82
$\begingroup$

I know what "surface area" means for:

  • a 2d shape
  • a cylinder or cone

but I don't know what it actually means for a sphere.

For a 2d shape

Suppose I'm given a 2d shape, such as a rectangle, or a triangle, or a drawing of a puddle. I can cut out a 1cm by 1cm piece of paper, and trace that piece of paper on the shape. Many full 1 cm squares will be traced on the shape, and there will likely be many partial squares traced on the edges of the shape. Suppose I can accept that I can "combine" the partial squares into full squares. Then I count the total number of full squares, to find the surface area.

For a cone or cylinder

I can convert a paper cone into two 2d shapes. The bottom of the cone is a circle. I can then cut the curved (ie not-bottom) part of the cone using scissors, and unfold that part into a flat 2d shape.

Similarly, I can convert a cylinder into flat 2d shapes: two circles and a rectangle.

For a sphere

But the above methods for understanding surface area don't work for a sphere. I can't lay a 1 cm by 1 cm piece of paper onto a sphere in a flat way. I can't even trace a square centimetre onto the sphere using that piece of paper!

People might say, "suppose you have an orange, and you peel the orange. Then you can lay the peel flat onto the table, into a flat 2d shape". But they're lying! The orange peel can never be mashed down perfectly flat onto the table!

So, I don't know what "surface area of a sphere" even means, if you cannot measure it using flat square pieces of paper!

What does "surface area of a sphere" even mean?

$\endgroup$
24
  • 11
    $\begingroup$ The one dimensional analog of your question is: what does it mean to talk about the perimeter of a circle? Can you see how constructing a parameterization of a straight line (in the colloquial sense) to a circle allows us to make sense of it? Can you adapt this to a sphere? $\endgroup$ Apr 24 '20 at 8:09
  • 11
    $\begingroup$ @DerekAllums The difference is that the cut string is not an approximation; when you pull it taught, you get a straight line and you can measure it. But cutting up a sphere always approximates, and the number of pieces governs the accuracy of the approximation. That leads to the need for calculating the limit (i.e. calculus). $\endgroup$
    – Barmar
    Apr 24 '20 at 16:11
  • 23
    $\begingroup$ Your backyard (if you have one) is on a curved surface (our globe). Does is strike you as problematic to talk about its area? $\endgroup$ Apr 24 '20 at 16:47
  • 12
    $\begingroup$ You say you know the surface area of a flat shape or one that can be flattened... but what is the area of a disk? You have the same issue here as with the sphere. You cannot tile a disc exactly with squares except as a limit with infinitesimal squares. $\endgroup$ Apr 24 '20 at 17:26
  • 18
    $\begingroup$ @eps If someone's asking about the nature of surface area, then I promise you ... showing them Banach-Tarski is NOT going to help them gain a better understanding. $\endgroup$
    – Brondahl
    Apr 24 '20 at 17:58

13 Answers 13

66
$\begingroup$

This is actually an interesting question. It involves how to define "area" on a curved surface. The examples you have provided are surfaces that are developable (can be flattened onto a plane) after a few cuts. And you can compute the flattened area. You can never do this to a sphere, because no matter how small a patch from a sphere is, it can never be flattened onto a plane. The idea is to break down the sphere to small patches such that each is flat enough and you compute the area as if it is flat, and then add up the areas of the patches.

Mathematically, suppose $S$ is a sphere. The above procedure is stated as:

  1. Break up $S$ into patches $P_1,\dots,P_n$, where each $P_i$ is a patch that is flat enough, and $n$ is the number of patches you have.

  2. Compute $\operatorname{Area}(P_i)$ as if each $P_i$ is flat. As suggest by levap, one way to do it is to project each patch onto one of its tangent planes. Note that I am not saying this is the only way to approximate a patch, and I am also not saying that one way that would seem correct at first glance would really be correct, see Update 2 for an example, there's also discussion about this in the comments.

  3. Use $\operatorname{Area}(P_1)+\dots+\operatorname{Area}(P_n)$ as an approximation of the area of $S$.

  4. If the patches are small enough, then the approximation should be a good one. But if you want better precision, use smaller patches and do the above again.

  5. This is to make the math precise, I can't guarantee that a third-grade student can understand this: As you take smaller and smaller patches, the value of the approximation above should tend to a fixed number, which is the mathematical definition of the area.

