# …but what does it look like?

It will soon be time to teach about finding the volumes of solid figures using integration techniques. Here is a list of links to posts that will help your students what these figures look like and how they are generated.

Visualizing Solid Figures 1 Here are ideas for making physical models of solid figures. These make good projects for students.

A Little Calculus is an iPad app that does an excellent job in helping students visualize many of the concepts of the calculus. Volumes with regular cross section, disk method, washer method, cylindrical shells are all illustrated.

The first illustrations show square cross sections on a semicircular base. The base is in the lower part and the solid in the upper. By using the plus and minus button (lower right) you can increase or decrease the number of sections in real time and see the figures change. The upper figure may be rotated by moving your finger on the screen.

The illustration below shows a washer situation.

The following older posts show how to use Winplot to generate and explore solid figures. Unfortunately, Winplot seems to have gone out of favor. I’m not sure why; it is one of the best. I still use it and like it. You may download Winplot here for free (PC only).

Visualizing Solid Figures 2 This post demonstrates how to use Winplot to generate solids with regular cross sections and solids of rotation.

Visualizing Solid Figures 3 The washer method is illustrated using Winplot. These post all relate to finding volumes by washers: Subtract the Hole from the Whole and Does Simplifying Make Things Simpler?

Visualizing Solid Figures 4 Using Winplot to see the method cylindrical shells. Note that this method is not tested on either the AB or BC Calculus exams, so you do not have to teach it. Many teachers present this topic after the exams are given. As a footnote you may also find Why You Never Need Cylindrical Shells interesting. (However, this is not the reason it is not tested on the AP Calculus exams.)

Visualizing Solid Figures 5 An exercise demonstrating how “half” can mean different things and shows that how the figures are generated makes a difference.

# Type 4 Questions: Area and Volume Problems

Given equations that define a region in the plane students are asked to find its area, the volume of the solid formed when the region is revolved around a line, and/or the region is used as a base of a solid with regular cross-sections. This standard application of the integral has appeared every year since year one (1969) on the AB exam and almost every year on the BC exam. You can be pretty sure that if a free-response question on areas and volumes does not appear, the topic will be tested on the multiple-choice section.

What students should be able to do:

• Find the intersection(s) of the graphs and use them as limits of integration (calculator equation solving). Write the equation followed by the solution; showing work is not required. Usually no credit is earned until the solution is used in context (as a limit of integration). Students should know how to store and recall these values to save time and avoid copy errors.
• Find the area of the region between the graph and the x-axis or between two graphs.
• Find the volume when the region is revolved around a line, not necessarily an axis or an edge of the region, by the disk/washer method.
• The cylindrical shell method will never be necessary for a question on the AP exams, but is eligible for full credit if properly used.
• Find the volume of a solid with regular cross-sections whose base is the region between the curves. For an interesting variation on this idea see 2009 AB 4(b)
• Find the equation of a vertical line that divides the region in half (area or volume). This involves setting up an integral equation where the limit is the variable for which the equation is solved.
• For BC only – find the area of a region bounded by polar curves: $A=\tfrac{1}{2}\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{\left( r\left( \theta \right) \right)}^{2}}}d\theta$
• For BC only – Find perimeter using arc length integral

If this question appears on the calculator active section, it is expected that the definite integrals will be evaluated on a calculator. Students should write the definite integral with limits on their paper and put its value after it. It is not required to give the antiderivative and if a student gives an incorrect antiderivative they will lose credit even if the final answer is (somehow) correct.

There is a calculator program available that will give the set-up and not just the answer so recently this question has been on the no calculator allowed section. (The good news is that in this case the integrals will be easy or they will be set-up-but-do-not-integrate questions.)

Occasionally, other type questions have been included as a part of this question. See 2016 AB5/BC5 which included an average value question and a related rate question along with finding the volume.

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see January 911, 2013 and “Subtract the Hole from the Whole” of December 6, 2016

Free-response questions:

• 2014 AB 2, 2013 AB 5.
• 2015 AB 2
• Variations: 2009 AB 4,
• 2016 AB5/BC5,
• 2017 AB 1 (using a table),
• Perimeter 2011 BC 3 and 2014 BC 5

Multiple-choice questions from non-secure exams:

• 2008 AB 83 (Use absolute value),
• 2012 AB 10, 92
• 2012 BC 87, 92 (Polar area)

Revised to add perimeter question 3-16-18,

Revised March 12, 2021

# Subtract the Hole from the Whole.

Sometimes I think textbooks are too rigorous. Behind every Riemann sum is a definite integral. So, authors routinely show how to solve an application of integration problem by developing the method starting from the Riemann sum and proceeding to an integral that give the result that is summarized in a “formula.” There is nothing wrong with that except that often the formula is all the students remember and are lost when faced with a similar situation that the formula does not handle. .

