How to properly define volume for beginner calculus students?

Question

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?

Answer 1

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.

Answer 2

$dV$ 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 $V = \int dV = \int A(x) dx$.

Answer 3

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.

Answer 4

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.

Answer 5

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.