Volume of solid of revolution

In Calculus I, we teach two different techniques for finding the volume of a solid of revolution:

  • Disks (or washers), in which the cross-section is perpendicular to the axis of revolution, and
  • Cylindrical shells, in which the cross-section is parallel to the axis of revolution.

Both of these could be expressed as either an integral with respect to x or as an integral with respect to y, depending on the axis of revolution. I won’t go into a full treatment of the procedure here; this can be found in places like http://www.cliffsnotes.com/math/calculus/calculus/applications-of-the-definite-integral/volumes-of-solids-of-revolution or http://mathworld.wolfram.com/SolidofRevolution.html or http://en.wikipedia.org/wiki/Disk_integration or http://en.wikipedia.org/wiki/Shell_integration.

A natural question asked by students is, “If I have the choice, should I use disks or shells?” The correct answer, of course, is “Pick the method that gives you the easier integral to compute.” But that’s not a very satisfying answer for novice students who’ve just been exposed to integral calculus. So, over the years, I developed a standard reply to this query:

That’s an excellent question, and it’s one of the classic conundrums faced by mankind over the years.

Should I choose Coke… or Pepsi?

McDonald’s… or Burger King?

Ginger… or Mary Ann?

Disks… or shells?

The answer is, it just takes a little practice and experience to determine which technique gives you the easier integral.

If you don’t get the cultural reference, here’s a reminder. As of 10 years ago, I could still tell this joke to college students and still get smiles of acknowledgement. But, given the passage of time, I’m not sure if this same joke would fly college students now.

Engaging students: Volume and surface area of pyramids and cones

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Angel Pacheco. His topic, from Geometry: finding the volume and surface area of pyramids and cones.

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How has this topic appeared in high culture (art, classical music, theatre, etc.)?


Show an example of the pyramid of Giza, give them dimensions of the pyramid as well as the dimensions of the blocks that were used to build it and have the students guess how many blocks it took to build it. The students can use this as a competitive edge to want to get the correct answer. Students will have to solve for the surface area of the pyramid and the area of the face of the block. There can also be an example where I will tell the students if the pyramid was fill of blocks and they’re given the dimensions of the pyramid and block. They then find the volume of both to determine how many blocks can fill in the pyramid.


I will then show an image of a Greek amphitheater and explain how it resembles a cone. I will give them dimensions of a Greek amphitheater and have them find the surface area and the volume of cone if the amphitheater was folded into a cylinder.


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How can this topic be used in your students’ future courses in mathematics or science?

Students will be reintroduce to the volume of a cone in multivariable calculus when they learn about triple integrals and the different forms of integrals, like Cartesian, Polar, and Spherical coordinates. Surface Area and Volume of both the shapes will be seen in architectural engineering whenever they come across an assignment or job that requires them to find how big the cone or pyramid is in their draft of a monument or building.

This topic can also assist the students in their Geometry class in high school as well as college level. In mathematics, it’s better if there is a stronger foundation build in the early ages. When students face volume and surface area of pyramids and cones, they will gain more knowledge of the concept as time progresses. It’s always good to start early. Talking to students about different shapes and their areas and volumes gives them perspective in geometry.


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How have different cultures throughout time used this topic in their society?

In Ancient Greece, there were famous scientists that contained vast amount of knowledge. For example, Thales of Miletus and Democritus were some of the scientists that used surface area and volumes of cones and pyramids. Democritus was one of the first to observe that cones and square pyramids were one third of the volume of a cylinder and prism, respectively if they have similar measurements. I would use this as an engagement because Greek mythology is pretty popular. This could be used to show students that the math they are doing today is similar to the math that was done in the past, ancient past.

In Ancient Egypt, square pyramids were used to create the famous pyramids of Egypt such as the Pyramid of Giza. Pyramids were used to idolize their kings. The Mayan Indians also used pyramids to idolize their leaders. Bringing up different examples of different cultures that talk about the shapes they see in class then it can grab their attention. The link below is a lesson that talks about surface area and volume of cones and pyramids. It seems as an effective tool to assess students if they understand the concepts of SA and Volume.

Source: http://www.cordonline.net/cci_bridges_pdfs/Bridges12_12-5.pdf


Area of a triangle: Vertices (Part 7)

Suppose that the vertices of a triangle are (1,2), (2,5), and (3,1). What is the area of the triangle?

