Visualization of surface area of a sphere

Solution 1:

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 $2 \times 2\pi = 4\pi$.

Solution 2:

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):

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.