P.S. For a visualization of this approximation, you can search online for sphere parametrization, or simply think of a football (soccer ball).


Update 1: Thanks to Leander, we have a visualization:

visual representation of the patches

One might notice that this visualization is slightly different from cutting up a sphere; it takes sample points on the sphere and attach triangles to these sample points. I want to remark that there is no essential difference between this and my method. The idea is the same: approximation.


Update 2: A comment (by Tanner Swett) mention that the method of using a polygon mesh may be flawed. Indeed, the example of Schwarz lantern shows that some pathological choice of the polygon mesh may produce a limit different from the surface area. The following explanation should be helpful:

As I have mentioned in step 2 above, if we are not careful with how we approximate the areas of the patches, the approximation may not work. The Schwarz lantern is an example where a careful choice of the approximating triangles can lead to the following result: Suppose $T$ is a triangle we use to approximate a patch $P$, then it is possible ${\rm Area}(T)/{\rm Area}(P)\to a\neq1$. To illustrate this, consider a single triangle on the Schwarz lantern: schlant

We assume the cyclinder has total height $1$ and radius $1$. We take $n+1$ axial slices, and on each slice $m$ points. The area enclosed by the red curves is a patch on the cylinder, and the triangle enclosed by the blue dashed lines is the one used to approximate the patch. Let $P$ and $T$ denote the patch and the triangle respectively. We see that the bottom edge of $P$ and $T$ has ratio $1$ as $m\to\infty$. What really makes a difference is the ratio of their heights. Suppose along the vertical direction the height of $P$ is $$h=1/n$$ Then the height of the triangle is $$h_T=\sqrt{1/n^2+a^2}$$ By a simple computation we know $a=1-\cos(\pi/m)\approx(\pi^2/m^2)/2$. Therefore, $$h_T/h=\sqrt{1+\frac{\pi^4n^2}{m^4}}$$ If $n$ has higher order than $m^2$, then the limit is bigger than $1$, and consequently ${\rm Area}(T)/{\rm Area}(P)\not\to1$.

This problem would have a smaller probability of occurring in practice. Imagine if you do cut the cyclinder into patches, you'd use $h$ instead of $h_T$ to estimate the area. But again, it is hard to make this (what approximation is acceptable) precise without using the language of calculus.

$\endgroup$
31
  • 3
    $\begingroup$ upon further reflection, i'm wondering if it makes sense to an elementary school student to say "yes, the orange peel doesn't lie flat. but we can kind of measure it as if it was flat. but if we break the orange peel into smaller bits, it would lie more flat. if we then measure the area of this now, the surface area will be just slightly bigger.", and then somehow find some metaphor that gives insight about doing the limit of this. $\endgroup$
    – silph
    Apr 24 '20 at 7:15
  • 2
    $\begingroup$ @silph If you want to try it yourself, do it with three oranges, but use big, medium and small-sized pieces depending on the orange, and notice there's a value you're getting closer to as the pieces are shrink. $\endgroup$
    – J.G.
    Apr 24 '20 at 7:18
  • 2
    $\begingroup$ How sure are you about the area of your puddle? Obviously you have some partially filled squares on your big grid, in order to figure out how much that is you will need to measure the area of them...so you use a smaller grid. Now your smaller grid will have some partially filled squares, rinse, repeat. For a physical puddle eventually you reach physical particles and you're done, but if your puddle was mathematically a fractal you could go forever. Yet your puddle clearly has an area, you can say it's no bigger than x and no smaller than y. (You can even do this for distances with rivers.) $\endgroup$ Apr 24 '20 at 15:47
  • 2
    $\begingroup$ @user3067860 it's interesting how my grade-three brain is able to "accept" this intuitive idea of a limit (ie, the area of a puddle tending towards a unique answer, when using smaller and smaller squares to measure it) for a 2d shape. that is, my grade 3 brain can "accept" placing a grid of 1cm x 1cm squares, and for each edge square, acceptng tht there is a unique correct "answer" to its area. but it's curious that, in contrast, my grade 3 brain has trouble visually accepting a similar idea for a 1cm x 1cm piece of paper, applied to sphere! i will have to think more about the intuitive chasm. $\endgroup$
    – silph
    Apr 24 '20 at 15:57
  • 1
    $\begingroup$ @user3067860 I can't really quantify this without introducing calculus on surfaces. Let's just say in real life you can trust your eyes to tell whether something is flat enough. $\endgroup$
    – trisct
    Apr 24 '20 at 16:13
55
$\begingroup$

Take a sphere (or any other shape), and paint it blue. The amount of paint required is just proportional to the surface area. This is a way to measure it.

