## My Brown Big Spiders

Professor: You have to learn to be able to play it blindfolded. The page, for God’s sake! The notes!

David: I’m sorry I was, uh, forgetting them, Professor.

Professor: Would it be asking too much to learn them first?

David: And-And then forget them?

Professor: Precisely.

from the movie Shine

If I want to find the volume and surface area of a sphere, I do it with calculus:

$V = \int_{r = 0}^R\int^{2\pi}_{\phi = 0}\int_{\theta = 0}^\pi r^2\sin\theta \textrm{d}\theta \textrm{d}\phi \textrm{d}r = \frac{4}{3}\pi R^3$

.

$S = \int_{\theta = 0}^\pi\int_{\phi = 0}^{2\pi} R^2 \sin\theta\textrm{d}\theta\textrm{d}\phi = 4\pi R^2$

This is correct, but I can’t use it with high school geometry students because they don’t know what an integral is, much less a Jacobian.

However, Archimedes came up with a beautiful way of discovering the volume and surface area of a sphere. He did it by relating the sphere to a known shape – a cylinder with a cone cut out of it.

He drew a picture like this:

On the left there’s a hemisphere with radius $R$. On the right, there’s a cylinder with radius and height both also $R$, so that the hemisphere would fit perfectly inside the cylinder. The cylinder has had a cone cut out from the top down tapering down to the center of the bottom. First, we’ll show that these two shapes have the same volume.

We imagine slicing the hemisphere horizontally at some certain height $h$. This would reveal a circle as seen in the picture. Call its radius $r$.

At the same height, we also slice the cylinder, leaving us with a disk. We’ll find the areas of this circle and disk.

The area of the circle is $\pi r^2$, which by the Pythagorean theorem is also $\pi (R^2 - h^2)$.

Looking at the cylinder, the outer edge of the disk has radius $R$ and the inner edge has radius $h$, so the area of the disk is also $\pi (R^2 - h^2)$.

Because every horizontal slice of the hemisphere has the same area as the corresponding horizontal slice of the drilled-out cylinder, they must have the same volume. The volume of the cylinder is its original volume minus the volume of the cone, or $\pi R^3 - 1/3 \pi R^3 = 2/3 \pi R^3$. Hence, the volume of a full sphere is

$V = 4/3 \pi R^3$

Next, we’ll show that the hemisphere has the same surface area as the outside of the cylinder (the cone is now unimportant).

Take a slice of the outside of the cylinder at height $h$ and of thickness $\textrm{d}h$. This forms a band around the cylinder whose area is

$\textrm{d}S = 2 \pi R \textrm{d}h$

Now slice the sphere at the same height with the same $\textrm{d}h$. This also forms a band. The band is a shorter distance around, but due to the slant of the edge of the circle, it’s also thicker. Let’s call the thickness of this band $\textrm{d}x$.

The area of the band around the hemisphere is the circumference at height $h$ multiplied by the thickness $\textrm{d}x$.

$\textrm{d}S = 2\pi\sqrt{R^2 - h^2}\textrm{d}x$

If we draw a tangent line on the sphere, it’s perpendicular to the radius. This gives us similar triangles.

So

$\frac{\textrm{d}x}{\textrm{d}h} = \frac{R}{\sqrt{R^2 - h^2}}$

Plugging back into the previous expression,

$\textrm{d}S = 2\pi\sqrt{R^2 - h^2}*\textrm{d}h * \frac{R}{\sqrt{R^2 - h^2}} = 2\pi R \textrm{d}h$

So the band around the outside of the cylinder and sphere have the same surface area, so the entire shapes have the same surface area. That makes the surface area of a sphere

$S = 4 \pi R^2$

This is a really lovely argument. The problem is pretty hard, but the solution is simple. (I’m not sure if this is quite how Archimedes did it. To be honest I never even met the guy. I learned the idea from this animation).

I was reviewing solid geometry with a high school junior the other day, so I showed her this argument (but only the volume part). I was proud of myself for offering this little example of how interesting mathematical ideas can be. At least, I was as we began.

“It’s all so complicated!” she moaned a few minutes later when I asked her to identify a certain quantity in our sketch.

Complicated? I had thought the argument was remarkably simple – just draw a sphere and a cylinder next to each other and you’re practically done. What could be simpler? Somehow my student was getting entangled in brambles I couldn’t even see.

I did not draw quite the same picture for her that I drew earlier in this post. I didn’t want to give it all away, so I drew something more like this and asked for $r$:

Finding $r$ is a simple application of something she knew well – the Pythagorean theorem. She didn’t see it, though, so I showed her this right triangle:

But then she didn’t see how long the new line I just drew was. It’s just $R$ because it’s a radius of the sphere, but although she knew that all radii of a sphere have the same length, she couldn’t easily identify the two lines as radii and call up the relevant information. So I showed her that step, too.

After a bit more prodding, she wrote down $r = \sqrt{R^2 + h^2}$, a mistake that comes from applying the Pythagorean theorem incorrectly. She knows better, and should have found $r^2 = R^2 - h^2$, but by this point she was already flustered from her earlier mistakes, confused about what we were trying to do, self-conscious, and generally unable to approach the problem equanimously.

When she realized she had applied the Pythagorean theorem wrong, her frustration mounted, and moments later, at my next question, I was shocked with, “It’s all so complicated!”