The volume of solid figure problems are developed from the idea that if a solid figure has a regular cross-section (that is, when cut perpendicular to a line, each face is similar – in the technical sense – to all the others). They are all squares, or equilateral triangles, or whatever. The last shape considered is usually a “washer”, that is, an annulus or two concentric circles. This is formed by revolving the region between two curves around a line. Authors develop a formula for such volumes: $\displaystyle \pi \int_{a}^{b}{{{\left( R\left( x \right) \right)}^{2}}-{{\left( r\left( x \right) \right)}^{2}}dx}$.

Now there is nothing wrong with that, but I like to give the students their chance to show off. They can usually figure out the answer without Riemann sums. Here is my suggestion. After students have had some practice with circular cross-sections (“Disk” method”) I give them a series of three volumes to find.

Example 1: The curve $f\left( x \right)=\sin \left( \pi x \right)$ on the interval [0, ½] is revolved around the x-axis to form a solid figure. Find the volume of this figure.

Solution: $\displaystyle V=\int_{0}^{1}{\pi {{\left( \sin \left( \pi x \right) \right)}^{2}}dx}=\frac{\pi }{4}$

Example 2: The curve $g\left( x \right)=8{{x}^{3}}$ on the interval [0, ½] is revolved around the x-axis to form a solid figure. Find the volume of this figure.

Solution: $\displaystyle V=\int_{0}^{1/2}{\pi {{\left( 8{{x}^{3}} \right)}^{2}}dx=}\frac{\pi }{14}$

These they find easy. Then, leaving the first two examples in plain view, I give them:

Example 3: The region in the first quadrant between the graphs of $f\left( x \right)=\sin \left( \pi x \right)$ and $g\left( x \right)=8{{x}^{3}}$ is revolved around the x-axis. Find the volume of the resulting figure.

A little thinking and (rarely) a hint and they have it. $\displaystyle V=\frac{\pi }{4}-\frac{\pi }{14}$

What did they do? Easy, they subtracted the hole from the whole. We discuss this and why they think it is correct. We try one or two others. And now they are set to do any “washer” method problem without another formula to memorize.

Extensions:

1. In symbols, when rotation around a horizontal line, if R(x) is the distance from the curve farthest from the line of rotation and r(x) the distance from the closer curve to the line of rotation the result can be summarized in the formula

$\displaystyle V = \int_{a}^{b}{\pi {{\left( R\left( x \right) \right)}^{2}}dx}-\int_{a}^{b}{\pi {{\left( r\left( x \right) \right)}^{2}}dx}$.

Notice, that I like to keep the $\pi$ inside the integral sign so that each integrand looks like the formula for the area of a circle. What the students need to know is to subtract the volume hole from the outside volume. With that                idea and the disk method they can do any volume by washers problem.

2. You should show the students how this equation above can be rearranged into the formula in their books,

$\displaystyle V = \pi \int_{a}^{b}{{{\left( R\left( x \right) \right)}^{2}}-{{\left( r\left( x \right) \right)}^{2}}dx}$.

This is so that they understand that the formulas are the same, and not think you’ve forgotten to tell them something important. It is also a good exercise in working with the notation. (see MPAC 5 – Notational fluency)

3. Next discuss what ${{\left( \pi R\left( x \right) \right)}^{2}}-\pi {{\left( r\left( x \right) \right)}^{2}}$ is the area of and how it relates to this problem. See if the students can understand what the textbook is doing; what shape the book is using.. Discuss the Riemann sum approach. (MPAC 1 Reasoning with definitions and theorems, and MPAC 5 Notational fluency)

4. With the idea of subtracting the “hole” try a problem like this. Example 4: The region in the first quadrant between x-axis and the graphs of $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=\sqrt{2x-4}$ is revolved around the x-axis. Find the volume of the resulting figure.

Solution:$\displaystyle V=\int_{0}^{4}{\pi {{\left( \sqrt{x} \right)}^{2}}dx}-\int_{2}^{4}{\pi {{\left( \sqrt{2x-4} \right)}^{2}}dx}=4\pi$

(Notice the limits of integration.)

