This is the problem I have been assigned:

A water tank has the shape of a paraboloid of revolution: its shape is obtained by rotating the parabola $y=x^2/4$, for $0\le x\le 4$, around the $y$-axis. Assume that the water in the tank is $3$ ft deep.

Find the work required to pump water out of the top of the tank, to lower the water level to $2$ ft. (The weight of water is $62.5$ lbs/ft${}^3$)

I know that the radius of this equation would be $2y^{1/2}$ and I have my integral set up as $$\int 4\pi y(62.5)y\,dy$$ but I am unsure if this is correct. Also I am not sure what my bounds should be in order to pump water out of the tank so the remaining amount is $2$ feet deep. Thanks!

  • $\begingroup$ I formatted the formulas in your question. See math notation guide. $\endgroup$
    – user147263
    Commented Dec 12, 2014 at 22:53

1 Answer 1


Consider a thin horizontal sheet of water, going from distance $y$ from the bottom of the tank to distance $y+dy$. The volume of this water is approximately $\pi x^2\,dy$. This water has to be lifted through distance $4-y$. The work done is approximately $$(62.5)(4-y)(\pi x^2)\,dy.$$ "Add up" (integrate) from $y=2$ to $y=3$. Note that $x^2=4y$, so we want $$\int_2^3 (62.5)(4-y)(4\pi y)\,dy.$$


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