$\endgroup$
23
  • 21
    $\begingroup$ One problem with this is in practice is that paint has a thickness. For a very large sphere, each 1cm^2 is sufficiently flat that it's nearly equivalent to painting 1cm^2 on a flat piece of paper. But if you paint a very small sphere, the curvature will make a difference. If your sphere has a radius of only 1mm, a paint thickness of 1mm makes a big difference. The ratio of paint needed to surface area increases the smaller your sphere is - in practice, it is not a constant ratio as you suggest. This doesn't solve the OP's "paper" problem that you cannot equate a curved area and a flat one. $\endgroup$ Apr 24 '20 at 15:39
  • 12
    $\begingroup$ @NuclearWang: if you want to play the fool, think that paper has thickness and cannot be folded because that would change its area, which is impossible. So a cylindre has no area. $\endgroup$
    – user65203
    Apr 24 '20 at 16:35
  • 13
    $\begingroup$ @trisct Yes, you can get asymptotically close to the correct answer as the paint shell becomes infinitesimally thin. But in practice, this method works no better than laying overlaying 1cm x 1cm squares onto the surface of the sphere, which is where the OP started - if the sphere is large compared to the square/paint, it will work "well enough". Your formulation is exactly correct, the surface area of a sphere is the volume/thickness of an infinitesimally thin shell around the sphere. Whether that's more intuitive than the sum of infinitely small flat patches of area is anybody's guess. $\endgroup$ Apr 24 '20 at 18:36
  • 3
    $\begingroup$ See also How do I experimentally measure the surface area of a rock? $\endgroup$
    – Brian
    Apr 24 '20 at 19:11
  • 2
    $\begingroup$ @YvesDaoust Gabriel's Horn is interesting. Here it says "The paradox is resolved by realizing that a finite amount of paint can in fact coat an infinite surface area — it simply needs to get thinner at a fast enough rate. (Much like the series 1/2^N gets smaller fast enough that its sum is finite.) In the case where the horn is filled with paint, this thinning is accomplished by the increasing reduction in diameter of the throat of the horn." $\endgroup$
    – user76284
    Apr 24 '20 at 22:34
22
$\begingroup$

A circle of radius $r$ has area $\pi r^2$ and perimeter $2\pi r$. If we run a very thin line of pencil around the perimeter of thickness $\delta$, the graphite area will approximate $2\pi r\delta$.

A sphere of radius $r$ has volume $\frac43\pi r^3$ and surface area $4\pi r^2$. If we cover the surface with a very thin layer of spray-paint of thickness $\delta$, the volume of paint lost from the can will approximate $4\pi r^2\delta$.

Note that in both cases there are two formulae, one for how much space is inside the shape, and how much of a different kind of space, with one lower dimension, is on the edge of the shape. Basically, the edge size is how quickly the interior size grows as the shape widens.

(Edited to link to somewhat more detailed explanations.)

$\endgroup$
19
$\begingroup$

Imagine a perfect sphere the size of the Earth, perfectly smooth, and that you've got a vast number of perfect little centimeter-square tiles and a large army of bored quarantined kids to lay them out and count them.

On that huge sphere, each tiny tile will seem to lay flat, and to fit perfectly with the tiles on all four sides, and cover the planet with no visible gaps; and after you tally them all you can say that the surface area of the earth is so-many square centimeters. It will be a very (very!) large number, but it'll be a definite number and that's the surface area.

For a smaller sphere, like a beach ball or an orange or a ping-pong ball, a square-cm tile isn't going to fit well at all. So use a smaller tile: one a mm square, or a micron, or Angstrom, or smaller. Give your kids tweezers and magnifying glasses and get them to work. Eventually you'll have the surface area of your sphere, in sq mm, or square Angstroms, or barns (yes, that's a unit of area!) or whatever.

So to conceptualize the surface area of a curved surface, just think smaller and smaller until your hypothetical tile is so much smaller than the curvature of the surface that it seems to lie flat and join perfectly with the tiles surrounding it. And get ready to count to very large numbers.

