(4) In the circle, the equation to which is If this be taken from x = a to x = 0, we find the area (6) The equation to the witch of Agnesi is xy4a2 - a b = πα παι. a Taking this from a 2a to a = 0, and doubling it on account of the symmetry on both sides of the axis of x, we find the whole area between the curve and its asymptote to be 4πα. A ▲ = − 2x (2 ax − x2)2 + 3. circ. area whose versine is + C, a 218 and the whole area included between the asymptote and the two branches of the curve is 3a2. This being only a differential equation, it is necessary to use an artifice for the purpose of effecting the integration. If we integrate fyda by parts we have Taking this integral from a = 0 to 2a, and doubling it, we find the whole area of the cycloid to be 3πa2, or three times the area of the generating circle. (9) The differential equation to the tractrix being and the whole area included between the curve and the positive (11) The equation to the evolute of the ellipse is This is best investigated by Dirichlet's method of evaluat ing definite integrals. See Chap. XI. (12) The equations to the companion to the cycloid are fyda xy-fxdy a2 (sin 0 - 0 cos 0) + C. = = The whole area is 2 a2, or twice the area of the generating circle. When an area is referred to polar co-ordinates r and 0, its value is given by the double integral ffrdrdė taken between proper limits. Integrating with respect to r we have A = 1 fr2 do + C; and if we suppose C = 0, the integral A = fr2 do, in which there is substituted for r its value in terms of given by the equation to the curve, is the value of the sectorial arca swept out by the radius vector. In taking the integral between the limiting values of 0, the same precaution must be observed as in the case of rectilinear co-ordinates, that the interval shall not contain a value of which causes r to vanish or become infinite. If we suppose to increase indefinitely, the same geometrical space will be repeatedly swept over by the radius vector at each revolution, so that, when the curve is not re-entering, the analytical area (if we may use the phrase) differs from the geometrical area to obtain the latter we must subtract from the analytical area that portion which has been previously swept over. Thus if we wish to find the geometrical area included between the values 0 and 4 of 0, and if we put This is the fourth part of the whole area of the curve, which is therefore equal to a3. 0 = In this case, if we had at once integrated from 0) = 0 to This anomaly would arise from our integrating through an interval in which r becomes zero. (14) Let the equation to the curve be r = a cos 0 + b, where a>b. The form of this curve is given in fig. 42. If we wish to find the area included within ODCAHG, it is sufficient to integrate from 0 to that value of causes to vanish, and then to double the result. a = cos-1 then the area ODCAHG is equal to } {(a2 + 2b2) a + 3b (a2 − b2)3} ; and the area OEBF is equal to } {(a2 + 2b2) (π − a) − 3b (a2 − b2)1}. which Let a, the curve becomes the common cardioid, and 3πα 2 (15) The equation to the conchoid of Nicomedes when referred to polar co-ordinates is r = a sec 0 + b, and its area is +b20} + C. {atan 0+2ab log tan (+) (16) The curve whose equation is r = a sin 30 and it is sufficient to find the area πα This is easily seen to be and has six loops (see fig. 49), inclosed by one of them. therefore the sum of the areas of the six loops is a2, or one half of the area of the circle which bounds them. 12 (17) The equation to the spiral of Archimedes is After n revolutions the analytical area swept out is a2 n2 (2π)3 6 but to obtain the geometrical area we must sub tract from it the area corresponding to (n − 1) revolutions, as the required geo which gives us (3n2 − 3 n + 1) (2π) a2 metrical area. In the same way we should obtain as the geometrical arca corresponding to (n + 1) revolutions, the a expression (3n2 + 3n+ 1) (2π), and the difference between 6 these or the space between the arcs after (n + 1) and after n revolutions is n (2) a', which is n times the space between the arcs after the first and second revolutions. (18) In the hyperbolic spiral The area swept out by the radius vector from 0 to r is ar, which is equal to the triangle formed by the radius, the tangent and the sub-tangent. If the equation to the spiral be given by a relation between p and r, we have where c = a + 2b, a and b being the radii of the fixed and generating circles respectively. Hence c (c2 — a2) 4a a2 − (r2 — a®)} |