v" = the vol. generated by BFC = §d'πr22. The sign of "" is or according as AG is on the same or opposite side of the center as BF. .'. v == }π (2 ar2 + d'r'2=d"r" 2). Let v = 217. Examples. = = 1. r 12 in., r3 in., " 10 in.; required v. 2. Two parallel planes divide a sphere whose diameter is 36 in. into three equal segments; required the altitude of each. Ans. 13.93 in.; 8.14 in.; 13.93 in. 218. Problem. To find the volume generated by the revolution of a circular segment about a diameter exterior to it. vol. generated by ADB. = vol. generated by ABC. v = v′ — v". D = A τα παρ2. G CI, perpendicular to AB. υ'' = {παρ ̓. v" p2) C 219. Examples. = 1. a = 5 in., c 8 in.; find v. Ans. 167.552 cu. in. 2. A sphere 6 in. in diameter is bored through the center with a 3-inch auger; required the volume reAns. 73.457 cu. in. maining. 3. Prove that the volume generated by the segment whose altitude is a and chord e is to the sphere whose diameter is c as a : c. 4. Prove that if c is parallel to the diameter about which it is revolved, the volume generated by the segment is equal to the volume of a sphere whose diameter is c. 220. Problem. To find the volume of a wedge. The base is a rectangle, the sides are trapezoids, the ends, triangles. Let e be the edge, I the length of base, b the breadth of base, and a the altitude. 7. Passing planes through the extremities of the edge perpendicular to the base, we have a triangular prism and two pyramids. These pyramids may fall within or without the wedge, or one or both of the pyramids may vanish. But in all cases the formula is the same. abe the volume of the prism. a (l — e) b = the volume of the pyramids. .'. v = & ab (21+e). = 221. Examples. 1. The edge of a wedge is 6 in., the altitude 12 in., the length of base 9 in., and the breadth of base 5 in.; what is the volume? Ans. 240 cu. in. 2. The edge of a wedge is 20 ft., the altitude 24 ft., the length of base 15 ft., the breadth of base 10 ft.; what is the volume? Ans. 2000 cu. ft. 222. Problem. To find the volume of a rectangular prismoid. The bases are parallel rectangles, the other faces are trapezoids. Let and b be the length and breadth of the lower base, l' and b' the length and breadth of the upper base, and a the altitude. ひ = Passing the plane as indicated, the prismoid is divided into two wedges. † ab (2 1 + l') t ab' (2 l' + 1) the vol. of wedge whose base is bl. the vol. of wedge whose base is b'l'. .. v=ta[b (2 1 + 1') + b' (2 l' + 1)]. 223. Examples. 1. The length and breadth of the lower base of a rectangular prismoid are 25 ft. and 20 ft., the length and breadth of the upper base are 15 ft. and 10 ft., and the altitude is 18 ft.; what is the volume? Ans. 5550 cu. ft. 2. The length and breadth of the lower base of a rectangular prismoid are 15 yds. and 10 yds., the length and breadth of the upper base are 9 yds. and 6 yds., and the altitude is 18 yds.; what is the volume? Ans. 1764 cu. yds. To find the dihedral angle included by the faces of a regular polyhedron. Conceive a sphere whose radius is 1 so placed that its center shall be at any vertex of the polyhedron. 224. Problem. The faces of the polyhedral angle will intersect the surface of the sphere in a regular polygon whose sides measure the plane angles that include the polyhedral angle, and whose angles are each equal to the required dihedral angle. Let ABCD be such a polygon, P the pole of a small circle passing through A, B, C, D, E. Join P with the vertices and with the middle of AB by arcs of great circles. or cos Let n denote the number of sides of the polygon, s = one side, and A a dihedral angle. 360° 2 n -) n 180° N .. sin A 180° ጎሪ By Napier's circular parts, we have sin (90° APQ) 180° or sin (90° coss cos (90° — 1⁄2 A). = cos 60° E and AQ=8. In the Tetrahedron, n = 3, and 8 = ... sin A D A = B 70° 31′ 42′′. ... sin A = sin A = cos 60° cos 45° = cos 45° 3, and 4, and cos 36° cos 30° S= 90°, A In the Dodecahedron, n = 3, and 8 = cos 60° cos 54° S = A = A In the Icosahedron, n = 5, and 90°, 60°, 109° 28′ 18′′. 108°, 116° 33′ 54′′. 60°, 138° 11′ 23′′. 225. Problem. To find the volume of a regular polyhedron. If planes be passed through the edges of the polyhedron and the center, they will bisect the dihedral angles and divide the polyhedron into as many pyramids as it has faces. The faces will be the bases of the pyramids, the center will be their common vertex, the line drawn from the center of the polyhedron to the center of any base will be perpendicular to the base, and will be the altitude of the pyramid. From the foot of the perpendicular draw a perpendicular to one side of the base, and join the foot of this perpendicular with the center. We thus have a right triangle whose perpendicular is the altitude of the pyramid, the base the apothem of one face of the polyhedron, the angle opposite the perpendicular onehalf the dihedral angle of the polyhedron. |