See Hirsch's Integral Tables, pp. 130, and 122. 2ar ; then the time of emptying the whole cone is 1 obtained. Let x = 701. Let the sluice be (as sluices usually are,) rectangular, two of its sides, (a) being parallel to the surface of the fluid. Also, let H, and h be the distances of these sides from the surface of the fluid, and supposing this finite oritice a x (H – h) to be composed of innumerable indefinitely small ones, a x dx, distant from the surface by x, the quantity of fluid run out in the time t will be denoted by Q = sadx x t X velocity 2at v.9 x (H) 3 on the supposition that the velocity through each small orifice is due to half its depth in the fluid. Now the reservoir being supplied at a given rate, suppose the supply to be the given quantity Q in the given time t; then H being also known by the question, we have 3 h = (H 2at ng and :: H - h, or the height required. 702. Let be the axis of the paraboloid, B the radius of its base, x the distance of any horizontal section from the vertex ; then X = 4y = mpir p being the parameter of the generating parabola. Hence (see 696) 703. Let a be the common altitude of the cylinder and cone, and r the radius of their common base; then since (696) Хdr 2 and X:X' :: an® : 704. Let a be the altitude of the cylinder, 2r the diameter of its base, and A the area of the aperture. Also let x be any variable height of the surface of the water ; then the quantity discharged in 1" is and, if the constant supply in 1" be Agm then the rate of evacuation is AV9 x (x - Nm) Also by the evacuation the surface descends with the velocity A and by means of the supply it ascends at the same time with the velocity A Х so that upon the whole, the descending velocity is AN 9 x (W x - - Nm). Hence the time through x is Х ✓m x {Wix + 7 m l. (W x – vm) + c } Ang and putting x = a and x = m, and taking the difference between the results, we have the time in which the influx is equal to the efflux, viz. 27r2 * {va- vm+1 mil. Va AV 9 m - Nm5 which, being infinite, shews that there can never exist an equality between the efflux and influx. 705. Let my be any circular section of the vessel whose altitude is x; and A the area of the orifice, then the velocity at the orifice is the equation to a parabola of the fourth order. If a be the given altitude, then the radius of the base is A V ga } { CT which gives the required dimensions of the vessel. 706. Let r be the radius of the sphere, and x the altitude of any common section parallel to the horizon of the sphere and cylinder; then (see 696) X = 7ro, and X' = my? = 7 (27X -- x?) Xdx and since t oC NO dx 7.3 4 2 :: 2r2 x : 3 Let x = 2r; then t: :: 15 : 8, the proportion required. 707. Here the area of the descending surface is (Py being a side of the generating square when at a distance x) X = (24) = 4y = = 4px? from the equation to the semicubical parabola. Hence (see 696) S4pxd. 2p Ag a being the altitude of the vessel. 1 2p 708. Let r be the radius of the sphere, and x the altitude of any horizontal section made by the water; then (696) X = kyo = 7(2rx — xo) VOL. II. 2 L and X' = my” = 7(r? – x2) according as the vertex or base is downwards. d') ::t:ť :: S(2r – x)dx : vix 41 2 2.23 : 2r? ixt' 3 Let x = x = 2r; then t :ť :: 8 : 3. 709. Let a be the axis, and 28 the base of either parabola, and supposing them to consist of an indefinite number of orifices 2y x dx, (y being the ordinate, and r the abscissa of the parabola measured from the surface of the fluid,) the quantity of water discharged in the time t, when the vertex is upwards is s2ydx x t x velocity at the orifice 2ydx 21fydx ga = 2t 9 fvdu = 1 N og and for the whole parabola Q=twpg X a. Again, the quantity discharged, when the vertex is downwards in the same time is Q: 2t fade 1 But in this case y = (4-1) :: Q' = 2t PO SV (au – x2) dx = 2t v p9 x Vers. - 1 see Hirsch's Tables, p. 150. CL Let c = a; then Q' = 2t pg X vers.-12 |