$\endgroup$
7
  • 5
    $\begingroup$ wow, my grade 3 brain is (surprisingly!) completely okay with this visualization. it builds upon my already-present grade 3 brain's acceptance of what a square centimetre is, and my acceptance that the Earth is a globe. what a sneaky way to get my brain to visualize very tiny units of measurement on a very large surface! this visualization, combined with the approximation idea in trisct's answer, of cutting an orange into big patches, then smaller patches (ie, while continuing to think in terms of unit cntimtre squares), gives enough for my grade 3 brain to undrstnd both limits and aprxmtn! $\endgroup$
    – silph
    Apr 24 '20 at 17:12
  • 3
    $\begingroup$ It's not quite as simple as that. Suppose we start by laying out a row of tiles along the equator, and work our way northwards. Every so often, we will have to reduce the number of tiles as the lines of latitude get shorter; and when we do this, there is no way to avoid significant mismatches, of maximal size (i.e. of the order of half a centimeter). This contradicts your claim that the tiles will seem to fit perfectly on all four sides. And this is not just a consequence of the way that I chose to lay out the tiles; it is a topological certainty....[cont'd] $\endgroup$
    – TonyK
    Apr 24 '20 at 19:32
  • $\begingroup$ ...Your idea is not without merit, by the way; it's just not quite as straightforward as it seems. $\endgroup$
    – TonyK
    Apr 24 '20 at 19:34
  • $\begingroup$ Perhaps, imagine a collection of hexagons and pentagons, arranged in a sort of geodesic configuration instead. the area of each of these can easily be calculated, and can also get progressively smaller. Or, measure the triangles between the rows as separate items. I imagine you could imagine the rest... Or just use triangles! $\endgroup$ Apr 24 '20 at 21:47
  • $\begingroup$ @TonyK That's a good point. In fact to fill the surface of a sphere as per CCTO's suggestion you need two types of tile: round ones and complementary-round ones. Squares wouldn't do the trick at all. Even with round ones you would have to decide how to position them relative to one another: there is no positioning pattern (e.g. vaguely hexagonal) that is "natural" or "ideal", and I think only some numbers would work at all. Plus, taking a simple example, with a 12 points evenly distributed throughout the sphere's surface, the "complementary" shapes would actually take up a lot of area. $\endgroup$ Apr 26 '20 at 13:58
13
$\begingroup$

I would first introduce approximating the area of a shape and pi via the method of exhaustion

Exhaustion of a Circle - Wikipedia

The area or circumference is approximately the average of the two, but not quite..

Once students understand this for a two-dimensional shape, it should seem clear both

  • pi exists and is a transcendental number
  • it is illogical to try and represent the area or circumference of a circle without it

With this out of the way, you could pose using Exhaustion with an N-faceted polygon (perhaps beginning with a cube inside a cube?). Ideally this will lead them to discover again that they will need pi to find the true surface area, while also subtly preparing them for calculus.

Plausibly you could purchase or fashion an object to show this, but I suspect some graphics simulation software will aid you (and also trivialize discovering the area of the contained and surrounding solids)

$\endgroup$
3
  • 2
    $\begingroup$ This strategy works well to describe a circle's circumference (though I rather doubt that it makes clear that $\pi$ is transcendental), but I'm not convinced it's a viable strategy for surface areas in any higher dimensions. The method of exhaustion makes explicit use of the existence of infinitely many regular polygons, and the higher dimensional analog is resoundingly false-- there are only finitely many regular polytopes in every dimension $n>2$. That isn't to say the method can't salvaged, but you quickly lose the intuitive appeal when invoking non-regular approximating polytopes. $\endgroup$
    – jawheele
    Apr 25 '20 at 20:26
  • 3
    $\begingroup$ Beyond that objection, I also feel this isn't a good approach to defining area (the OP's interest), the reason being that the intuitive feel of this picture is precisely the same as that of this well-known false proof that $\pi=4$: math.stackexchange.com/questions/12906/… (the same method can start with an inscribed square and seem to demonstrate $\pi = 2 \sqrt{2}$). If one's only conception of the definition of circumference came from this picture, you'd be hopeless in discerning why your picture works and the false proof above doesn't. $\endgroup$
    – jawheele
    Apr 25 '20 at 20:37
  • $\begingroup$ Erratum: I believe I spoke too quickly when claiming that starting with an inscribed square can seem to demonstrate $\pi = 2 \sqrt{2}$, though the objections still stand. $\endgroup$
    – jawheele
    Apr 27 '20 at 19:19
11
$\begingroup$

All the solid shapes whose surfaces you are able to understand are finitely rectifiable -- that is, you can think of a finite number of transformations such that the areas (if we agree that areas are additive) can be transformed from covering a solid to lying entirely in one plane.

