But if a be the axis of the paraboloid, and 8 the radius of its base, P+Q=× circumscribing cylinder of the whole para "pa" is the volume of any part of the parabo прохо 2 loid corresponding to the axis x, it is easily shewn that the axis of the frustum is 676. Let s, s' be the specific gravities of the fluids; M, M' their magnitudes; also let S be the specific gravity required; then by the question, the magnitude of the mixture is (M + M') n Again, the sum of the weights of the ingredients = weight of the mixture; 677. Let s, s', S be the specific gravities of iron, the fluid and water respectively, s', as the problem requires, being supposed less than S. Let also r be the radius of the sphere, a the thickness of the shell, and a the required depth (we'll call it) of the shell. Then (Vince's Fluxions) the volume of the exterior segment is and that of the interior segment is m' = x {r — a. (x — a}2 — (x—a)"} = x(x—a)". (3r—2a — x) Hence the bulk of the iron is 3 3 = 3 {x3-3x2 + (6ar — 4a2) x + 2a3}. Now, since the vessel just floats the weight of it and its contents weight of the water it displaces (Vince, Prop. XIV.) ..ms+m's' = MS which gives by substitution {x2 (3r — x) — (x − a)2. (3r — 2a − x)} s + whence by approximation, or otherwise, the required value of x may be found. 678. The centres of gravity of the three surfaces pressed will be in the same point, so that the pressures will be as the surfaces, or as But if x be the depth required, and a the axis of the paraboloid, 680. Let a be a side of the cube; then the distance of the centre of percussion or centre of pressure of the face which is loose from the surface of the fluid, is (Vince's Flux. p. 130,) 3 - Hence the locus of this centre is the straight line in the face parallel to the surface, and distant from it by 3 a. 4 But there is no reason, from the symmetry of the vessel, why the centre should be on one side of the middle of this line, rather than on the other. It is, therefore, in the middle of this line; hence the point of application of the required force. Moreover the pressure to the face is s being the specific gravity of the fluid. Hence the force to be applied in a direction to the face, and at the point, as found above, is a3s 681. Let r, r', be the radii of any two of these circles, d, d' the depths of their centres of gravity; then the pressures are as 682. Let a be the axis, ß the radius of the base of the hollow part of the paraboloid; then its volume is p being the parameter of the generating parabola. Hence the quantity of fluid put into it is and if x be the depth of that fluid, we have Again, let r be the radius of the sphere, s its specific gravity, and S the specific gravity of the fluid; then if Q be the part mersed, and P the other part, we have QP+Qs: S im Hence the portion of the paraboloid occupied up to the surface of the water is now if y denote the depth of the water; which gives y, and . yx, or the height required. 683. Let R, r, be the radii of the generating circles of the unequal cycloids; then since the equation to the latter referred Hence the distance of its centre of gravity from the vertex is and the distance of the centre of gravity of the former curve from its vertex is, in like manner, Now, when at rest, the condition is that the line joining the centres of gravity of the part out of the fluid, and of the part immersed, shall be vertical, that is, to the surface of the fluid. Hence we may find the inclination of the line joining the extremities of the rod to the surface of the water, and therefore the position required. For the distance between the centres of the bases of the component cycloids is zr + %R = % . († + R). .. if the line joining the centres of gravity cut this distance at an angle 0, and divide it into the two portions |