=6×3×24 cub. ft. = 1975 = cub. ft. 61 cub. ft. = 61 ft. 1242 in. Ex. Let the dimensions of a rectangular parallelopiped be 10.5 inches, 2.05 ft., and 3.27 yds. Find the solid content, or volume, and also the surface. feet, Reducing the dimensions to the same unit, viz. 10.5 =2×{3·27×3+2·05}×· +2×3·27×3×2.05, 12 And the surface =20·755+40·221=60·976 sq. ft. Next, let the parallelopiped be oblique, as ABCDEFGH, which has for its base the parallelogram EFGH, but not rectangular. Through D draw the plane DHKL perpendicular to AD or EH, and through A draw AEMI also perpendicular to AD, meeting CB and GF produced in I and M. Then AK is a rectangular paGrallelopiped. Also, since AI, AB are equal, and parallel to, DL and DC respectively; and since AE DH; the volume of ABIEFM is equal in all re = spects to that of DCLHGK; and taking these equal volumes from the whole figure, we have the remainders equal, viz. the oblique parallelopiped EC is equal to the rectangular one AK. But the volume of AK = base HM × AE, .. vol. of oblique parallelopiped = area of the base ×ht. 288. To measure the volume, and surface, of a prism. DEF. A Prism is a solid bounded by plane surfaces, of which two (called the ends of the prism) are equal, similar, and parallel figures of any number of sides, and the rest parallelograms. 1st. Let the base of the prism be triangular. Suppose a plane to be made to pass through AE, CG, the opposite edges of the oblique parallelopiped of the last article; then that solid will be divided into two equal prisms with triangular bases; and the volume of each will therefore be equal to half the volume of the parallelopiped; that is, it will be equal to the product of the height and the area of its triangular base. 2ndly. Let the base be a polygon. The prism may be divided into other prisms, having triangles for their bases by planes through pairs of edges; and, by the last case, the volume of each is equal to the product of the height and the area of its triangular base; therefore the volume of the whole prism is equal to the product of the height and the area of the whole base. 3rdly. Let the prism be oblique. Take an upright prism of the same base and height; suppose it to be divided into any number of equal thin prisms by sections parallel to the base; and let them be placed, as in the figure, so that they may assume the form of an oblique prism; then, if the number of sections be indefinitely increased, and therefore the thickness of each portion correspondingly diminished, the volume of all the small pieces will not differ appreciably from that of the oblique prism. .. vol. of the oblique prism = sum of the vol3. of all the small prisms, = the vol. of the upright prism, = base × perp". height.* Hence, in all cases, the vol. of a prism = base × perp. ht. If the surface be required, since a prism is bounded by plane surfaces which are parallelograms, and by two triangular or polygonal ends, the whole surface is therefore obviously obtained by measuring all these plane surfaces according to the methods already given for measuring any plane surfaces whatever, and taking their sum. Ex. Let the height of an upright prism upon a triangular base be 10 ft., and the sides of the base be 3, 6, 7, feet, respectively; find the volume, and surface, of the prism. By (230) the area of each triangular end = 8.95 ft. nearly. .. volume of prism=10×8.95=89.5 cub. ft. Also, the three rectangular parallelograms bounding the prism have the same length, 10 ft., and the sum of all these =10×(3+6+7)=160 sq. ft. Also, area of the ends = 17.9 sq. ft. .. whole surface=177.9 sq. ft. 289. To measure the volume, and convex surface, of a cylinder. * A pack of cards, pushed out of the perpendicular slightly, and equally from the lowest to the highest card, furnishes a good illustration of this case. If the straight line joining the centres of the two ends of the cylinder be perpendicular to the base, the cylinder is a right cylinder; but if it makes any other angle, it is termed an oblique cylinder. A right cylinder has already been defined in (281). The surface of a right cylinder has been found in (283), from its being capable of being unwrapped, and measured as a plane surface. Also, since the cylinder, whether right or oblique, may be considered as a prism, upon a polygonal base with an indefinitely large number of sides, therefore, its volume, as in the case of a prism, is equal to the product of its height and the area of the base. But if the cylinder be oblique, its convex surface may be found as follows: Suppose the surface be unwrapped, as in (283); then it will be found to be a portion of a circular ring, of which the two circular arcs are similar and equal, having their extremities joined by equal and parallel straight lines. Now this plane area may be supposed to be divided into an indefinite number of small equal parallelograms, by lines parallel to the ends; and the area of each will be equal to its length x its height. Hence the whole area is equal to the product of the length, and the sum of all these small heights, i. e. the convex surface is equal to the product of the length of the slant side of the cylinder, and the circumference of a section at right angles to the slant side. By another method the same result may be obtained: Let two sections of the oblique cylinder be made at right angles to the axis, and passing through the extreme opposite points in the circumferences of the two ends; the portions of the cylinder thus cut off will evidently be equal and similar in all respects; and if one of them be removed, and placed at the other end, neither the convex surface, nor the volume, of the cylinder will be altered. But we shall then have a right cylinder, whose height is equal to the slant side of the oblique cylinder, and base one of the sections before-mentioned; and its sur = face its height x circumference of base (283); therefore, for the oblique cylinder, surface required = slant side × circumference of section at right angles to axis. Also, volume of oblique cylinder = slant side area of section at right angles to axis. 290. To measure the volume, and surface, of a pyramid. DEF. A Pyramid is a solid bounded by plane surfaces, of which one (called the base of the pyramid) is any rectilineal plane figure, and the rest are triangles converging to one point as a common vertex. If the pyramid be placed with its vertex upwards, and a perpendicular from that vertex upon the base falls upon the centre of the base, or of any regular curve circumscribing the base, the figure is termed a right pyramid. If the perpendicular does not so fall, it is termed an oblique pyramid. A 1st. When the base of the pyramid is a triangle. D F E Let FABC be the pyramid, either right or oblique, having the triangle ABC for its base. Through the points A and C draw the lines AD, CE, equal and parallel to BF; and let planes pass through DA, FB, and FB, EC; and a third plane through the points D, E, F; we shall thus form a triangular prism ABCDFE, of which the proposed pyramid is a part, and having the same base and height as the prism. B From this prism take away the pyramid FABC; there remains a pyramid whose base is ACED, and vertex F. Through D, F, C, draw a plane; the pyramid FACDE will be thereby divided into two triangular pyramids, FDAC, FDEC, which are equal, because they |