Why did this happen? Why did I so horribly misjudge the difficulty of the exercise?

The other day I read this comment on an essay on teaching

I used to teach English as a second language. It was a mind trip.

I remember one of my students saying something like “I saw a brown big spider”. I responded “No, it should be ‘big brown spider'”. He asked why. Not only did I not know the rule involved, I had never even imagined that anyone would ever say it the other way until that moment.

Tutoring has been exposing my own brown big spiders – the little steps and bits of knowledge that I take for granted – for years. I’ve rarely stopped to notice it.

Just to follow each step in the Archimedes argument, you must make an enormous number of mathematical connections behind the scenes in your mind. Here’s a partial list:

• A “sphere” is a round three-dimensional object like, a ball
• Every point on the surface of a sphere is the same distance from the center
• The “surface” of the sphere means its outside edge, or skin
• A “point” is a little dot with no size at all. It simply marks a place.
• You can represent three-dimensional figures in two dimensions with certain types of drawing.
• The point of doing this drawing is to make things easier to visualize.
• A “hemisphere” is half a sphere – the top half in this case
• A “cylinder” is basically a tube with constant width.
• The center of the bottom of the hemisphere is the same point as the center of the sphere it came from.
• The height of the hemisphere is the same as the distance from the center to the edge horizontally.
• This means that the cylinder drawn is twice as wide as it is tall.
• The volume of a cone is one third the area of its base times its height.
• The volume of a cylinder is its base times its height
• The area of a circle is $\pi$ times the square of its radius

And so on. I only stopped writing so that I’d eventually finish the rest of this post. Each item I added to that list sparked off several new ones I hadn’t considered.

Try writing your own list and you’ll quickly be overwhelmed by the exponentially-proliferating leaves on your conceptual tree. We didn’t even get close to things like the Cavalieri’s principle.

The items on my brown big spider list are not supposed to be mathematical facts so much as cognitive patterns the reader is required to have. For example, mathematically a point is not, “a little dot with no size at all,” as I called it. It’s a primitive notion and has no definition. The list still calls a point a dot, though, because the mathematically-accurate description isn’t helpful to a student, and isn’t they way most people think of it even when they’ve already learned geometry well.

When I started writing the list, I found myself wanting to say, “A sphere is a set of all points equidistant…”, but that’s no good. It uses the significant brown big spiders of “set” and “equidistant”, as well as the general idea of giving mathematical definitions, something most high schoolers don’t yet understand well. Then I wanted to say, “A sphere is a shape that’s symmetric with respect to rotations about any axis…” but this has all the same problems.

Ultimately, I chose “a sphere is a ball.” It’s imprecise, but it’s the way you think about a sphere before you’ve packaged the concept away so tightly you don’t need to think about it any more. Anyone who tells you a sphere is the two-dimensional manifold $S^2$ is someone who has forgotten how much they actually know about spheres. They’ve forgotten it in the good way, of course – the way David was supposed to forget the notes to Rachmaninoff. Unfortunately, I experience a crippling side effect when I forget things this way. I forget that other people haven’t yet forgotten them.

This forgetting is the psychological phenomenon of “chunking“. The most famous example involves chess players. Give expert chess players a position from a game between grandmasters and they can easily memorize the positions of thirty pieces. Give them pieces strewn randomly about the board and they’ll remember just a few – no more, in fact, than your average Joe who knows little more about chess than what the real name of the horsey is.

A position from a real game has lots of meaning, if you’re an expert. If you’re an expert you extract order from the position automatically, without consciously processing every detail. The entire task must seem quite simple to a grandmaster. Similarly, the experienced mathematician sees all the important properties of the sphere and the cylinder and the cone without having to list them out one by one, and the process is so automatic they don’t even realize they’re doing it.

In “Simple” Isn’t “Easy”, I learned not to judge the difficulty of new ideas by how simple they are, but by how familiar to the student. Despite this, I have continued to make a similar mistake when dealing with ideas the students have already learned.

“Learned” isn’t “chunked”. My student understood the meaning of “hemisphere” and the formula for the volume of a cone, but she still needed conscious effort to recall and wield those bits of knowledge. Each sat in its own corner in her mind, accessible only by dint of concerted effort, and certainly not ready to flow into a flood of beautiful ideas.

I was trying to dictate a soliloquy for her to transcribe, but I was assuming that because she could see the letters on her keyboard, should could touch-type. It turned out that the effort to hunt-and-peck was so great, all the artistry of the speech was lost.

I want to watch out for my brown big spiders in the future. I want to be more patient when they are discovered and more studious in cataloging, remembering, and working with them. Most of all, I want to look back later, and remember my students forgetting them.

### 6 Responses to “My Brown Big Spiders”

1. Visualizing Elementary Calculus: Introduction « Arcsecond Says:

[…] Arcsecond « My Brown Big Spiders […]

[…] Mark Eichenlaub: Visualizing Elementary Calculus: Introduction, Visualizing Elementary Calculus: Trigonometry, A Non-mathematician’s Non-apology, My Brown Big Spiders […]

About volume of the sphere: Neat but it assmes the formula for the volume of the cone. I do not see how to do it without calculus except in simple cases.

4. Mark Eichenlaub Says: