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See Hirsch's Integral Tables, pp. 130, and 122.


; 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
=at sdx vgx

2at v.9 x (H)


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

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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)

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703. Let a be the common altitude of the cylinder and cone, and r the radius of their common base; then since (696)


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and X:X' :: an® :

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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

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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

Ngx x

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and by means of the supply it ascends at the same time with the velocity


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so that upon the whole, the descending velocity is

AN 9

x (W x - - Nm).

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Hence the time through x is



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.


* {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.

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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

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the equation to a parabola of the fourth order.

If a be the given altitude, then the radius of the base is


V ga }


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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


..t: | :: po : S(2r x) V xdx


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:: 2r2 x :


Let x = 2r; then

t: :: 15 : 8, the proportion required.


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)



Ag a being the altitude of the vessel.



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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

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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'


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


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.


Let c = a; then

Q' = 2t pg X


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