Traditionally, this is done by the method of cylindrical shells, but you don’t need that. You could divide the region into two parts with a vertical line at x = 2 and use disks on the left and washers on the right, but you don’t need to do that either. Just subtract the hole from the whole.

.

# Visualizing Solid Figures 5

To end this series of posts on visualizing solid figures, we will look at a problem that relates to how volumes of solid figures are formed. It has 5 parts which will be presented first. Then the solution will be given.

Consider the region in the first quadrant bounded by the graphs of the parabola $y={{x}^{2}}$ and the line $y=9$ both for $0\le x\le 3$.

This region is revolved around the y-axis to form a solid figure.

1. Use the disk/washer method to find the volume of this figure.
2. Use the method of cylindrical shells to find the volume of this figure.
3. Use the disk/washer method to find a number j , such that when x = j the volume of the figure is one-half that of the original figure.
4. Use the method of cylindrical shells to find a number k , such that when x = k the volume of the figure is one-half that of the original figure.
5. The answers to parts a) and b) should be the same, but the answers to parts c) and d) are different. Explain why they are different.

Solutions:

1. The volume by the disk/washer method is

$\displaystyle V=\int_{0}^{9}{\pi {{x}^{2}}dy}=\int_{0}^{9}{\pi ydy}=\frac{81}{2}\pi \approx 127.235$

2. The volume by the method of cylindrical shells is

$\displaystyle \int_{0}^{3}{2\pi x\left( 9-y \right)dx}=\int_{0}^{3}{2\pi x\left( 9-{{x}^{2}} \right)dx}=\frac{81}{2}\pi \approx 127.235$

3. The value is be found by solving the equation for j:

$\displaystyle \int_{0}^{{{j}^{2}}}{\pi y\,dy}=\frac{81}{4}\pi$, so $j\approx 2.52269$

4. The value is be found by solving the equation for k:

$\displaystyle \int_{0}^{k}{2\pi x\left( 9-{{x}^{2}} \right)dx}=\frac{81}{4}\pi$ and $k\approx 1.62359$

5. The reason the values are not the same is this. Think of the revolved parabola as a bowl. If you pour water into the bowl until it is half full, the bowl looks like the figure below. The water is pooled in the bottom of the bowl as you would expect. This is what happens when using the disk/washer method. The washers stack up starting in the bottom of the bowl until it is half full.

On the other hand, the method of cylindrical shells sort of wraps the water in layers (the shells) around the y-axis. Picture the water being sprayed on the y-axes and frozen there. Each new layer (shell) increases the amount and you end up half of the total volume arranged as a cylinder with a rounded (paraboloid shaped) bottom as shown in the figure below. Both bowls contain the same amount of water, arranged differently.

# Visualizing Solid Figures 3

Volume by “Washers”

Today I will show you how to visualize not just the solid figures but the disks and washers used in computing the volume using Winplot. The next post will show how to draw shells.

Winplot is a free program. Click here for Winplot and here for Winplot for Macs.

For the example we’ll use the situations from the 2006 AP calculus exams question AB1 / BC 1. The students were given the region between the graphs of y = ln(x)  and yx -2. In the first part they were asked to find the area of the region. To do that they first had to determine, using their calculator, where the curves intersect. The x-coordinates of the intersections  are x = 0.15859 and x = 3.14619.

In part (b) they were asked to find the volume of the solid formed when the region was rotated around the horizontal line y = -3 . The volume is found by using the disk/washer method. Here is how to show the washers using Winplot. This gets a little complicated so I will mark each step with a bullet

• Starting in the 2D window, graph the two functions as shown in the previous post.. When entering the equations click the “lock interval” box and enter 0.159 for “low x” and 3.146 for “high x.”
• Next we will enter a Riemann sum rectangle which we will be able to move, and, once rotated, will appear as the washer. Go to Equa > Segment > (x,y) and in the box enter the endpoints of the vertical segment between the two graphs in terms of B: x1 = B, y1 = ln(B), x2 = B, and y2 = B – 2. Click “ok.”
• Go to the Anim button, choose “B” (Anim > Individual  > B).
• Enter the left value 0.15859 and click “set L”, and enter 3.14619 and click “set R.” (Remember how to do this, as we will do it again.)
• You may now move the “Riemann rectangle” (which, of course, is very thin, approaching 0) across the region.

