The volume of the solid generated by its revolution round the axis of y, or the base of the curve is V = a sdy (2a – x) = 7 a sd0 (1 + cos 0); which taken from 0 = 0 to 0 = 7, and doubled, gives as the whole volume of the solid generated V = 3a. If the curve revolve round the axis of x, V = sydx = 7 a jde oʻ sin ; which taken from 0 = 0 to 0 = 7, gives as the volume of the whole solid generated V = = a (71 4) a?. Sect. 4. Quadrature of Surfaces. 'd x d x dy the limits of and y being determined as in the cubature of solids. (1) In the sphere where x? + y + z* = r’, we have dx so that S=r ( Integrating with respect to y, we have Y (juč – xo)?? = } #rx + C, and, supposing S to vanish when x = 0, S = rx. S This is the part of the surface included within the positive axes, and if we multiply it by 4 we have 271 x as the surface of a zone of the sphere, the height of which is æ : it is therefore equal to the corresponding zone of the circumscribing cylinder. The whole surface of the sphere is 4cm? or four times the area of a great circle. (2) The axes of two equal right circular cylinders intersect at right angles, find the area of the surface of the one which is intercepted by the other. The equations are x? + = a, a + y = a*; dz dx dx dy dx dy Here 0; z? dx dy therefore S = . Integrating with respect to y from y = 0 to y = (a – xo)!, S ='a fo“ dx = a”; and the whole surface, being eight times this, is 8a”. dx dy Stat (3) Circumstances being the same as in Ex. (7) of the last section, to find the area of the intercepted surface of the sphere. The equations to the surfaces being a 2 + y2 + x = a, xo? + age = a X, a SS – ')}" Transforming into polar co-ordinates r and 0, we have rdr de S= a (a– the limits of , being 0 and a cos 0, those of being O and . Therefore S = a* f** (1 – sin 6) = a* (1 – sin ) = a? (17 – 1). The area of the surface of the octant of the sphere is Ta; and therefore the area of the surface of the octant which is not included in the cylinder is equal to a’, or the square of the radius of the sphere. If the sphere be pierced by two equal and similar cylinders, the area of the non-intercepted surface is 8a', or twice the square of the diameter of the sphere. This is the celebrated Florentine enigma which was proposed by Vincent Viviani as a challenge to the mathematicians of his day in the following form : “ Inter venerabilia olim Græciæ monumenta extat adhuc, perpetuo quidem duraturum, Templum augustissimum ichnographia circulari Almæ GEOMETRIÆ dicatum, quod testudine intus perfecte hemisphærica operitur : sed in hac fenestrarum quatuor æquales areæ (circum ac supra basin hemisphæræ ipsius dispositarum) tali configuratione, amplitudine, tantaque industria, ac ingenii acumine sunt extructæ, ut his detractis superstes curva Testudinis superficies, pretioso opere musivo ornata, tetragonismi vere geometrici sit capax. Acta Eruditorum, 1692. (4) Under the same circumstances to find the area of the intercepted surface of the cylinder. The element of the circumference of the base of the dx cylinder being 2 (a x - x*)?' dr x2 and the whole area of the intercepted surface of the cylinder is 4 a’, or equal to the square of the diameter of the sphere. If a solid be generated by the motion of a plane parallel to itself, the surface may be found by a method similar to that used for finding the volume. If u be the periphery of the generating plane, 8 the arc of the curve made by a plane perpendicular to the generating plane S = suds. a we have a S = So (ax – x“) 2 1 (5) Under the same circumstances as in Ex. (11) of the last section, to find the surface of the intercepted solid. If s be the arc of the circle passing through () and perpendicular to PQ, the area of the element P Qpq is lds, and the area of the surface AOB is sids. Now 1 = 2p cosec a = 2 (a– x®)} cosec a, adx and ds = (a’ – x)} therefore S = 2a cosec a f* dx = 2 a' cosec a ; and the whole surface is 16 a2 cosec a. (6) Under the same circumstances as in Ex. (12) of the last section to find the area of the convex surface of the part of the cylinder cut off. s being the element of the circumference of the base, and s the element of the surface. xdx (a’ – x”)? S = 2a tan a {a – (a’ – x?)!}; and the whole convex surface is 2 a’ tan a. When a curve surface is formed by the revolution round the axis of x of a curve the equation to which is y=f(x), the area of the surface is given by the integral, dy S = 27 Sdxy d x) (7) For the paraboloid of revolution we have g° = 4 m 2 ; m} (x + m) + C. (8) For the prolate spheroid we have 62 a? and S = 2 пъ sdæ (a’ – e*x"), where e a’ - 6 a? Integrating, we have παο S sin-1 If the whole surface be required, this is to be taken from - a to x = + a, so that 2 a ab {sin-le +e (1 – e')}}. S = e In the oblate spheroid we have for the whole surface (9) The equation to the cycloid (the base being the axis of wis dy 2a dic ; dy = 27 (2a) sdy (2a - y) Integrating this and extending the integral over the whole surface, we find |