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I'm interested in opinions based on experience about how to introduce volume for beginner calculus students. Below I present some observations and specific questions.

  1. In Stewart's book, the volume of a solid $S$ that lies between $x=a$ and $x=b$ is defined by $$V=\int_a^b A(x)\ dx,$$ where $A(x)$ is the area of the vertical cross-sectional area of $S$ through $x$. This definition comes from vertical slicing.

  2. In the examples of the said book, the following formula that comes from horizontal slicing for the volume of a solid $S$ that lies between $y=c$ and $y=d$ is also used: $$V=\int_c^d B(y)\ dy,$$ where $B(y)$ is the area of the horizontal cross-sectional area of $S$ through $y$.

  3. It seems the said book gives no explanation of the fact that both formulas give the same result when applied to the same solid.

Do you introduce volume for beginners calculus students as defined in Stewart's book?

  • If so, how do you deal with the fact that both formulas give the same number? Do you ignore this fact or assume that it is obvious? Why?
  • If not, what is your approach? Do you give a definition or just integral formulas? Why?
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    $\begingroup$ How does this differ from properly defining area? In a textbook such as Stewart, there will be an informal definition of area using triangles or rectangles. Then the working definition is by way of a definite integral for a function $y=f(x)$. If $f$ is monotonic with inverse $x=g(y)$ then we can write another definite integral for the same area in terms of $g(y) dy$. The fact that the two are the same is not formally justified. $\endgroup$
    – user52817
    Commented Aug 27, 2023 at 16:21
  • $\begingroup$ @user52817 In the case of the area, it is given a function $y=f(x)$ and we define "the a area between the graph of $f$ and the x-axis". The x-axis belongs to the defined object. In the case of the volume, it is given a solid $S$ and we define "the volume of $S$". The definition uses the x-axis, but (contrary to what happens with the area) the x-axis does not belongs to the defined object. So, the volume should not depend on the position of the axis. $\endgroup$
    – Pedro
    Commented Aug 27, 2023 at 22:10
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    $\begingroup$ Pedro: OK but this completely side steps the importance of Fubini's Theorem en.wikipedia.org/wiki/Fubini%27s_theorem See the short section on Riemann integrals. $\endgroup$
    – user52817
    Commented Aug 27, 2023 at 22:21
  • $\begingroup$ @Pedro All of these, length, area, volume involve definitions which can be based on covering the relevant set by line segments, squares, cubes (respectively) of decreasing size. However, it is important to remember that the intro calc sequence is not a real analysis class. $\endgroup$
    – Adam
    Commented Aug 27, 2023 at 23:40
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    $\begingroup$ If they already understand using calculus to calculate area, it should be straight-forward to explain that volume is similar, just adding a third dimension. $\endgroup$
    – Barmar
    Commented Aug 29, 2023 at 15:24

5 Answers 5

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It depends somewhat on the style of the course, but the majority of calculus students do not need a formal definition of volume or area, in my experience. They have studied geometry and (usually) physics, and they understand these notions on an intuitive level. That's enough, at this stage. (Of course, one can remind students about the concept by recalling some formulas for the volume of polyhedra, etc.)

My approach could be described as "we understand what volume is, and here are several ways to calculate it." All the various formulas that appear in the text are regarded as tools for calculating the volume, rather than definitions of volume.

Formal/logical questions like "what really IS volume?" are of interest primarily to students who specialize in mathematics, and they are by far in the minority out of students studying calculus.

But no matter the style of the course, the fact that different orders of integration lead to the same answer for the area/volume should certainly be emphasized, as this is one of the most interesting and useful aspects of multivariable integration.

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    $\begingroup$ I can't fault the "shut up and integrate" approach as the most practical in a lower-division course of non-mathematicians. That said, the fact that volume can be computed by slicing into generalized cylinders while surface area cannot should give some pause about how "obvious"/"intuitive" these concepts really are... $\endgroup$
    – user168715
    Commented Sep 1, 2023 at 16:44
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$dV$ represents a tiny bit of $V$.

$V = \int dV$ says that you can find the volume by adding up all the tiny bits of volume. This is why it is called an "integral;" you need to integrate all the little pieces together into a whole.

Now we just need a formula for $dV$.

Each formula for $dV$ comes from a way of creating the tiny bits of $V$.

There are lots of ways to slice a volume $V$ into tiny bits $dV$. We will learn two ways in this class:

  • You could slice it horizontally: $dV = B(y) dy$, so $V = \int dV = \int B(y) dy$, or
  • You could slice it vertically: $dV = A(x) dx$, so $V = \int dV = \int A(x) dx$.
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  • $\begingroup$ So, don't you give an explicit definition of volume? $\endgroup$
    – Pedro
    Commented Aug 27, 2023 at 19:05
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    $\begingroup$ @Pedro Wouldn't the students already know what volume is, more informally, from geometry and life in general? $\endgroup$ Commented Aug 28, 2023 at 13:15
  • $\begingroup$ @RadvylfPrograms Yes, but I think that at higher education level we should be able to prevent students from having the same feeling as Grothendieck, who said: What was least satisfying to me in our [high school] math books was the absence of any serious definition of the notion of length (of a curve), of area (of a surface), of volume (of a solid). Until read it carefully, I thought that the Stewart's definition was satisfactory. However, it seems it cannot be a definition, just a formula... $\endgroup$
    – Pedro
    Commented Aug 28, 2023 at 14:19
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    $\begingroup$ @Pedro I expect you will find very few students with similar feelings as Grothendieck. In the off chance you encounter one? I would simply be ready to recommend resources -- internet resources, books, other classes, whatever -- which might be more to their liking. $\endgroup$
    – academic
    Commented Aug 28, 2023 at 18:53
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    $\begingroup$ @Pedro I'm not sure if my philosophy of mathematics is more or less popular, but I'm pretty sure I'm not the only one who sees length, area, and volume as values that can be calculated from what are essentially arbitrary metrics defined on metric spaces - the mapping of those spaces to the real world notwithstanding. Put more simply: volume is a number and we've decided how we can calculate that number for a given real or abstract mathematical object. In this philosophy, the formula to calculate a volume is the "serious" definition of volume. $\endgroup$ Commented Aug 28, 2023 at 20:57
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How to properly define volume for beginner calculus students?

