Well, crap. I missed this post in my first read through. It’s straightforward enough that I don’t think I needed to offer a different way of saying the same thing.
You can make also make anything with more than two terms FOILable with abuse of associativity!
(a+b+c)(d+e+f) = ((a+b)+c)(d+(e+f))
For instance.
This would take forever to do, of course, comparatively, but it works!
On the broader issue of mnemonics, I have found that I dislike them more and more the longer I teach.
Kids like them too much. Somehow knowing a mnemonic–especially a “cute” one–feels more clever than actually understanding. It feels like gaming the system, like “getting away with something”. So they cling to the mnemonic and actively avoid learning the concept because they don’t want to lose that “Oh, I know a trick!” thrill. The kid that slogs through actually remembering how something works and how to do it each time gets faster and faster, until he can solve problems much faster than the mnemonic kid.
It’s like someone has a GPS locator and is smug because they can get to places faster than the guy who has to try to remember the directions. But a year later, the guy with the GPS is still taking the one slow route his machine tells him, where as the other guy knows five different ways to get there and can switch between them effortlessly as the situation warrants. Any time “saved” by mnemonics is lost later because of this.
But the real problem with the mnemonic is that when there is a problem, the kid can’t see it. This was a huge issue for me when I taught economics: kids wanted ways to remember what the curves look like. But if you rely on a mnemonic, it never looks “wrong” when you make a careless mistake.
Now, I do use a couple mnemonics for various things. But I think they have very limited use, and teachers need to be aware of these potential costs.
In twenty-odd years of mucking about with maths and finally taking a degree in it, I’ve never heard this before. I have the strong urge to sit down with a pencil and paper and go through it instead of just taking your word for it 
but that is really, really cool. Thank you!
No problem; I’m glad it’s of use to someone. But please don’t just take my word for it; see it for yourself!
At latitude L, basically by definition, latitudinal length is shrunk by a factor of cos(L) from that at the equator, which is to say, shrunk by a factor of cos(L) from its projection outwards (directly away from the polar axis) onto the cylinder with the equator as its circumference.
While also, at latitude L, longitudinal lines run at a tilt of L from the polar axis, so that, again basically by definition, longitudinal length is stretched by a factor of cos(L) from its projection.
(In case you are curious what the resulting “map of the world” looks like, here you go.)
I have a huge problem with this with the “number of days in a month” poem (30 days hath September etc etc), you can swap month names around willy nilly and the poem still flows really well, making it completely useless.
Yes, I have a mental picture of what’s going on and that’s about where I’d got to.
And oddly enough I’d known about the Lambert projection and I know more or less what the resulting map looks like (weird!) but I hadn’t made the leap about the enclosing cylinder having the same surface area as the sphere.
I’ve always wondered if my way of envisioning a sphere’s surface area actually worked, or if I was just rationalizing bad math:
The surface area is the shell of a sphere. Imagine pushing a sphere through a plane, you’ll get a point that enlarges to a circle of the sphere’s radius, and then going back to a point and blinking out of existence.
What does a “point that enlarges to a circle” sound like? The area of a circle! pir^2. But that’s only for half of it, it then blinks out of existence again, which is just another pir^2.
Now we have:
2pi*r^2
Okay, but the surface area is 4pi*r^2?
Well, by only pushing it in one direction, we’re only getting the longitudinal (or latitudinal) circles. To get the whole shell, we also need the other one. So you have to rotate the sphere on an orthogonal axis and push it through again. Luckily, being a sphere, this area calculation is exactly the same, which yields another 2pir^2. Add the two shells together, and you get 4pir^2.
I’ve never seen this mentioned anywhere, so I assume there’s a reason it’s wrong, but I can’t really intuit why. One possible flaw is that I’m double-counting the midsection in my integration. It also doesn’t generalize well to an ellipsoid, which is probably a red flag. (Though it does seem to work for a hemisphere not counting the base).
I can’t make sense of that. You have indeed counted all the latitudinal circles in your 2pi*r^2 count, and that already hits all of the surface; there’s no reason to go back and now count longitudinal circles (which, mind you, do not behave like a point that enlarges to bigger and bigger circles and then shrinks back down again; all longitudinal circles have the same size).
But the problem is that you’ve counted all the latitudinal circles weighted uniformly over the distance from the polar axis (this is implicit in your reasoning “a point that enlarges to a circle is how one fills the interior of a circle, so these should be taken to add up to the area of a circle”; as one sweeps over a disk this way, each piece of latitude moves out perpendicularly to itself directly away from the polar axis), when actually, you need to weight them uniformly over the latitude; i.e., the longitudinal distance from the equator (because as one sweeps over the surface of the globe with lines of latitude, each piece of latitude moves out perpendicularly to itself along lines of longitude). These are very different weighings, which is why your answer was off (until the nonsensical “Now we have to count the longitudes instead of the latitudes” correction, which is just wishful thinking to reach the value you already knew to be correct).
You’re totally right on the wishful thinking part, because I couldn’t visualize how it didn’t generate the whole shell, but I assumed it had to be missing because of the 2 rather than 4.
Is there a way to correct for the calculation while keeping the “push through a plane” visualization, or is the only sane way to salvage it to switch to integrating over angles in spherical coordinates?
In calculus-y presentation, the correct integral to set up for the surface area of a sphere would be of (circumference of latitudinal circle such-and-such) * d(longitudinal distance from equator at latitudinal circle such-and-such). This could be done as, for example, (circumference of latitudinal circle at longitudinal distance x from equator) * dx = 2πr * cos(x/r radians) dx.
However, you’ve instead integrated (circumference of latitudinal circle at distance x from polar axis) * |dx| = 2πx |dx|. Which is the right integrand to get the area of a circle, but not the right integrand to get the surface area of a sphere.
You don’t have to do it in spherical coordinates, but you do have to account for the rate at which you move longitudinally.
That is, “push through a plane” is absolutely, perfectly fine, as long as you keep track of the rate at which the circles formed by the intersection of the sphere and the plane sweep out area. This rate will be the circumference of the circles times the rate at which they move perpendicularly to themselves (along the surface of the sphere), and the latter factor is more subtle than it would be if the circles were simply radiating out in a flat plane.
As another illustration of why this rate matters: a cone pushed through a plane will also produce the same cross-sectional circles (starting as a point and becoming bigger and bigger), yet the surface area of a cone need not be the same as that of a hemisphere of the same radius. Indeed, different cones with the same base will have different surface areas (a very short cone has essentially just the area of its base circle, while a very tall cone looks like a very tall cylinder near the base and can thus be made to have arbitrarily large surface area).
A different presentation of your approach and its salvaging: you’ve correctly noted that if we project each hemisphere of a globe down onto its base plane, we get a disk. However, you’ve incorrectly implicitly supposed this projection to be area-preserving (it preserves latitudinal length, but squashes longitudinal length). If we instead account for exactly how area-distorting this is on each patch of the disk, we can get the correct surface area of the globe.
Specifically, at any point on a globe, the ratio between longitudinal length and its projection is equal to (1 - (distance from polar axis/radius)[sup]2[/sup])sup[/sup].
So the ratio between the area of a hemisphere and the area of a disk of the same radius is the (areally-uniform) average value of (1 - (distance from center/radius)[sup]2[/sup])sup[/sup] over a disk.
This average value must be larger than the minimal value (1 - 0[sup]2[/sup])sup[/sup] = 1. In fact, it turns out to be exactly 2. And thus it legitimately serves as the correction factor you wished for: the area of a hemisphere is 2 * the area of a disk of the same radius.