A.3 Context 2: Total Force on a Water Tower's Walls

On this screen we will start to compute the total outward force that the water inside a water tower exerts on the container's walls. This will involve a bit of Physics (that we'll explain) and then developing (you guessed it) Sums we can use to calculate overestimates and underestimates of the total force. This scenario represents another common use of setting up an integral that we're using to help you understand what an integral is.

Water towers on the New York City skyline

The water towers that dot the New York City skyline have received attention for various reasons, including:

Millions of Americans rely on water towers for their everyday water use, ranging from smaller towers that sit on top of urban buildings, to larger towers that act as an entire rural town's supply. Suppose you want to build a water tower that a small community can rely on. Of course you would need to construct the tower such that its walls can withstand the outward force that the water inside exerts on them. We're going to (of course) use Calculus to calculate the total outward force that the column of water exerts on the tank's walls.

In particular, let's consider a water tower that has 5-foot radius and 15-foot height, and is filled to the top with water.

Your goal is to find an approximation, good to within a 500-pound error,
for the total outward force that the water exerts on the tower's walls.

Information you need for this scenario

Pressure and Force relationship:

Array of equal-length arrows showing uniform pressure P on a rectangle of area A. Text says: Total force on wall F = P • A.

Pressure is defined as force per unit area. If the force is constant (uniform) across a surface of area A, then the resulting force on that surface is $𝐹 = 𝑃 \cdot 𝐴$ For pressure we'll be using units of $[l b / f t 2],$ and for area we'll be using units of $[f t 2]$ . Then the resulting force has units of [lbs].

Surface area of a cylinder:

A cylindrical ring of height h and radius r; also a rectangular strip of the same surface area with height h and width 2 pi r. Both have surface area A = 2 pi r h.

The surface area A of a cylindrical ring of radius r and height h is $𝐴 = 2 𝜋 𝑟 ℎ$ (Rather than memorizing this formula, imagine putting a single vertical cut into the cylindrical ring and then unrolling it. You then have a rectangle of length equal to the ring's circumference, $2 𝜋 𝑟,$ and with the same height h. The the ring and the rectangular strip thus have the same surface areas: $𝐴 = 2 𝜋 𝑟 ℎ .$ )

FAQ: "Don't we have to include the surface area of the bottom of the cylinder?"Answer: We don't, because our task is to find the outward force on the walls of the tower. We're not concerned with the force on the bottom surface.

Pressure due to water as depth varies:

Water tower showing depth d below the water's surface, and arrows depicting the water's outward pressure on the tower's walls. Text says that the pressure at depth d beneath the surface is P = 62.4*d pounds per square-foot.

At water depth d, the weight of the water above causes the pressure at that depth. Hence the deeper you go, the greater the pressure due to the greater weight of the water above pushing down. The water's pressure is in direct proportion to the depth d beneath the water's surface, according to $𝑃 (𝑑) = 62.4 \cdot 𝑑$ where d is measured in [ft] and P is still measured in [ $l b f t 2$ ]. Hence at d = 1 foot beneath the water's surface, the pressure is $62.4 l b f t 2;$ at d = 2 feet the pressure is $124.8 l b f t 2;$ and so forth.

The interactive graph below shows you the pressure due to the water as a function of depth below the water's surface.

This pressure at a given depth acts in all directions: upward, downward, and most importantly for us here, outward, pushing against the tower's walls.

Quick Question: Metal bands around water towers

Single water tower, showing metal bands spaced closer together toward the bottom of the tower than at the top.

Quick Question: What's the reason steel bands encircle a water tower? Why are the bands closer together near the bottom of the tank than near the top?

Part I: 5-term approximation

Recall that our goal is to find the total outward force that the water exerts on the tower's wall, to within an error of 500 lbs. Let's again use our Approximation Questions to frame our work:

Questions for Approximations

1. What are you approximating? Why can't you find the exact value immediately?

2. How will you generate an overestimate for the value we're after? An underestimate?

Since this is our first time dealing with a slightly complex scenario, we're going to step you through the process here. The key move is the same as before: we're going to replace the entirely of the wall's surface (the entire interval of interest) with a set of subintervals.

Specifically, as a first calculation we're going to imagine breaking the tower's cylinder into a set of 5 vertically stacked rings as shown, thus giving us 5 subintervals to work with. We could have chosen more or fewer; 5 is just a convenient number since the tower is 15-feet tall, and so each of the five rings is 3 feet tall.

Left figure: the water tower is replaced by five stacked rings, each of height 3 feet. The top of the first is at d = 0; the top of the second at d = 3; the top of the third at d = 6, and so forth. The bottom of the fifth ring is at d = 15, corresponding to the bottom of the tower. Right figure: Five strips, each with arrows representing a uniform pressure across its surface. The pressures are labeled P-subscript-1, P-subscript-2, and so forth.