I'd say that the Apostol's approach, which I found after looking through several books, is the best. His definition of volume is axiomatic. Here is a simpler adaptation:

Definition: We define the volume of a three-dimensional solid $S$ to be a real number $v(S)$ that express the "amount of space occupied by $S$", having the following properties:

  • $v(S)\geq 0$ for all solid $S$.
  • $v(S_1\cup S_2)=v(S_1)+v(S_2)-v(S_1\cap S_2)$ for all solids $S_1$ and $S_2$.
  • Cavalieri's principle: If $S_1$ and $S_2$ are two solids such that $\operatorname{area}(S_1\cap \alpha)\leq \operatorname{area}(S_2\cap \alpha)$ for every plane $\alpha$ perpendicular to a given line, then $\operatorname{volume}(S_1)\leq \operatorname{volume}(S_2)$.
  • If $R$ is a rectangular parallelepiped with edges having lengths $a$, $b$ and $c$, then $v(R)=abc$.

The key point is the Cavalieri's principle, which allows us to relate areas to volumes without double integrals.

In this context, it is proved that $$v(S)=\int_a^b A(u)\,du,$$ for any $u$-axis, where $A$ is the cross-sectional area for the $u$-axis.

The generic $u$-axis can be replaced by the usual $x$-axis or $y$-axis. In other words: horizontal as well as vertical slicing are allowed, that is, the usual formulas $$V(S)=\int_a^b A(x)\,dx\quad\text{and}\quad V(S)=\int_c^d B(y)\,dy$$ are valid.

Apostol's approach uses step functions. However, under suitable hypothesis, it's possible to adapt the argument for the context of Stewart's definition of integral. See the proof here.

Remark: An axiomatic definition is not mandatory. Given that Cavalieri's principle is assumed to be true, the said proof (of the independence of the axes) validates Stewart's definition. Maybe this is the simplest possible approach.

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    $\begingroup$ To be fair, Stewart actually defines volume in this fashion, using cross sectional areas by parallel planes. Also, Cavlieri is a special case of Fubini. And this approach simply reduces the original question of "properly" defining volume to that of "properly" defining area. $\endgroup$
    – user52817
    Commented Aug 29, 2023 at 22:43
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    $\begingroup$ @user52817 In Stewart, the integral is the definition. In Apostol, the integral is a theorem and the definition is axiomatic. They are different approaches. The case of the area can be properly dealt analogously, by means of an axiomatic definition (unfortunately, a definition of area by means of the integral does not work). $\endgroup$
    – Pedro
    Commented Aug 29, 2023 at 22:52
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    $\begingroup$ The Apostle definition assumes we have a definition of what we mean by a three-dimensional solid. There is something circular here. The Apostle definition is similar to the definition of a measure $\mu$ on a $\sigma$-algebra $\Sigma$. en.wikipedia.org/wiki/Measure_(mathematics) $\endgroup$
    – user52817
    Commented Aug 29, 2023 at 23:52
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    $\begingroup$ The danger with an axiomatic approach like this is that it immediately raises the question of consistency of the postulated properties and, if I remember it right, the list you provided is inconsistent without restricting the class of sets unless you believe that AC is false. For beginner's calculus, I would say that Jordan's definition of volume on the class of sets with negligible boundaries is about the right idea (on par with Riemann integration) but I find even that quite hard to comprehend for the "generation Z" students... $\endgroup$
    – fedja
    Commented Aug 30, 2023 at 22:51
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In a third-semester calculus course one typically defines double-integration over a suitable family of plane regions, by approximating the region by small rectangles as size tends to zero. In such a setting, one proves Fubini's theorem, to the effect that the double-integration equals the iterated integral (again over suitable domains). If you are only interested in a rotation of $\frac{\pi}2$, then such a double application of Fubini is sufficient to establish that the integral (in particular, the area) will be the same.

If you want more general rotations, probably one would have to do a bit of more advanced differential geometry and introduce at least 1-forms and 2-forms, so one can speak of the area form and its invariance under rotations. But possibly one can take a shortcut via polar coordinates and show that the expression remains the same under integration.

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As a former student, I would prefer an actual example of calculating a solid volume.
My favorite one is calculating the formula for the volume of pyramid, say hexagonal one. You can do that by horizontally slicing the shape into infinitely many flat sections and integrating over that. Just use function for the area of each slice, dependent on the value of h, the height.
It has the added benefit of explaining where these weird 1/3 in the formula came from.
You can also show that number of walls does not affect the volume in any way - a three sided pyramid has the same formula as twenty sided one, and in fact the same as a cone.

You can end the example by stating they no longer have to memorize the formula for the volume, since they can derive it themselves.

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