The volume of the solid generated by its revolution round the axis of y, or the base of the curve is = = πа3 fd0 (1 + cos 0)2; which taken from 0 to 0, and doubled, gives as the whole volume of the solid generated If the curve revolve round the axis of x, SECT. 4. Quadrature of Surfaces. The general expression for the surface of a solid is the limits of x and y being determined as in the cubature of solids. (1) In the sphere where + y2 + x2 = r2, we have which when taken from y = 0 to y = (r2 − x2)1 gives and, supposing S to vanish when a = 0, This is the part of the surface included within the positive axes, and if we multiply it by 4 we have 2πra as the surface of a zone of the sphere, the height of which is a it is therefore equal to the corresponding zone of the circumscribing cylinder. The whole surface of the sphere is 4πr 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 Here x2 + z2 = a2, x2 + y2 = a2; 2 dz and S = ssdæ dy{1 - ƒ ƒdx dy {1 + (-2)2 + (12)}'. therefore S = a * dx dy Integrating with respect to y from y = 0 to y = (a2 — x2)}, and the whole surface, being eight times this, is 8 a2. (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 x2 + y2 + x2 = a2, x2 + y2 = ax, Transforming into polar co-ordinates r and , we have the limits of being 0 and a cos 0, those of being 0 and T. Therefore π. S = a2 ft" (1 - sin 0) = a2 (π- 1). The area of the surface of the octant of the sphere is a; 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 8a2, 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 ALME GEOMETRIE dicatum, quod testudine intus perfecte hemisphærica operitur: sed in hac fenestrarum quatuor æquales area (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. (4) Acta Eruditorum, 1692. 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 and the whole area of the intercepted surface of the cylinder is 4a2, 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, s the arc of the curve made by a plane perpendicular to the generating plane S = fuds. (5) Under the same circumstances as in Ex. (11) of the last section, to find the surface of the intercepted solid. If be the arc of the circle passing through O and perpendicular to PQ, the area of the element PQpq is lds, and the area of the surface AOB is flds. Now 1 = 2p cosec a = 2 (a2 - x2) cosec a, therefore S = 2a cosec a f" dz = 2a2 cosec a; and the whole surface is 16a 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. When x = 0, S=0 and Ca, therefore S = 2a tan a {a - (a2 − x2)1} ; and the whole convex surface is 2a tan a. When a curve surface is formed by the revolution round the axis of of a curve the equation to which is y = f (x), the area of the surface is given by the integral, dy S = 2x [day {1 +(1/2)"}! (7) For the paraboloid of revolution we have therefore S4πm3 [dx (x + m)} = 8 п m3 (x + m)3 + C. |