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A spherical tank of radius $12$ feet is $40$ feet above the ground. How much work is done in pumping water into the tank until it is full?

I obtained $$ w= \int_{16}^{40}[12^2-(40-y)^2y] \, dy. $$ Is this correct? I've been trying to figure this out for the last 2 hours.

Thank you!

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  • $\begingroup$ Do you think the bottom is 40 feet above the ground or the center is 40 feet above the ground? Presumably we are pumping from ground level. $\endgroup$ – Ross Millikan Dec 17 '14 at 1:34
  • $\begingroup$ I suggest you look at the edit dustin made, Latex is really powerful, and the symbols are fairly easy to render. $\endgroup$ – Trevor J Richards Dec 17 '14 at 1:36
  • $\begingroup$ It's not clear on how the water is getting into the tank...it could just be dumped in from the top $\endgroup$ – ClassicStyle Dec 17 '14 at 1:37
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    $\begingroup$ When you say the tank is 40 feet above ground, are you referring to the center? And as @TylerHG suggested, you probably want to include that the water pump doing the work is at ground level. Is this all correct? $\endgroup$ – Trevor J Richards Dec 17 '14 at 1:37
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You have nothing to indicate the density of the water nor the acceleration of gravity. It looks like you are assuming that the top of the tank is 40 feet above the ground. The center of the tank is then $28$ feet above the ground. We can consider the water between height $y$ and $y+ \Delta y$ to be a disk of radius $r$, so of volume $\pi r^2 \Delta y$ We have $r^2=12^2-(28-y)^2$ so the work done to lift the water in the disk is $\pi (12^2-(28-y)^2)y g\rho$, where $g$ is the acceleraion of gravity and $\rho$ is the density of water. The integral becomes $$\int_{16}^{40}\pi(12^2-(28-y)^2)yg\rho\ dy$$ enter image description here

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  • $\begingroup$ I like your picture. The calculation will be a little simpler if we do $$\int_{-12}^{12} \pi (12^2 - y^2) (y+28) g\rho dy$$ $\endgroup$ – Sungjin Kim Dec 1 '17 at 2:45
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The way to do this is think about the work needed to move just one horizontal slice (with thickness $dy$ up to height $y$ from ground height (assuming that the pump is at ground level). I will assume also that the height of the top of the tank is $40$ feet up, as that seems to be what your solution suggests.

Then the tank has $y$-values fro $16$ to $40$. The radius of a horizontal slice at height $y$ is $\sqrt{12^2-(y-28)^2}$. Therefore the volume of the slice is $\pi\left(\sqrt{12^2-(y-28)^2}\right)^2dy$. Work equals force times distance. The force acting on this water is the mass of the water (which equals $\pi\left(\sqrt{12^2-(y-28)^2}\right)^2dy$ times $\delta$, the density of water) times the acceleration due to gravity (denoted $g$). The distance needed to move the water is the height of the slice (or $y$). Thus the integral should read

$\displaystyle\int_{y=16}^{40}\delta*y*g*\pi\left(\sqrt{12^2-(y-28)^2}\right)^2dy$.

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  • $\begingroup$ Water density of $1$ is not consistent with the usual units that include feet. $\endgroup$ – Ross Millikan Dec 17 '14 at 2:33
  • $\begingroup$ @RossMillikan Oops, didn't see feet, and was thinking in meters. I have fixed the solution. $\endgroup$ – Trevor J Richards Dec 17 '14 at 2:34
  • $\begingroup$ It probably warrants comment that the $16$ comes from $40-2\cdot12=40-24=16$. $\endgroup$ – Ian Dec 17 '14 at 2:39
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To check your answer, consider an arbitrarily small "cell" of water inside the tank after it is filled. If the water in that cell has mass $\mathrm d m$ and is at height $y$ above the ground, the amount of energy you have to expend to raise it from the ground to its final position is $y\;\mathrm d m.$

For each such cell of water above the level of the center of the tank, there is an equal-volume cell of water inside the tank that is the mirror-image reflection of the first cell through the horizontal plane in which the center of the tank lies. And vice versa, for each cell below that plane there is a mirror image of that "cell" in the tank above the plane. The entire tank can be partitioned into pairs of tiny "cells" reflected through that central plane.

Let $h$ be the height of the center of the tank above the ground, and consider a pair of cells of which the upper cell is at height $h + \Delta y.$ The other cell is at height $h - \Delta y.$ The total potential energy you must invest in those two cells is

$$ (h + \Delta y)\,\mathrm d m + (h + \Delta y)\,\mathrm d m = 2h\;\mathrm d m.$$

That is, the energy invested in those two cells (of combined mass $2\,\mathrm d m$) is the same as the energy required to lift a mass $2\,\mathrm d m$ to a height $h$ from the ground.

We can therefore simplify the calculation by replacing every such pair of cells with one such mass, that is, we take the entire mass of water inside the sphere and substitute an exactly equal mass at a height of exactly $h.$

Your answer will be correct if it equals $mh,$ where $m$ is the mass of water in the tank.

I suspect your instructor did not intend you to use this method, so I recommend to either keep this as a mere check on the result you get by the intended methods, or at most show it as an alternative method after deriving the result by those other methods.

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