With this picture in mind, we can develop the same two steps as on the preceding screen to generate our overestimate and underestimate for the total force on the tower's walls:

Determine the approximate total outward force on each individual ring, $𝐹 𝑖$ : $𝐹𝑖=(representative pressure on ring \#𝑖's wall)⋅(surface area of ring \#𝑖)=𝑃𝑖⋅𝐴𝑖$
Sum the force on all of the rings: $𝐹 t o t a l o n w a l l s = 𝐹 1 + 𝐹 2 + 𝐹 3 + 𝐹 4 + 𝐹 5 = 𝑃 1 \cdot 𝐴 1 + 𝑃 2 \cdot 𝐴 2 + 𝑃 3 \cdot 𝐴 3 + 𝑃 4 \cdot 𝐴 4 + 𝑃 5 \cdot 𝐴 5$

Figure of just Strip 2. Its top has depth d = 3, and its bottom has depth d = 6. Its height is h_2 = delta-d = 3.

In order to rewrite the areas $𝐴 1,$ $𝐴 2$ and so forth in terms of other quantities, let's focus on just one strip, Strip #2, as shown in the figure. (You could choose whichever strip you like; we happen to have chosen #2.) Notice that the strip's height, $ℎ 2,$ is the same as the change in depth from the top of the strip to the bottom of the strip: $ℎ 2 = Δ 𝑑 .$ In fact, each of the five strips has the same height $ℎ 1 = ℎ 2 = . . . = ℎ = Δ 𝑑 = 3 f t,$ and the same area $𝐴 1 = 𝐴 2 = . . . = 2 𝜋 𝑟 Δ 𝑑 .$ Hence we can rewrite the total force on the tower's walls as $𝐹 t o t a l = 𝐹 1 + 𝐹 2 + 𝐹 3 + 𝐹 4 + 𝐹 5 = 𝑃 1 \cdot 𝐴 1 + 𝑃 2 \cdot 𝐴 2 + 𝑃 3 \cdot 𝐴 3 + 𝑃 4 \cdot 𝐴 4 + 𝑃 5 \cdot 𝐴 5 = 𝑃 1 \cdot 2 𝜋 𝑟 Δ 𝑑 + 𝑃 2 \cdot 2 𝜋 𝑟 Δ 𝑑 + 𝑃 3 \cdot 2 𝜋 𝑟 Δ 𝑑 + 𝑃 4 \cdot 2 𝜋 𝑟 Δ 𝑑 + 𝑃 5 \cdot 2 𝜋 𝑟 Δ 𝑑 = 2 𝜋 𝑟 [𝑃 1 \cdot Δ 𝑑 + 𝑃 2 \cdot Δ 𝑑 + 𝑃 3 \cdot Δ 𝑑 + 𝑃 4 \cdot Δ 𝑑 + 𝑃 5 \cdot Δ 𝑑] = 10 𝜋 [𝑃 1 \cdot Δ 𝑑 + 𝑃 2 \cdot Δ 𝑑 + 𝑃 3 \cdot Δ 𝑑 + 𝑃 4 \cdot Δ 𝑑 + 𝑃 5 \cdot Δ 𝑑] ◂$ where in the last line we used the fact that r = 5 feet.

FAQ: "You factored out the $2 𝜋 𝑟 = 10 𝜋$ from every term. Why didn't you also factor out the $Δ 𝑑$ ? Or make the substitution $Δ 𝑑 = 3 f t ? ?$ "
Answer: Those are great questions. The honest answer is that we're building up a new tool here, and we won't see fully how it works until further down on this screen. For now, know that in order to improve our approximation later we need a series of terms that we're summing, and each term must be the product of (1) a representative value $𝑃 𝑖,$ and (2) a subinterval's width or duration $Δ 𝑑 .$ Hence we must leave the $Δ 𝑑$ in each term. Furthermore, if we decide later to use fewer or more subintervals, then the value of $Δ 𝑑$ changes. That's an easy change to track if we leave it as $Δ 𝑑$ rather than substituting the value we happen to have for the number of rings we're currently using. By contrast, the radius of the tank doesn't change no matter how many subintervals we use, so we can safely make the substitution $𝑟 = 5 f t$ now.

Now that we have an expression for the 5-term approximation of the total force, let's work to develop both an overestimate and underestimate. We'll start with the overestimate.

Overestimate

Let's begin with Step 1 above, and write an expression for an overestimate of the force on just one subinterval. We again arbitrarily choose to focus on Strip #2.

We have an initial choice to make: what pressure do we use as our single constant, representative pressure across the entire surface of Strip 2? Since the pressure changes linearly from the top to the bottom of the strip, we know our choice will be either the pressure at the top or at the bottom (as opposed to somewhere in between).

