A water tank is shaped like an inverted cone with height of 6 meters and base radius of 1.5 meters.

a.If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank?

b. Is it true that it takes half as much work to pump the water out of the tank when it is filled to half its depth as when it is full?

  • $\begingroup$ $\approx 0$. Just siphon out the water with a hose. $\endgroup$ – John Joy Sep 10 '15 at 1:07
  • $\begingroup$ it says in my book you should get something like $\int_0^6 \rho g \pi \frac{y^2}{16}(6-y)dy$ as the formula but I dont know how they got the $\frac{y^2}{16}$ $\endgroup$ – MD_90 Sep 10 '15 at 1:09
  • $\begingroup$ See this pdf, pp. 25-26 . $\endgroup$ – Tony Piccolo Sep 10 '15 at 11:35
  • $\begingroup$ Notice that at the very top, that $y=6$ ,an that the radius (at the top) is 1.5. In other words, the radius is $\frac{1}{4}$ of the hieght. This relationship remain constant (by similarity) no matter what $y$ is. i,e, $\frac{y}{4}$ is the radius. $\endgroup$ – John Joy Sep 10 '15 at 13:36

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