If, then, G be the centre of gravity of the parabolic area *34. THE CENTRE OF GRAVITY OF A SURFACE OF REVOLUTION. Any surface of revolution BAC is evidently symmetrical about its axis of revolution AD, its centre of gravity is therefore in that axis. Let the moments be measured from a plane passing through A and perpendicular to the axis AD, and let x and y be co-ordinates of any point P in the generating curve APB of the surface, and s the length of the curve AP. Then M being taken to represent the area of the surface, and being supposed to be made up of bands parallel to PQ, the area AM of each such band is represented (see p. 44.)* by 2ry As, and its moment by 2πμαν Δ8, EXAMPLE. To determine the centre of gravity of the surface of any zone or segment of a sphere. Let B1AC, represent the surface of a sphere, whose centre is D, and whose radius DP is represented by a, and the arc AP by s. Then x=DM = DP cos. PDM=a cos. s.—, y=PM=DP sin. PDM a where S, and S, are the values of s at the points B, and B2, where the zone is supposed to terminate. If S, 0, or the zone commence from A, then *35. THE CENTRE OF GRAVITY OF A SOLID OF REVOLUTION. Any solid of revolution BAC is evidently symmetrical about its axis of revolution AD, its centre of gravity is therefore in that line; and taking a plane passing through A and perpendicular to that axis as the plane from which the moments are measured, we have only to determine the distance AG of the centre of gravity, from that plane. Now, if x and y represent the co-ordinates of any point P in the generating curve, and M the volume of the portion PAQ of this solid, then, conceiving it to be made up of cylindrical laminæ parallel to PQ, the thickness of each of which is Ar, the volume of each is represented by Ar, and its moment by zury. EXAMPLE. To determine the centre of gravity of any solid segment of a sphere. Let B1AC, represent any such segment of a sphere whose centre is D and its radius a. Let x and y represent the co-ordinates AM and MP of any point P, ≈ being measured from A; then by the equation to the circle y2=2ax-x3, *Sxy2dx == X2 Also, M=T = ƒy3dx=7f (2ax—x2)dx=¤(ax,? — }x;"), 0 If the segment become a hemisphere, r,= a, ... G1=ja, F 36. The centre of gravity of the sector of a circle. Let CAB represent such a sector; conceive the are ADB to be a polygon of an infinite number of sides and lines, to be drawn from all the angles of the polygon to the centre C of the circle, these will divide the sector into as many triangles. Now the centre of gravity of each triangle will be at a distance from C equal to the line drawn from the vertex C of that triangle to the bisection of its base, that is equal to the radius of the circle, so that the centres of gravity of all the triangles will lie in a circular arc FE, whose centre is C and its radius CF equal to CA, and the weights of the triangles may be supposed to be collected in this arc FE, and to be uniformly distributed through it, so that the centre of gravity G of the whole sector CAB is the centre of gravity of the circular arc FE. Therefore by equation (23), if S', C1, and a', represent the arc FE, its chord FE, and its radius CF, and S, C, a, the similar arc, chord, and radius of ADB, a1Ci then CG= = ; but since the arcs AB and FE are ST similar, and that a'a, .. C'=3C and S'S. Substituting these values in the last equation, we have 37. The centre of gravity of any portion of a circular ring or of an arch of equal voussoirs. Let B,C,C,B, represent any such portion of a circular ring whose centre is A. Let a represent the radius, and C, the chord of the arc B,C1, and S, its length, and let a,, C, similarly represent the radius and chord of the arc B.C, and S. the length of that arc. G1 G Also let G, represent the centre of gravity of the sector AB,C1, G, that of the sector ABC, and G the centre of gravity of the ring. Then AG, × sect.AB ̧C ̧+AG × ring B ̧C,B,C,=AG, x sect.AB,C,. Now (by equation 32), AG,=§¶;C), AG,=3";C;; AG1= also sector AB,C,S,a,, sector AB ̧C ̧=}S ̧¤ ̧, = S ... ring B1C1C,B ̧=sect. AB,C,—sect. AB ̧C ̧=}S ̧a ̧—}S,α, If NL represent any plane area, and AB be any axis, in the M N RS same plane, about which the area is made to revolve, so that NL is by this revolution made to generate a solid of revolution, then is the volume of this solid equal to that of a prism whose base is NL, and whose height is equal to the length of the path which the centre of gravity G of the area NL is made to describe. L For take any rectangular area PRSQ in NL, whose sides are respectively parallel and perpendicular to AB, and let MT be the mean distance of the points P and Q, or R and S, from AB. Now it is evident that in the revolution of NL about AB, PQ will describe a superficial ring. Suppose this to be represented by QFPK, let M be the centre of the ring, and let the arc subtended by the angle QMF at distance unity from M be represented by, then the area FQPK equals the sector FQM |