# Work required to pump water out of tank in the shape of a paraboloid of revolution

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!

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

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 $$\int_2^3 (62.5)(4-y)(4\pi y)\,dy.$$