Check Question 1: Pressure choice for overestimate

We're currently developing the overestimate for the force on Strip 2. The single value we want to use as the uniform pressure across Strip 2's entire surface is therefore:
(Tap on the correct choice on the image below, either $𝑃 t o p = 𝑃 (3)$ or $𝑃 b o t t o m = 𝑃 (6) .$ )

With that correct choice we now have the overestimate for the force on Strip 2: $𝐹 2, o v e r e s t i m a t e = 𝑃 (6) \cdot 𝐴 2 = 𝑃 (6) \cdot 2 𝜋 𝑟 Δ 𝑑 ◂$

Image of the 5 strips with the top and bottom depth of each labeled. Text shows the pressure equation P(d) = 62.4 d. To the right of each strip is its representative pressure equaling the pressure at the bottom of the strip: P-subscript-1 = P(3) = 187.2, P-subscript-2 = P(6) = 374.4, and so forth.

We can then apply the same reasoning to write the overestimate for the force on each of the strips, and then sum those to find the total overestimated force. Since we're using the pressure at the bottom of each strip for our overestimate, we have $𝑃 1 = 𝑃 (3),$ $𝑃 2 = 𝑃 (6),$ and so forth: $𝐹 t o t, o v e r e s t i m a t e = 10 𝜋 [𝑃 1 \cdot Δ 𝑑 + 𝑃 2 \cdot Δ 𝑑 + 𝑃 3 \cdot Δ 𝑑 + 𝑃 4 \cdot Δ 𝑑 + 𝑃 5 \cdot Δ 𝑑] = 10 𝜋 [𝑃 (3) \cdot Δ 𝑑 + 𝑃 (6) \cdot Δ 𝑑 + 𝑃 (9) \cdot Δ 𝑑 + 𝑃 (12) \cdot Δ 𝑑 + 𝑃 (15) \cdot Δ 𝑑] = 10 𝜋 [187.2 \cdot 3 + 374.4 \cdot 3 + 561.6 \cdot 3 + 748.8 \cdot 3 + 936 \cdot 3] = 10 𝜋 [8424] = 264648 l b ✓$ That's a lot of force on the walls of this 10-foot wide and 15-foot tall water tower. No wonder they use those steel bands around them! Granted, this is an overestimate; we'll do the underestimate in a minute.

Check Question 2: Graphical Representation of the overestimate

Below is a graphical representation of the 5-term overestimate sum between the brackets [...] above (leaving out the factor of $10 𝜋$ in front): each rectangle represents one of the $𝑃 𝑖 \cdot Δ 𝑑$ terms.

Tap the rectangle that represents the term $𝑃 (15) \cdot 3 f t .$

As Check Question 2 highlights, since pressure is an increasing function of depth, for our overestimate here we computed a right sum, $𝑆 𝑅,$ using the pressure value at the right endpoint of each subinterval as shown in the pressure versus depth graph.

Let's pause to highlight that we now know the actual force on the tower's walls is less than our overestimate: $𝐹 t o t a l, a c t u a l < 𝑆 𝑅 ⏞ ¯¯ ⏞ ¯¯ ⏞ 264648 l b ◂$ Now that we have an upper bound on the true force on the tower's walls, let's compute an underestimate so we have a lower bound as well.

Underestimate

Let's turn now to compute the underestimate of the total force on the tower's walls. The calculation is the same as above, except now for the representative pressure on each strip we use the pressure at its top rather than at the bottom.

𝐹 t o t, u n d e r e s t i m a t e = 10 𝜋 [𝑃 1 \cdot Δ 𝑑 + 𝑃 2 \cdot Δ 𝑑 + 𝑃 3 \cdot Δ 𝑑 + 𝑃 4 \cdot Δ 𝑑 + 𝑃 5 \cdot Δ 𝑑] = 10 𝜋 [𝑃 (0) \cdot Δ 𝑑 + 𝑃 (3) \cdot Δ 𝑑 + 𝑃 (6) \cdot Δ 𝑑 + 𝑃 (9) \cdot Δ 𝑑 + 𝑃 (12) \cdot Δ 𝑑] = 10 𝜋 [0 \cdot 3 + 187.2 \cdot 3 + 374.4 \cdot 3 + 561.6 \cdot 3 + 748.8 \cdot 3] = 10 𝜋 [5616] = 176432 l b ✓

Even our underestimate is a lot of force on those tower walls.

Check Question 3: Graphical representation of underestimate

Below is a graphical representation of the 5-term overestimate sum between the brackets [...] above (again leaving out the factor of $10 𝜋$ in front): each rectangle represents one of the $𝑃 𝑖 \cdot Δ 𝑑$ terms.

Tap the rectangle that represents the term $𝑃 (6) \cdot 3 f t .$

Check Question 3 highlights that our underestimate is a left sum, $𝑆 𝐿,$ using the pressure at the leftmost point of each subinterval.

