Solving Water Tank Homework: Find Work & Remaining Depth

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Gauss177
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Homework Statement


A tank full of water has the shape of a parabloid of revolution with shape obtained by rotating a parabola about a vertical axis.
a) If its height is 4 ft and the radius at the top is 4 ft, find the work required to pump the water out of the tank.
b) After 4000 ft-lb of work has been done, what is the depth of the water remaining in the tank?

Homework Equations


m = density*volume

The Attempt at a Solution


I don't know how to do part (b). This is what I have for (a):
I labeled the radius of cross section as Ri (ith subinterval)
Ri/(4-Xi) = 4/4
Ri = 4-Xi
Volume of ith layer of water = pi(4-Xi)^2 dx
Mass of ith layer of water = 62.5pi(4-Xi)^2 dx
Force to raise ith layer = (9.8 m/s^2)(62.5pi(4-Xi)^2 dx
W to raise ith layer = 612.5pi*x*(4-x)^2 dx
Total work = Integral of 612.5pi*x*(4-x)^2 dx on [0, 4]

The answer is not right, so can anybody tell me what I did wrong and how to fix it? Also, how would you do part (b)?

Thanks :smile:
 
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You seem to have got the wrong shape for the tank. The equation of any parabola is y = ax^2 (with y vertical and x horizontal) and the problem says y = 4 when x = 4 so you can find the value of a.

Your "Ri = 4-Xi" seems wrong - that would be a cone, not a parabola.

In your "W to raise ith layer" you are not using consistent units - you used g in m/s^2.

For part (b), just find the work to pump out the water to depth D (a similar integral to the first part).
 
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I see. What do you do with the 'a' if I plug in 4 for both x and y? I get a=1/4, but not sure where to go with that.

I thought of it as a cone, so that was wrong. How would you find the radius then for a parabola?
 
thanks. But instead of using m/s^2 for acceleration of gravity, what should the units be? The problem uses feet, so is it ft/s^2? ft/h^2? I'm not sure.