However, there's no reason to think that this will always be possible for all surfaces which clearly posses an area (albeit intuitively). That would be tantamount to the naive assumption of the Pythagoreans that all quantities can be measured using only integers and ratios of integers -- hence their intense shock on discovering the irrationality of the diagonal of a square!

The general lesson is that elementary methods aren't always sufficient to capture everything we would like to capture -- we have to extend our elementary methods and notions in a way that accommodates objects that wouldn't fit in the earlier scheme, while still preserving their logical character. This is precisely the triumph of the infinitesimal calculus over finite calculus. Many things may be done (with more and more difficulty) with only the latter, but sooner or later one has to admit that one cannot escape using infinitesimal analysis, for even some very basic things.

So, again, the point boils down to an extension, a rising I'll say, from finite methods to infinite methods. That one has to do this doesn't mean that the objects that only admit of infinite methods don't have the properties analogous to those objects that can be conquered using only finite methods -- after all, those old objects can be consistently analysed using the new infinite methods as well.

So, how to understand the surface of a sphere? Accept that it may be impossible to rectify in only a finite number of steps, and accept therefore that you would need infinitely many operations to completely rectify it. Accept that this is not strange, since in the end you'll have a definite quantity for your area. Finally, since we have only finite brains, how do you think of this process -- just cut the sphere into smaller and smaller pieces (one way is to go along the longitudes), and continue ad infinitum. As you continue this process, you see that the strips become thinner and thinner, and so more and more rectifiable, although they still contain a tinge of curvature. This curvature will never disappear after any finite number of thinning steps, but it gets arbitrarily smaller, so that we know that it approaches a definite rectified form. This is the approach of limits. In the approach of infinitesimals, one would say that after infinitely many such operations the strips become infinitely thin, and flexible, so that the curvature may be completely removed.

Then the sum of the areas of all these strips, gives the area. In the limit approach, you'd have to approximate the area of each strip at each stage of the process, and note that the approximations get arbitrarily close to a certain quantity, which is the desired area.

$\endgroup$
1
  • $\begingroup$ Precisely. The only way for an "elementary school brain" to grasp the meaning of the surface of a sphere (and think about it non-casually) is to progress beyond "elementary" territory, or at least glimpse that other territory, which may either excite, baffle or appall. Or maybe all three. $\endgroup$ Apr 26 '20 at 14:24
7
$\begingroup$

There is a conceptually simple way to think of this: Build a hollow sphere out of some rigid material such as metal, or plastic. This material will have some thickness, say $d$. Suppose its inner radius is $r$ and its outer radius is $R$ (so we have $R=r+d$).

Now get out your kitchen scales and weigh the thing. Suppose it mass is $W$ grams; and suppose further that the weight of a unit square of your stiff material is $w$ grams. Then the surface are of the sphere is about $W/w$.

I say "about", because of the finite thickness $d$ of the spherical shell. But we know that the inner surface area is less than $W/w$ and the outer surface area is greater than $W/w$. And in the limit, as the thickness $d$ tends to zero, this value $W/w$ will tend to a limit, which is the area of the outer curved surface.

$\endgroup$
6
$\begingroup$

Consider a "convex polyhedron":

enter image description here

You can start with a simple pyramid or cube, but, as the polyhedron gets more and more complex, it can be made more and more like a sphere. At each step along the way you can measure the dimensions of each flat surface, add the surface areas together, and come up with an estimate of the surface area of the equivalent sphere. As the polyhedron is made with more and more pieces, it becomes a closer approximation to the sphere.

There is this mathematical concept known as a "limit" where the approximation, after an infinite number of refinements, essentially becomes a sphere, and the sphere's surface area is determined.

