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I help mentor some really young, bright kids in mathematics. We were looking at geometric properties of various shapes, and one of the kids noted that the surface area of a sphere $S = 4\pi r^2$ contains the equation for the area of a circle $A = \pi r^2$.

She was a bit confused why the factor of $4$ was mysteriously there. I told her I'd get back to her.

I know how to prove the formula using calculus, but I spent a long time trying to find an elementary way of doing it.

Does anyone know of a way of proving the first equation using almost no advanced mathematics$^1$? This seems unlikely, so as a separate question, does anyone know of a good visualization to show the relation between $S$ and $A$?

The naive approach of taking four circles and showing you can "place them" on a sphere is clearly wrong (you can't just place four circles on a sphere), but I'm not sure what the alternative is.

$^1$These kids have a working knowledge of variable manipulation, basic geometry, and I guess combinatorics?

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    $\begingroup$ I don't know whether this will qualify, but there is the beautiful method of Archimedes from The Sphere and the Cylinder, in which a sphere is projected to a bounding cylinder. $\endgroup$ – André Nicolas Jun 20 '16 at 22:07
  • $\begingroup$ Area is preserved, so the area of a sphere is $(2r)(2\pi r)$. $\endgroup$ – André Nicolas Jun 20 '16 at 22:15
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    $\begingroup$ I know it is not exactly the question asked, but Cavalieri proved that the volume of a hemisphere equals the volume of cylinder less the volume of a cone. Because each has a cross sections of the same area. $\endgroup$ – Doug M Jun 20 '16 at 22:24
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    $\begingroup$ See also math.stackexchange.com/a/782265 (and the other replies there). Hint at half an orange peeled off, then the skins flattened on the table. $\endgroup$ – dxiv Jun 21 '16 at 1:23
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    $\begingroup$ @dxiv Peeling an orange and showing you can reconstruct it into four circles isn't a bad idea. I might have to run to the store and practice on some oranges, but it'll probably be a worthy investment (and I'll get my vitamin C for the day). $\endgroup$ – anonymouse Jun 21 '16 at 1:43
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One way to proceed is to make use of the well-known (well, it should be well-known) property of a sphere: If you inscribe a unit sphere within a right cylinder, and slice them "horizontally" (i.e., perpendicular to the axis of the cylinder) the corresponding strips of the sphere and of the cylinder have equal areas.

That this is true can be seen by examining the strips in the limit. Each strip of the sphere has smaller radius than the corresponding strip of the cylinder, by an amount equal to the cosine of the "latitude" of the strip, but by the same token, the sphere's strip is wider than the corresponding strip of the cylinder, by an amount equal to the secant of the latitude. The two factors cancel each other out.

Since the entire cylinder, neglecting the ends (which don't correspond to any portion of the sphere), has height $2$ and circumference $2\pi$, its area—and therefore the area of the sphere—is $2 \times 2\pi = 4\pi$.

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  • $\begingroup$ I think this is a good way to visualize it, but without them having a solid foundation in trigonometry, it sounds really hand-wavy to say the smaller radius is cancelled out by thicker strips. Still, this does help me think of some ideas, so I appreciate the answer! $\endgroup$ – anonymouse Jun 21 '16 at 1:47
  • $\begingroup$ @anonymouse: I used the terminology of trigonometry, but the basic argument can be formulated in terms of similar triangles. If you draw out the cross section, I think that will make matters clearer; otherwise, I can create a drawing if you think that will help. $\endgroup$ – Brian Tung Jun 21 '16 at 2:01
  • $\begingroup$ FWIW, Wikipedia says this is the way Archimedes did it. $\endgroup$ – Qiaochu Yuan Jun 21 '16 at 3:22
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    $\begingroup$ @anonymouse, look at this done with pictures: maa.org/sites/default/files/pdf/upload_library/22/Ford/… $\endgroup$ – hkr Jun 21 '16 at 17:49
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I just got done teaching the class again and it went extremely well, so I wanted to post my methods here. This isn't technically an answer, but more of what I did which works as an answer.

All credit for my fundamental idea goes to @dxiv and the link he posted in the comment to my question.


I gave each kid half of a cored orange (essentially a hemisphere) and told them to use it as a stencil to trace out a circle. They determined the radius of this first circle to be $1.6 \text{ inches}$.

I argued that the surface area is a measure of how much paint we'd need to cover the top of the half-orange. I then argued that if we smushed the orange into a flat disk, the amount of paint we need obviously wouldn't change (if we painted every point pre-smush, no point would be unpainted post-smush), so the surface area wouldn't change.

They smushed the orange and, again, used it as a stencil to trace out a circle. They found the radius of this circle to be $2.25 \text{ inches}$.

The area of our original circle, $\pi r^2$, is about $8 \text{ inches}^2$. The area of our new circle is about $16 \text{ inches}^2$. So, the surface area of the hemisphere is $2\pi r^2$, and the surface area of the full sphere is $4\pi r^2$.

They were extremely happy with this demonstration, but I obviously wasn't - it wasn't quite rigorous. So to round out the lesson, I showed them the following diagram (reproduced here in Paint):

enter image description here

Under the smush operation, I said that the distance between the red and blue points should not change. This would mean that our second circle should have a radius that is off by a factor of $\sqrt2$ from our first circle. Lo and behold:

$$1.6\sqrt2 \approx 2.25$$

They all left pretty satisfied, so I think the demonstration worked, sans the fact it was a bit hand-wavy.

