Assume a cylindrical tank of radius [latex]4[/latex] m and height [latex]10[/latex] m is filled to a depth of [latex]8[/latex] m. How much work does it take to pump all the water over the top edge of the tank?

The first thing we need to do is define a frame of reference. We let [latex]x[/latex] represent the vertical distance below the top of the tank. That is, we orient the [latex]x\text{-axis}[/latex] vertically, with the origin at the top of the tank and the downward direction being positive.

Using this coordinate system, the water extends from [latex]x=2[/latex] to [latex]x=10.[/latex]

Therefore, we partition the interval [latex]\left[2,10\right][/latex] and look at the work required to lift each individual “layer” of water.

In pumping problems, the force required to lift water to the top of the tank is equal to the weight of the water, overcoming gravity. Given that the weight-density of water is [latex]9800 N/m^3[/latex], or [latex]62.4 lb/ft^3[/latex].

Calculating the volume of each layer gives us the weight. In this case, we have,

[latex]V=\pi {(4)}^{2}\text{Δ}x=16\pi \text{Δ}x[/latex]

Then, the force needed to lift each layer is,

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

Based on our choice of coordinate systems, we can use [latex]{x}_{i}^{*}[/latex] as an approximation of the distance the layer must be lifted.

Then the work to lift the [latex]i\text{th}[/latex] layer of water [latex]{W}_{i}[/latex] is approximately,

Adding the work for each layer, we see the approximate work to empty the tank is given by

This is a Riemann sum, so taking the limit as [latex]n\to \infty ,[/latex] we get

The work required to empty the tank is approximately [latex]23,650,000 J[/latex].

Assume a tank in the shape of an inverted cone, with height [latex]12[/latex] ft and base radius [latex]4[/latex] ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is [latex]4[/latex] ft.

How much work is required to pump out that amount of water?

Show Solution

As we did in the example with the cylindrical tank, we orient the [latex]x\text{-axis}[/latex] vertically, with the origin at the top of the tank and the downward direction being positive (step 1).

The tank starts out full and ends with [latex]4[/latex] ft of water left, so, based on our chosen frame of reference, we need to partition the interval [latex]\left[0,8\right].[/latex] Then, for [latex]i=0,1,2\text{,…},n,[/latex] let [latex]P=\left\{{x}_{i}\right\}[/latex] be a regular partition of the interval [latex]\left[0,8\right],[/latex] and for [latex]i=1,2\text{,…},n,[/latex] choose an arbitrary point [latex]{x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].[/latex] We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).

From properties of similar triangles, we have

[latex]\begin{array}{ccc}\hfill \frac{{r}_{i}}{12-{x}_{i}^{*}}& =\hfill & \frac{4}{12}=\frac{1}{3}\hfill \\ \hfill 3{r}_{i}& =\hfill & 12-{x}_{i}^{*}\hfill \\ \hfill {r}_{i}& =\hfill & \frac{12-{x}_{i}^{*}}{3}\hfill \\ & =\hfill & 4-\frac{{x}_{i}^{*}}{3}.\hfill \end{array}[/latex]

Then the volume of the disk is

[latex]{V}_{i}=\pi {(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 2).}[/latex]

The weight-density of water is [latex]62.4 lb/ft^3[/latex], so the force needed to lift each layer is approximately

[latex]{F}_{i}\approx 62.4\pi {(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 3).}[/latex]

Based on the diagram, the distance the water must be lifted is approximately [latex]{x}_{i}^{*}[/latex] feet (step 4), so the approximate work needed to lift the layer is

[latex]{W}_{i}\approx 62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 5).}[/latex]

Summing the work required to lift all the layers, we get an approximate value of the total work:

[latex]W=\underset{i=1}{\overset{n}{\text{∑}}}{W}_{i}\approx \underset{i=1}{\overset{n}{\text{∑}}}62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\text{(step 6).}[/latex]

Taking the limit as [latex]n\to \infty ,[/latex] we obtain

[latex]\begin{array}{cc}\hfill W& =\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{Δ}x\hfill \\ & ={\displaystyle\int }_{0}^{8}62.4\pi x{(4-\frac{x}{3})}^{2}dx\hfill \\ & =62.4\pi {\displaystyle\int }_{0}^{8}x(16-\frac{8x}{3}+\frac{{x}^{2}}{9})dx=62.4\pi {\displaystyle\int }_{0}^{8}(16x-\frac{8{x}^{2}}{3}+\frac{{x}^{3}}{9})dx\hfill \\ & =62.4\pi {\left[8{x}^{2}-\frac{8{x}^{3}}{9}+\frac{{x}^{4}}{36}\right]|}_{0}^{8}=10,649.6\pi \approx 33,456.7\hfill \end{array}[/latex]

It takes approximately [latex]33,450[/latex] ft-lb of work to empty the tank to the desired level.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “2.5 Physical Applications” here (opens in new window)