$\endgroup$
5
$\begingroup$

If the surface area of a sphere is $1\text{cm}^2$, that means that if you cut a sphere into very very very tiny pieces, so tiny they are almost perfectly flat, then the total area of those pieces will be very very very close to $1\text{cm}^2$.

$\endgroup$
1
  • 2
    $\begingroup$ And, moreover, the tinier (and thus flatter) you make the pieces, the closer the area would be to $1$ square centimeter. $\endgroup$ Apr 24 '20 at 6:51
4
$\begingroup$

First, I'm a new contributor, so try not to jump on me! :-)

Secondly, the o/p has asked how this problem might be explored in terms of elementary school math. I'm sure, at least, that we've all been there! Maybe we can approach this in terms of elementary grade math?

I was taken with the idea suggested of painting the surface area, and working out how much paint was required to paint the sphere's complete surface.

If we knew how much paint we started with, and how much we had left afterwards, we could calculate the surface area of the sphere if we measured the thickness of the layer of paint now coating the sphere.

We might go mad and measure the diameter of the sphere before painting it, and after painting it, in order to use good old elementary grade subtraction to calculate the added diameter of the sphere with its fresh coat of paint. That would tell us the thickness of the coat of paint.

How about then looking at the problem from a new point of view? Still with our pot of paint, how about we actually dunk the sphere in it, in order to coat it with paint? And doesn't that suggest an additional test? How would it be if we measured the amount (volume) of paint displaced by the sphere?

Perhaps the paint pot might be completely full, so that immersing the sphere in the paint would cause paint to be displaced from the pot, and thus it could be measured as it flowed into a measuring vessel held beneath the pot, so that the volume of liquid displaced by the sphere would thus be measured. That would also give us a measurement of the volume of the sphere, which must be equivalent to the volume of liquid displaced.

Seems to me I learned about Archimedes in elementary school! Our grade school teacher's favorite joke was that 'Eureka' is Greek for 'this bath is too hot'!

Once we know the volume of the sphere, along with such certain (measured) properties as its radius and its circumference, we can make some calculations of its surface area. Perhaps if we made a series of such experiments, with spheres of differing surface area, we could eventually use simple multiplication or division to arrive at the well-known formula of 4 Pie R Squared.

And nothing has to be flattened onto a plane. :-)

$\endgroup$
3
$\begingroup$

You already got great answers. I wanted to emphasize that already for flat surfaces you are accepting to approximate your area by small rectangles. And I think it's clear to you that there will always be a small error, that you can diminish but never get rid of (unless you do calculus, and that's one of its magical traits).

With the sphere it's not different really. The leap you need to make is to accept that, instead of "missing area" just by the sides of your rectangles, now you will be "missing area" by not being able to set your paper rectangles flush against the surface. But it should be clear that, the smaller the rectangle, the better the approximation.

One visualization that might help is to draw a circle with some plotting app (Desmos, to name one) and start zooming in. You will see that the more you zoom, the more the circle looks like a line. With the sphere, a 3d version of that phenomenon happens.

$\endgroup$
2
$\begingroup$

This is a very good question, with very good answers, so I'll just chime in with a comment. Some years ago, a researcher came to me asking how to compute the surface area of a coffee bean. I responded that it is very hard to get a really good answer. Like others have said, you need to get a triangulation of the surface, and then add the areas of the triangles. But if there are lots of small bumps on the coffee bean, it is hard to get a good approximation.

A similar question is "how long is the coast of England", which was originally asked by Benoît Mandelbrot. The trouble is, as you focus closer and closer in to the surface, the answer gets longer and longer.

Now if he had asked for the volume of the coffee bean, that would have been easy. Dunk it in liquid, and see how much spills over. I could have told him to paint the coffee bean, and see how much paint he had to use, but it is hard to apply an even coat when the surface is bumpy.

$\endgroup$
0
$\begingroup$

Archimedes showed that the surface area of a cylinder (without the top and bottom) is equal to that of the inscribed sphere. Further, the areas cut off by any planes perpendicular to the cylinder's axis are also equal. This makes intuitive sense as follows. The angle at which the sphere "recedes" at any "latitude" gives you MORE surface area than the cylinder slice. However, the smaller radius of the slice of the sphere at that "latitude", gives you LESS surface area than the cylinder slice. By drawing some triangles, I was able to convince myself that the MORE and the LESS offset each other exactly.

Cylinder and sphere slices equal in area

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.