Ex. 3. To determine the area contained between two successive convolutions of the spiral of Archimedes; fig. 30. Let the general form of the equation to the spiral be rap, being the whole angle through which the radius vector has revolved; and let SA, SB, SC, severally be the values of the radius vector after n-1, n, and n+1 complete revolutions, so that SA = 2(n-1)πα, Sв = 2n па, SC = 2 (n+1) π а; let PSA = 0; therefore SP = {2(n−1)π+0}a, SР1 = {2nπ+0}a, which values it is convenient to represent by r and r1; the problem is to determine the area of APB, BP, C, C, which is expressed by the following definite integral : therefore the area generated in the first revolution of the radius vector is 83 a2; and hence that generated in the nth revolution is n times that generated in the first. 230.] If the equation to the curve is given in terms of r and p, instead of finding the equivalent expression in terms of r and 0, and then integrating as in the preceding articles, it is more convenient to pursue the following course: Let r dr de be integrated first in respect of r; and supposing the limits of r to be the radius vector of the curve and 0, we have, and 0, being the limits of 0, where r and r。 are the radii vectores of the bounding curve corresponding to 0 and 0, respectively; in which expression p must be replaced by its value in terms of r, and the r-integration then effected. Ex. 1. To find the area contained between the involute of the circle and two limiting radii vectores; see fig. 31. The equation to the curve is r2-p2 = a2. Ex. 2. To find the area contained between an epicycloid and its base-circle during one revolution of the generating circle; see fig. 42, Vol. I. By (9), Art. 268, Vol. I, the equation to the curve is therefore the area contained between the pole and the curve and as the area of the circular sector which is included in the above expression is ab, the area included between the circle and In reference to the expression for the area which is given in (12) it is to be observed, that the triangle OSP, which is ex r2 do pressed by > 2 is the e-differential of the area. But the area 1 = r2de = pds; and if A is a sectorial area, Α A = 1Jpds; which is another expression for the area. 231.] The method of the present section is also applicable to the following problem : To find the area contained between a curve, its evolute, and any two limiting radii of curvature. Р In fig. 32 let OPQB be the plane curve on which P and Q are two consecutive points, P being (x, y), and PQ being an infinitesimal arc, and therefore equal to ds; let Pn be the radius of curvature at P, and be denoted by p: then the area of the infinitesimal triangle PпQ is equal to and as the required area is the sum of all these, we have (15) in which p and ds must be expressed in terms of a single variable, the limits of integration being assigned by the conditions of the problem. Ex. 1. To determine the area contained between the parabola, whose equation is y2 = 4ax, its evolute, the radius of curvature at the vertex, and any other radius of curvature. and therefore the area contained between the curve, the evolute, and the radii of curvature at the vertex and at the extremity of Ex. 2. To find the area contained between the cycloid, its evolute, and two given radii of curvature. In fig. 32, let o, the starting point, be the origin; then SECTION 3.-The Quadrature of Surfaces of Revolution. π 232.] In fig. 33, let APQ be a plane curve, and suppose it to generate a surface of revolution by revolving about a line or in its own plane, a'r'q' being its position, when half a revolution has been performed; and let the equation to AP be y = f(x); let OM = ∞, MP = y, PQ: ds; P and Q will, in a complete revolution, describe circles whose radii are respectively y and y+dy, and therefore the paths traversed severally by P and Q are 2y and 2 (y+dy): supposing the curve to be continuous and to have no points of inflexion between P and Q, the element PQ will describe a circular band whose breadth is ds, and the circumferences of whose bounding circles are 2πy and 27 (y+dy); the area therefore of the convex surface of this band is intermediate to 2 yds and 2π(y+dy) ds; and neglecting the infinitesimal of the second order, the convex surface of the infinitesimal band is equal to 2 yds; and therefore, as it is an infinitesimal bandelement of the surface, the surface =√ π (17) the integral being of course definite, and the limits being given by the conditions of the problem. As y = f(x) is the equation to the generating plane curve, dy = f'(x) dx; and the surface = 2′′ ƒƒ(x) {1+ (ƒ′(x))2}1 dx ; which is the form convenient in most cases; other be explained in the sequel. Ex. 1. To find the surface of a sphere. (18) processes will The equation to the generating curve is 2+ y2 = a2, so that y ds = a dx. a = Απα dx = 4π a2. .. the surface of the sphere Hence also a zone of a sphere contained between two planes perpendicular to the axis and at distances x, and a from the 0 centre is equal to Ex. 2. To determine the surface of the paraboloid of revolution. Ex. 3. To find the area of the surface described by the revolution of a cycloid about its base. Here x = a(0-sin 0), y = a (1—cos 0); so that = 32 = a2 * {(cos)2 - 1} d. cos πα 3 πa2. 0 2 Ex. 4. To determine the area of the surface described by the revolution of the tractrix about the axis of x. The differential equation to the tractrix is dy .. the whole surface = = [2 my ds - a dy; Ex. 5. The convex surface of a cone, whose generating line is ay-bx = 0, is πb (a2 + b2)3. Ex. 6. Find the equation to the plane curve which by its revolution about the x-axis generates a surface, the area of which is proportional to (1) the extreme abscissa; (2) the extreme ordinate. |