And it is manifest, from the symmetrical form of the figure, with regard to CA and CB, that y = x. Ex. 4. A sector of a circle ACB (fig. 39) revolves round one of its radii AC through a given angle (B), and generates a solid, the density at any point of which varies as the (n)th power of its distance from the centre C; to find the centre of gravity of the solid. Since the solid is perfectly symmetrical with regard to a plane passing through AC, and bisecting the angle ß, the centre of gravity must be in that plane. Let CA be the axis of x, and a line in the plane BCA at right angles to AC, the axis of y; the axis of ≈ being at right angles to both; ..zy tan Let a = AC, a= 2 BCA, 0= ECA, 80= FCE, CP=Cp=r, PQ=pq = dr, u = the density at the distance 1 from C. Then the area of the parallelogram Qp =r80.dr; and when the sector revolves about AC, this parallelogram generates a volume for P's distance from AC is r sin 0, and in revolving through the angle ẞ, the length of its path is r sin 0. B. The density of this volume and therefore the mass of the element generated by Qp = μr". Brdr.sin 0.80; ..the mass of the solid uß for +2 sin 0. = Again, the moment of the elementary mass with respect to In order to find z, we must divide the volume generated by the revolution of the parallelogram Pq into elements; to this end, let there be two planes passing through AC and inclined to the plane BCA, at the angles and +84 respectively; then the portion comprehended between them will be equal to the volume generated by Pq, in revolving through an angle do, and therefore is rsin 0.8p.ro.dr = r2dr.sin 080.80. And the density of this element is ur", and therefore its mass is ur+dr.sin 080.80, and its distance from the plane ABC is r sin 0. sin o, as is evident from the construction; and therefore its moment with respect to the plane xy = ur2+3Sr.sin3080.sin pdp; therefore the moment of the solid with respect to the plane xy therefore the moment of the solid with respect to the plane wy Ex. 5. Find the centre of gravity of a cone, the density at every point of which varies as the square of its distancefrom a plane through the vertex parallel to the base. Ex. 6. Find the centre of gravity of the eighth part of a sphere, the density at any point, whose distance from the centre is s, being proportional to GULDIN'S PROPERTIES. 184. The surface generated by a curve line, which revolves about a fixed axis, is equal to the product of the length of the curve line by the length of the path described by its centre of gravity. For let AB (fig. 33) be the curve line, and Ox the line about which it revolves through an angle ; then using the same notation as in Art. 174, the point P describes an arc =y0, consequently the arc PQ describes a zone, of which the length is ye ultimately, and the breadth = ds; hence the area of the zone is ultimately 0yds; and therefore the area of the whole surface generated is = Σ (0yds) = 0 fyds; the integral being taken between the limits corresponding to x = OC, x = OD. y But if be the distance of the centre of gravity of the arc AB from the axis Ox, we have shewn in Art. 174, that 7. (arc AB) = fyds, between the same limits; hence the surface generated = 07. (arc AB). Now Oy is the length of the path described by the centre of gravity, consequently the last equation expresses the property to be proved. 185. The volume generated by a plane area, revolving about a fixed axis in its own plane, is equal to the product of the area into the length of the path described by its centre of gravity. Let A be the revolving area; dA a portion of it so small that it may be all considered to be at the same distance y from the axis. Then if be the angle through which the area revolves, SA will describe a volume which may be considered to be a thin cylinder bent into the form of a portion of a ring. |