At first blush, this doesn’t fall under any of the categories of SSS, SAS, or ASA. And we certainly aren’t given a base b and a matching height h. The Pythagorean theorem could be used to determine the lengths of the three sides so that Heron’s formula could be used, but that would be extremely painful to do.

Fortunately, there’s another way to find the area of a triangle that directly uses the coordinates of the triangle. It turns out that the area of the triangle is equal to the absolute value of

\displaystyle \frac{1}{2} \left| \begin{array}{ccc} 1 & 2 & 1 \\ 2 & 5 & 1 \\ 3 & 1 & 1 \end{array} \right|

Notice that the first two columns contain the coordinates of the three vertices, while the third column is just padded with 1s. Calculating, we find that the area is

\left| \displaystyle \frac{1}{2} \left( 5 + 6 + 2 - 15 - 4 - 1 \right) \right| = \left| \displaystyle -\frac{7}{2} \right| = \displaystyle \frac{7}{2}

In other words, direct use of the vertices is, in this case, a lot easier than the standard SSS, SAS, or ASA formulas.

A (perhaps) surprising consequence of this formula is that the area of any triangle with integer coordinates must either be an integer or else a half-integer. We’ll see this again when we consider Pick’s theorem in tomorrow’s post.

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There is another way to solve this problem by considering the three vertices as points in \mathbb{R}^3. The vector from (1,2,0) to (2,5,0) is \langle 1,3,0 \rangle, while the vector from (1,2,0) to (3,1,0) is \langle 2,-1,0 \rangle. Therefore, the area of the triangle is one-half the length of the cross-product of these two vectors. Recall that the cross-product of the two vectors is

\langle 1,3,0 \rangle \times \langle 2,-1,0 \rangle = \left| \begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ 1 & 3 & 0 \\ 2 & -1 & 0 \end{array} \right|

\langle 1,3,0 \rangle \times \langle 2,-1,0 \rangle = -7{\bf k}

So the length of the cross-product is clearly 7, so that the area of the triangle is (again) \displaystyle \frac{7}{2}.

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The above technique works for any triangle in \mathbb{R}^3. For example, if we consider a triangle in three-dimensional space with corners at (1,2,3), (4,3,0), and (6,1,9), the area of the triangle may be found by “subtracting” the coordinates to find two vectors along the sides of the triangle and then finding the cross-product of those two vectors.

Furthermore, determinants may be used to find the volume of a tetrahedron in \mathbb{R}^3. Suppose that we now consider the tetrahedron with corners at (1,2,3), (4,3,0), (6,1,9), and (2,5,2). Let’s consider (1,2,3) as the “starting” point and subtract these coordinates from those of the other three points. We then get the three vectors

\langle 3,2,-3 \rangle, \langle 5,-1,6 \rangle, and \langle 1,3,-1 \rangle

One-third of the absolute value of the determinant of these three vectors will be the volume of the tetrahedron.

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This post has revolved around one central idea: a determinant represents an area or a volume. While this particular post has primarily concerned triangles and tetrahedra, I should also mention that determinants are similarly used (without the factors of 1/2 and 1/3) for finding the areas of parallelograms and the volumes of parallelepipeds.

This central idea is also the basis behind an important technique taught in multivariable calculus: integration in polar coordinates and in spherical coordinates.
In two dimensions, the formulas for conversion from polar to rectangular coordinates are

x = r \cos \theta and y = r \sin \theta

Therefore, using the Jacobian, the “infinitesimal area element” used for integrating is

dx dy = \left| \begin{array}{cc} \partial x/\partial r & \partial y/\partial r \\ \partial x/\partial \theta & \partial y/\partial \theta \end{array} \right| dr d\theta

dx dy = \left| \begin{array}{cc} \cos \theta & \sin \theta \\ -r \sin \theta & r \cos \theta \end{array} \right| dr d\theta

dx dy = (r \cos^2 \theta + r \sin^2 \theta) dr d\theta

dx dy = r dr d\theta

Similarly, using a 3 \times 3 determinant, the conversion dx dy dz = r^2 \sin \phi dr d\theta d\phi for spherical coordinates can be obtained.