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  • $\begingroup$ Kudos for putting a nice idea into action! I hope they weren't too disappointed when they thought they were getting oranges but instead got orange peels :P $\endgroup$ – pjs36 Jun 21 '16 at 16:14
  • $\begingroup$ Is there any way to make the "smush operation" rigorous? $\endgroup$ – Hrhm Jun 21 '16 at 16:15
  • $\begingroup$ @Hrhm After teaching I had to go to my research lab, and ever since then the thought of making the smush operation more rigorous has been the only thing on my mind. $\endgroup$ – anonymouse Jun 21 '16 at 16:17
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    $\begingroup$ @pjs36 I bought a bunch of oranges and had to eat all of them for breakfast this morning so they wouldn't go to waste. Right now I'm running on a cup of coffee and eight oranges. $\endgroup$ – anonymouse Jun 21 '16 at 16:31
  • $\begingroup$ Wonder if some (non calculus) argument could be made that half the spiral peeled off by youtube.com/watch?v=9atdIyTe0DY "flattens" to a disk of radius $r \sqrt{2}$. $\endgroup$ – dxiv Jun 21 '16 at 17:50
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Reading @NominalAnimal 's informative nonanswer suggests this almost answer.

I would start thinking about spherical trigonometry with your fifth graders. Examples show that the sum of the angles of a triangle is always greater than $\pi$ (radian measure, of course, which you've introduced if they don't know it). Then motivate Girard's Theorem - the area of a spherical triangle is the spherical excess: the sum of the angles - $\pi$.

Then each of the four spherical triangles that you get by blowing up the inscribed tetrahedron has area $\pi$, so the area of the sphere is $4\pi$.

This isn't quite a proof, because you may need to use the area of the the sphere to prove Girard's Theorem, but it is an interesting digression and an intuitive argument.

I hang out with some good fifth graders and think this is well worth doing. I will try it in the fall.

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  • $\begingroup$ This sounds like a really good idea for a lesson, to be completely honest. Most of these kids are not the best geometers, so they usually bug me to do more combinatorics and "number theory" stuff with them. But I think this would be a good digression to help motivate why geometry on a sphere is useful (and clearly different from geometry on a plane). $\endgroup$ – anonymouse Jun 21 '16 at 19:45
  • $\begingroup$ @anonymouse Do hyperbolic geometry too. The Poincare model of the hyperbolic plane is within reach. The metaphor I use is that the disk is a mud puddle and the going gets stickier/slower as you near the edge, hence geodesics bend toward the center. The sum of the angles of a triangle is always less than $\pi$; it can even be 0. That means you can tessellate with all the $n$-gons, see Escher patterns. The wikipedia page is good. $\endgroup$ – Ethan Bolker Jun 21 '16 at 20:11
  • $\begingroup$ These are really good recommendations that I never considered - have you worked with young students a lot before? I'm a bit jealous of the opportunities my students have in this program, because it's definitely something that didn't exist when I was growing up. $\endgroup$ – anonymouse Jun 21 '16 at 20:28
  • $\begingroup$ @anonymouse Let's take this discussion offline, out of the comments. It's pretty easy to find my email address. $\endgroup$ – Ethan Bolker Jun 22 '16 at 0:06
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This is definitely not an answer, but if you intend to use the "four parts of an orange peel" analogy, this might be relevant.

This might be entertaining for some, especially those who enjoy confusing people with numerical coincidences (like Randall Munroe's XKCD comics 217 and 1047).

I was thinking about the regular simplex in 3D -- the regular tetrahedron.

If you put a regular tetrahedron inside the unit sphere, i.e with the four vertices of the tetrahedron on the unit sphere, the six edges of the tetrahedron are $\sqrt{8/3}$. Each of the four faces is an equilateral triangle (each side being, of course, said $\sqrt{8/3}$). The area of each triangular face is therefore $\sqrt{4/3} = 2/\sqrt{3}$.

The area of the unit sphere is $4\pi$, so the area of each triangular quadrant on the unit sphere -- as if the tetrahedron was puffed up into a sphere -- is $\pi$.

(The edges of each puffed-up face is an arc of a great circle.)

The ratio between the area of the unit sphere quadrant and the tetrahedral faces is $\pi:2/\sqrt{3} \approx 2.72070$, which is within one thousandth of $e \approx 2.71828$.
(With three significant digits, both round to $2.72$.)

This is a coincidence; the ratios in lower and higher number of dimensions is quite different.

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I think the idea of trying to do this without calculus is misguided. Instead, try to understand the steps in the calculus. The surface area formula is derived from the volume formula so maybe the question should be: can I get the volume formula from the formula for circle area $A_c=\pi r^2$? First, break down the sphere into disks (centered on the x-axis). Then using cylinder volumes we have: $$V=\int_{-r}^{r}\pi(r^2-x^2)dx=\frac{4}{3}\pi r^3$$ We can do this because for cylinder volume $V=\pi R^2h$ where $h$ is height $R=y$ and so $R^2=y^2=r^2-x^2$. To understand how this relates to the surface area consider the end of the FTC proof, which goes like this: $$\frac{dA}{dx}=y$$ $$dA=y·dx$$ $$A=\int y·dx$$ where $A$ is the area function for $y$. Comparing the above equation to the previous integral we can see that we should be able to differentiate $V$ to produce something equivalent to $y$ i.e. something that encompasses and spatially defines itself. What could that be for the volume? It would have to be the surface area!

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    $\begingroup$ I really appreciate this answer. I thought about this when my student asked the question, but I didn't really understand how to resolve the explanation. I could start explaining piling disks of various radii, but I don't think this forms a convincing visual argument. If at any point I write down an integral symbol, the students are likely to be out of their element (they're fifth- and sixth-graders, after all). $\endgroup$ – anonymouse Jun 21 '16 at 0:21

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