MENSURATION 25 The surface and volume of a regular polyhedron whose edge is a is obtained by multiplying the surface and volume of a similar polyhedron whose edge is 1 by a? and a 3 respectively. SECTION II. SOLID S CONTAINED BY SURFACES WHICH ARE NOT ALL PLANES. A. The Cylinder. 80. Volume of a Cylinder (fig. 50). Inscribe in the cylinder a polygonal prism of which the number of sides may be increased indefinitely. Then in the limit the base of the prism becomes the base of the cylinder, and the volume of the prism the volume of the cylinder. Now by 67 we have volume of prism = area of base x altitude ; hence volume of cylinder = area of base x altitude. 81. Surface of a Right Cylinder. As above, in the limit the base of the prism becomes the base of the cylinder, and the surface of the prism the surface of the cylinder. Now the lateral surface of prism = perimeter of right section x length = perimeter of base x length, in the case of a right prism ( 68); hence lateral surface of right cylinder = circumference of base x length. Fig. 51. B. The Cone. 82. Volume of a Cone (fig. 51). Inscribe within the cone a pyramid of which the number of sides may be indefinitely in creased, then in the limit the base of the pyramid becomes the base of the oone and the volume of the pyramid the volume of the cone. By 70 volume of pyramid = J base x altitude, and hence volume of cone = ^ base x altitude. 83. Volume of the Frustum of a Cone. From 73 we find that the volume of the frustum of a pyramid hence, since in the limit the frustum of the pyramid becomes the frustum of the cone, we have volume of conical frustum = ^h(A. 1 + X/AjAj^- A 3 ) , where A x and A 3 are the areas of the terminating planes of the frustum. Let the terminating planes be circles of radii i and r 3 , then volume of frustum Again, by the same section we have volume of frustum of pyramid = ^h(A 1 + 4A 2 + A 3 ) , therefore volume of conical frustum = -Jir/i (? + 47-j; + r), where r 2 is the radius of the circular section parallel to the ter minating planes and equidistant from them. 84. Surface of a Right Cone. The lateral surface of a regular pyramid is by 71 = perimeter of base x slant height ; hence, since in the limit the surface of the pyramid becomes the sur face of the cone, the lateral surface of a right cone is equal to half the circumference of its base multiplied by the slant height. Thus the lateral surface of a right cone of slant height I and the radius of whose base is r is equal to x 2-n-r x l = -rrrl, and whole surf ace = lateral surface + area of base = irrl + irr- Again, if h, the altitude of the cone, be given, we have Z-VAa + r 5 , and therefore whole surface = irr( V/t" + r 2 + r) . 85. Surface of the Frustum of a Right Cone. The lateral surface of the frustum of a regular pyramid is equal to the product of the slant height and the perimeter of its middle section ( 74) ; hence in the limit we find that the lateral surface of the frustum of a right cone is equal to the product of its slant height and the cir cumference of the section equidistant from its parallel faces. Let r l and r 3 denote the radii of the ends of the frustum, and I the length of the slant height, then P2 = (fi + ?*3) = radius f middle section, and therefore lateral surface = 27rr 2 x I = 2ir x ^(r t + r 3 ) x I = itl^ + r 3 ) , and whole surface = TIT? + irlfa + r 3 ) + irrl . If h, the altitude of the frustum, be given, we have C. The Sphere. 86. Surface of a Spherical Zone. Let AB (fig. 52) be a small arc of the sphere, and let AA , BB be perpendicular to the axis XX , to find the surface of the zone generated by the arc AB. Join AB, and draw OP perpendicular to AB, BD parallel to XX 7 , and PP parallel to AA or BB . The chord AB generates the frustum of a cone, whose lateral surface = 2irPP x AB. But, since the triangles ABD and OPP are similar, AB^OP BD~PP therefore area of conical frustum = 27T.OP. BD = 2ir. OP. A B . Similarly the area of the frustum generated by BC = 2ir.OQ.B C / . But in the limit when the chords AB, BC, &c., are indefinitely diminished, the perpendiculars OP, OQ, &e., become each = r, and hence by summing all the areas we get in the limit area of zone = 2irr x (projection of arc on axis of revolution). Hence the convex surface of a segment of a sphere is equal to the circumference of a great circle multiplied by the height of the seg ment or zone. 87. Surface of a Sphere. The whole sphere being a zone whose height is 2r, we obtain at once surface of sphere = 2irr x 2r = 4ir?- 2 ; or the surface of a sphere is equal to four great circles. The total surface of the cylinder circumscribing the sphere of radius r is 67rr 2 , hence the surface of the sphere = f surface of cir cumscribing cylinder. 88. Surface of a Lune, a Spherical Triangle, and a Spherical Polygon. It is shown in spherical trigonometry that (a) the area of a lune included between two great circles of a sphere of radius r, and whose inclination is 6 radians, is 26>r 2 ; () the area of a spherical triangle whose angles are A, B, C is (A + B + C - 7r)r 2 ; (7) the area of a spherical polygon of r sides is {P - (r - 2)ir}r 2 , where P is the sum of its angles. 89. Measurement of Solid Angles. A convenient unit for the measurement of plane angles is the "radian." If we assume that each unit of surface of a sphere subtends the same solid angle at the centre, we can deduce a very convenient unit for the measurement of solid angles. This unit, which has received the name " stera- dian," we define to be the solid angle subtended at the centre of a sphere by a portion of the surface whose area is r 2 . 90. Number of Steradians in an Angle. Let A be the angle at the centre of a sphere, and let S be the portion of the surface of the sphere which it intercepts, then number of steradians in A S
For example, if A be a plane solid angle, S = a hemisphere = lirr- ; hence the number of steradians S 2ir?- 2 in a plane solid angle = - = = 2ir , y.2 y2 and therefore the number of steradians in the solid angle at apoint = 4ir. This solid angle is sometimes called a steregon. Hence, if we can find the surface sub tended by any solid angle, we can always find its magnitude in terms of the unit solid angle. A Fi _ co 91. Volume of a Sp7icrc.Let ABC A (fig. 53) be the quadrant of a circle, draw DB and DC tangents to it then, if AD be joined and the whole figure be conceived as
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