Next we will produce the 3D images.

• As we did in the last post click on One > Revolve surface… Enter the values shown below. (The “arc start” and “arc stop” value are the x-values of the intersection points. Attach an “@S” to the “angle stop” as shown.)

• Click “see surface.”
• In the 3D window that appears click Anim > S and you will be able to revolve the curve. Make the “set L” value -2pi by typing the value in the box and clicking “set L,” leave “set R” at 2pi. Adjust the value to 0 by typing 0 and “enter.”
• Adjust the viewing widow with the 4 arrow keys and the Page Up and Page down keys. Add the axes with Ctrl+A.
• Return to the “surface of revolution” window and choose the second function from the drop-down box at the top. not change anything else. Click “see surface” and the second curve will be added to the graph.

Next we graph the “washer:”

• In the surface of revolution box, select the segment in the drop-down box at the top change the “angle stop” to 2pi@R. Click “ok.”
• Then in the 3D Inventory window for this file select the segment and click “edit.”
• Change the “low t” value to 0 and the “high t” value to 1. Change the “u hi” to 2pi@R. Click “ok.” The window should look like the one below.

Finally, in the 3D window:

• To show the line y = –3, in the 3D window go to Equa > 2. Parametric and enter the values shown in the box below and click “ok.” A short segment at y = -3 will appear in the 3D window.
• In the 3D window go to Anim > Individual and open a slider for “B” and for “R.”
• For the “B” slider make “set L” = 0.15859 and the “set R” to 3.14619 (the intersection values).
• For the “R” slide make “set L” to 0 and “set R” to 2pi.
• Adjust the R and S sliders to 0 and the B slider to its minimum value.
• Save everything just to be safe. The extension will be “.wp3.” Later you can open this file from the 3D window, but it will no longer be in touch with the 2D window even if you save that.

That should do it.

Move first the B slider, then the R slider, then the B slider again and finally the S slider to explore the situation.

In the video at the top you will see this example with these things happening in order.

• The Riemann rectangle moving in the plane using the B slider
• The Riemann rectangle rotated into a washer using the R slider.
• The washer moving through the curves using the B slider again.
• The two curves rotated part way using the S slider
• The washer moving through the solid using the B slider.
• The solid rotated with the 4 arrow keys.

The next post will show how to do a similar animation for the cylindrical shell method.

I posted a challenge question on March 28, 2013. Only one person, “April,” replied with an explanation and it was a very good and succinct answer. You may read it here.

The conundrum resulted from trying to calculate when a parabolic bowl was half full. Using the “disk” method there was one answer; using the “shell” method resulted in a different answer. Both are correct; how can that be?

The explanation is this:

When you half fill the bowl in the usual way, by pouring water into it, the water accumulates in the bottom until the bowl is half full, as shown below. This is the result you get using the “disk” method.

But filling the bowl by “shells” is different. Think of spraying the y-axis with paint.  When the paint dries, spray another layer, and then another. Each layer is a “shell.” The paint accumulates as “shells” around the y-axis until the bowl is half full. The paint forms a cylinder with a rounded (parabolic) bottom that fills half the bowl, as shown below.

Geometry Counts

# Challenge

A student was asked to find the volume of the bowl-shaped figure generated when the curve y = x2 from x = 0 to x = 2 is revolved around the y-axis. She used the disk method and found the volume to be $\displaystyle \int_{0}^{4}{\pi ydy}=8\pi$. To check her work she used the method of cylindrical shells and found the same answer: $\displaystyle\int_{0}^{2}{2\pi x\left( 4-{{x}^{2}} \right)dx}=8\pi$.

The second part of the question asked for the value of x for which the bowl would be half full. So she first solved the equation $\displaystyle \int_{0}^{k}{\pi ydy=\frac{1}{2}\cdot 8\pi }$  and found $k=2\sqrt{2}$. This is a y-value so the corresponding x-value is $\sqrt{2\sqrt{2}}\approx 1.682$. She again checked her work by shells by solving $\displaystyle \int_{0}^{k}{2\pi x\left( 4-{{x}^{2}} \right)dx=\frac{1}{2}\cdot 8\pi }$ and found that $k=\sqrt{-2\left( \sqrt{2}-2 \right)}\approx 1.082$. (This is the x-value.)

Both computations are correct. Can you explain to her why her answers are different? Use the comment box below to share your explanations. I will post mine in a week or so.