The problem presents to us a paraboloid tank whose height is h and whose radius is r. As a preliminary step, let us rst solve the volume of such a paraboloid. But before this, we nd the equation for the parabola, noting the fact that at x = r, y = h and for simplicity we have it centered at the origin, we get
y= h 2 x . r2
Now we can carry out the integration, which is that of a solid of revolution using cylindrical shells. Doing this, we note that the radius of the ¢ b
V V V V = 2πxf (x) dx
¢
=
a r
2πx
0
h 2 x dx r2 r4 4
= =
2π
h r2
1 πhr2 . 2
1. First, we are asked for the work that must be done to pump out the water. The work is simply,
W W W = mgh = = (ρV ) gh 1 πρghr2 . 2
The density of water is ρ = 62.4lb/ft3 and we are given the parameters, r = 4ft and h = 4ft. Substituting these, we get
W = 2.0074 × 105 ft-lb.
2. Now, given a value for the work, we solve for the height. Rearranging the previously derived equation, we get
h= 2W . πρgr2
Suppose the original height and radius was h0 and r0 . The equation for the parabola would be
y= h0 2 2x . r0
1
At this new height h, the corresponding r can be solved accordingly. We get
h = r2 = h0 2 2r r0 2 hr0 . h0
Substituting this value to our equation and isolating h, we get
h = h = h = 2W πρgr2 2W h0 2 πρg hr0 2W h0 2. πρg r0
Now we can solve for the height given that W = 4000ft-lb, ρ = 62.4lb/ft3 , r = 4ft and h = 4ft. Substituting, we get
h = 0.56ft.
But this is just the height of the water that went down. What the problem asks for is the current depth. This means we subtract this from the original to get d = 3.44ft.