Summarizing our work so far, from our 5-term sums we know $𝑆 𝐿 ⏞ ¯¯ ⏞ ¯¯ ⏞ 176432 l b < 𝐹 t o t a l, a c t u a l < 𝑆 𝑅 ⏞ ¯¯ ⏞ ¯¯ ⏞ 264648 l b$ Whew: that was a lot of work to develop those approximations! But now that we have them, and understand the process by which we found them, we can move on to the next question in our Approximation Framework.

Part II: Focus on the error bounds

Recall the third Question for Approximation:

3. The difference between your current overestimate and underestimate values is your "error bound." What is its current value?

Before we work to improve our approximations, let's consider a way to visualize the error bounds graphically.

Thinking Question: Visual representation of the error bound

Which graph below depicts regions that represent the error bound for each subinterval? (More than one answer is correct; choose the one you prefer.)

Multiple choice answers. (a) Same graph as above representing the underestimate. (b) Same graph as above representing the overestimate. (c) Both graphs overlaid, with the extra difference between them shown in a lighter shading. (d) Small rectangles representing the difference between the two graphs.

Let's consider now our fourth and final Approximation Question:

4. How can the error bound be made smaller than any predetermined value?

You probably already have the answer in mind: we'll increase the number of terms, n, in our two sums. Before we do that, let's develop the sigma notation we need to represent these sums — much better than actually writing out more and more . . . and more . . . terms by hand.

Activity: Develop sigma notation for the overestimate

Drag the appropriate dark-shaded pieces below up to the blank rectangles to represent correctly the sum for the overestimate. Recall that since we're using $𝑛 = 5$ subintervals, $Δ 𝑑 = 3$ feet.Note: You won't use all of the pieces below.

Activity: Develop sigma notation for the underestimate

Drag the appropriate dark-shaded pieces below up to the blank rectangles to represent correctly the sum for the underestimate. Recall that since we're using $𝑛 = 5$ subintervals, $Δ 𝑑 = 3$ feet.Note: You won't use all of the pieces below.

Increasing n

Now that we've developed the sigma notation, let's put it to use. The graphs immediately below show the rectangles that represent the overestimate (on the left) and the underestimate (on the right) approximations. Initially there are $𝑛 = 5$ subintervals as we worked out above, but you can use the slider or text-entry box below the graphs to increase that number, up to $𝑛 = 1000 .$ Both the $Σ$ -sums and the graphs will update accordingly. (There are a lot of real-time calculations and graph updates going on behind the scenes, so expect a slight pause and some screen-jumping as you increase n.)

Below that you'll see a graph that represents the error bounds as we developed them in Choice D in the Question immediately above.

Please familiarize yourself with how these sums and graphs work; we'll continue with a question below.

Whichever view you use, notice that as you increase n, the small rectangles that represent the error become smaller and smaller. That's of course the point of increasing the number of terms: to decrease the size of the error on each subinterval. Make sure you zoom in on the error-bound graph so you can still view some of the rectangles clearly and thereby see how they're becoming smaller because the overestimate and underestimate for each subinterval become closer and closer to each other. We are essentially squeezing the true value between these two estimates.

Now, time to address the goal of this entire screen, which we can do readily using the calculations we've set up above. Note in particular that the equation immediately above shows the error bound for the current value of n, and so you can determine the minimum value of n necessary to produce the desired error bound of 500 pounds.

Question: Minimum number of subintervals to meet tolerance goal

Conclusion and Reflection

Rather than our usual "Upshot" here, we're going to ask you to revisit the Approximation Questions and answer them for yourself since they are the basis for our work here. You can reveal our answers if you'd like; some of them repeat what we wrote above.

Questions for Approximations

1. What are you approximating? Why can't you find the exact value immediately?

2. How will you generate an overestimate for the value we're after? An underestimate?

3. What is the difference between your current overestimate and underestimate values is your "error bound." What is its current value?

4. How can the error bound be made smaller than any predetermined value?

We hope you're starting to become more comfortable with how to set up a sum that represents an overestimate, and underestimate, of an accumulation of some quantity; in this case, that quantity happens to be the total force on a water tower's walls. On the next screen we'll examine one more scenario where we need to accumulate something that has a varying rate-of-change, this time focusing on an object's motion.

For now, what are your thoughts about the material on this screen? What questions do you have, or suggestions for improvement? (We know there's a lot going on here, and would like to make things clearer for future students so we could use your help!) Please head on over to the Forum and let us know.

A.3 Context 2: Total Force on a Water Tower's Walls

Information you need for this scenario

Pressure and Force relationship:

Surface area of a cylinder:

Pressure due to water as depth varies:

Part I: 5-term approximation

Questions for Approximations

Overestimate

Underestimate

Part II: Focus on the error bounds

Increasing n

Conclusion and Reflection

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