My multivariate calculus students are not entirely happy with me. I am not entirely happy with them. The results of this week's exam on multiple integrals were—shall we say?—a trifle disappointing. Perhaps you will not be surprised to learn that I had great expectations for this exam. I had fussed over its preparation at length and I had been rather pleased with the way it progressed through the various coordinate systems (rectangular, polar, cylindrical, and spherical). I knew that some students would nevertheless insist on using suboptimal choices of coordinate system for some problems (and they did), but I keenly anticipated a generally good outcome.
That's what I get for trying to be helpful. My students swore they would have done better on the exam if I had not “tricked” them. They advanced as Exhibit A the “misleading” figure that had accompanied one of the exercises. During the hours of grading, I had marveled over what my students did to that particular problem. Let's work through their complaint and see if we can figure out what's at the bottom of it.
The exam problem in question required a given function to be integrated over a very specific three-dimensional region. Here is how I described it:
Let Q be the cone bounded by z = r and z = 2.Okay, it's a cone with the pointy end at the origin and its axis coincides with the z axis. In the margin of the exercise, I included a generic figure of a cone in the correct orientation. It was not to scale and I left it to my students how best to label it. As shown, the figure was labeled generically: a for the radius of the circular base (is it okay to call it a base if it's on top instead of on the bottom?), h for the height, and (just for the wild ones who want to use spherical coordinates) the semi-vertex angle is labeled α.
To my surprise, most of the answers worked out by my students contained a and h in them. For some reason, they did not realize that they knew (or should know) the dimensions of the cone Q. A few had grasped that z = 2 is a horizontal plane and thus the height of the cone must be h = 2. Even these people, however, couldn't figure out the value of a despite the helpful cone formula z = r. What was going on?
This is what you taught us
A few students looked at the graph of the cone and promptly wrote down a description of the region in terms of cylindrical coordinates (good choice!):
0 ≤ θ ≤ 2π
0 ≤ z ≤ h
0 ≤ r ≤ az/h
That was, of course, fine as far as it went, but I had given them specific equations. Could we perhaps please use those? No! Never! We have a picture to look at and we will use it, damn it! Don't try to stop us!
As I questioned them, I began to understand a bit more about what had occurred. They had (several of them) memorized the cylindrical coordinate definition of a cone, but it was just memorized, not understood. I've seen this many times before, but more typically among algebra students than in the ranks of a multivariate calculus class. Perhaps the recent spring break had caused them to lose their momentum and fall back on computational gimmicks in lieu of actual understanding. But perhaps I should have seen it coming. Students had asked me several time for “formulas” for converting the limits of integration of multiple integrals from one coordinate system to another. This was a bad sign I had insufficiently appreciated. In our discussion of the exam problems, I tried to get them to go back to basic principles.
“I'm afraid that you're hoping for a step-by-step computational procedure for transforming integrals. You can do that to some degree with integrands, where it's mostly a matter of substituting for one set of variables in terms of another, but don't expect that to work for the limits of integration. It's much, much better to rework the description of the region of integration. Go back to basics when you need to. Remember similar triangles?” On the board I sketched a triangular cross-section of a typical cone.
“Remember how we got the formula for a cone in cylindrical coordinates in the first place? z tells you how high you are and r tells you how far you are from the z axis. If you know the cone's radius is a and its height is h, then r and z have to be in the same proportion.”
r/z = a/h
When we solve for z, of course, we get z = hr/a. Since I had given them z = r in the statement of the problem, it should be pretty clear that a/h = 1, so a = h. What's more, since z = 2 provides the height of the cone, as previously noted, we must have a = h = 2. The value α = π/4 falls out immediately from these results, in case anyone cared to look at the isosceles right triangle.
They were right
In a way, my students were right. I have dutifully provided them with the basic tools to describe a region like a cone in cylindrical coordinates, but I had not worked with them sufficiently to instill an understanding of the patterns behind the formulas. When they saw the unidentified a and h in the illustration for the exam problem, they instantly fixated on the formal description that I had previously helped them to master (or at least memorize). Of course, it was their responsibility to work with the tools in their hands until they became adept with them (that's why I give them assignments to do at home and quizzes to take in class), but I could have drawn their attention more strongly to the process by which we built the descriptions and emphasized the flexible ideas that drove the process rather than the formulas on which it was based.
As I said at the beginning, we're unhappy with each other. They're on notice that I expect more from them than ritual formula application. And I'm on notice that once again the ideas in my head don't magically transfer into theirs. The follow-up quiz will probably show that the exam problem was a learning experience for most of my students—at least as far as cones are concerned. If I want more trouble, I could change the region to a sector of a sphere.
But that's a different problem and a different story.