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There is a cylindrical tank lying on its side filled completely with liquid. The diameter of the tank is 4ft, and the length of the tank is 10 ft.

Here is a terrible drawing of it but close enough:

enter image description here

I have to set up an integral to find the work required to pump all the liquid to a level 1 ft above the top of the tank.

So I have my formula:

Work = π (weight density) ∫ (area)(displacement)(thickness)

Someone else did this problem and obtained a bounds of 2 to -2, a displacement of (3-y), an area of 2x, and a thickness of dy. Where did the 2 to -2, (3-y), and 2x come from? This is probably intuitive but I don't see it.

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    You need the integral because this is homework, but because of the symmetry of the cylinder you can check the answer by looking at how much work is needed to raise the weight of the liquid by 3 feet.2011-05-20

1 Answers 1

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(Where did $\pi$ come from?)

Look at the tank straight on, so that you only see the circular side. Imagine that circle on the plane, with the center at $(0,0)$. Since the circle has diameter $4$, the bottom of the circle is at $(0,-2)$, and the top of the circle is at $(0,2)$. You want to lift the liquid to one foot above the top of the tank, so you are trying to lift it to the line $y=3$.

Now, take a horizontal slice of that circle of thickness $\Delta y$, at height $y_i$, where $y_i$ is somewhere between $-2$ and $2$. This slice corresponds to a volume of liquid that is approximately equal to the area of this slice, times $10$ (to account for the length of the tank).

To find the area, note that the circle has equation $x^2+y^2 = 4$, so that the $x$ coordinate is given by $x=\pm\sqrt{4-y^2}$. So the width of the slice is going to be $2x=2\sqrt{4-(y_i)^2}$. The thickness is $\Delta y$. So the volume of the slab of liquid is approximately equal to $20\sqrt{4 - y_i^2}\Delta y\text{ cubic feet.}$ Now, you want to lift it all the way to $y=3$; how far from $y=3$ are you? Since the slab is at height $y_i$, the distance to $y=3$ is $3-y_i$.

So the work done in lifting this one slice of liquid is approximately: $\begin{align*} \text{Work}&=\text{Weight}\times\text{Displacement}\\ &= \left(\text{Volume}\times\text{Density}\right)\times\text{Displacement}\\ &\approx \left(20\sqrt{4-y_i^2}\Delta y\right)\times\text{Density}\times(3-y_i)\\ &= \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y. \end{align*}$ (Be sure to get the units right; the density should be in pounds per cubic feet, in which case the units of work will be ft-lbs. )

Now, to get the total work, you slice up the tank into these thin slices, figure out the work for each, and add it all up: $\text{Work}\approx \sum_{i=1}^n \text{Density}\times 20(3-y_i)\sqrt{4-y_i^2}\Delta y.$ Taking the limit as $n\to\infty$, this becomes an integral. The integrand is $\text{Density}\times 20(3-y)\sqrt{4-y^2}\,dy.$ (The $\sqrt{4-y^2}$ is the value of $x$ at the slice).

What are the limits of integration? Going back to the picture, what are the possible values of $y$? Because $y$ denotes where you are in the tank, and the tank extends from $y=-2$ to $y=2$, then the limits of integration go from $-2$ to $2$.

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    Thank you, couldn't have asked for a better explanation. Visualizing the circle on a plane really helped.2011-05-20