the unit of solidity. As its sides are linear units, its faces are the squares of those linear units, that is, they are superficial units. If the side be one inch, the face will be one square inch, and the whole body one cubic inch; if the side be a foot, the face will be a square foot, and the whole solid a cubic foot. To measure a solid we seek how many of these units may be contained in the solid, or into how many cubes of equal magnitude with that taken for the unit, the solid to be measured may be divided. As we used the word area to denote a surface considered as measured; in a like manner we use the word solidity to denote the magnitude of a solid or its bulk. The terms volume and solid contents have a similar import. The Prism and the Cylinder. 196. If the parallelograms composing the convex surface of a prism are perpendicular to the base, it is called a right prism; otherwise it is called an oblique prism. If the bases of a prism be parallelograms, then all its faces will be parallelograms, and the prism is called a parallelopiped. If all the faces are rectangles it is called a rectangular parallelopiped. 197. We will first seek the method of finding the solidity of a rectangular parallelopiped. Let us suppose its base to be 4 feet long and 3 feet broad, then the area of this base will be 4 × 3: 12 square feet. If the parallelopiped be 1 foot thick or high, it is apparent that it may be divided into 12 lesser solids, each of which will measure 1 cubic foot; consequently the whole parallelopiped will measure 12 cubic feet; which is equal to 4 × 3 × 1, that is, to the product of its three dimensions. If the parallelopiped is 2 feet high, then there may be two layers of the small solids, that is, twice as many as before, and the solidity of the parallelopiped will be 4 × 3 × 2, that is, the product of its three dimensions. Hence we infer that the solidity of a rectangular parallelopiped may be found by multiplying together its three dimensions; or, since the product of its length and breadth is the area of the base, it may be expressed shortly; The solidity of a rectangular parallelopiped is equal to the product of its base by its altitude. The altitude of a prism is a perpendicular let fall from one base to the other, or to the other produced. In a right prism the altitude is equal to each of the upright sides. In this oblique prism (fig. 110) the line EG is the altitude. 198. The cube is a rectangular parallelopiped, of which the three dimensions are the same, consequently the solidity of a cube may be found by multiplying one side into itself twice. For example, if the side of a cube |