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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?

Community
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silph
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  • The fact that you are unable to "manufacture" a sphere doesn't mean that it does not have a surface and that this surface does not have an area. –  Apr 24 '20 at 06:52
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    @YvesDaoust i might argue that if the definition of "area" that i was taught only talks about covering surfaces with pieces of paper that are all a unit square big, then the surface of a sphere doesn't have an area; i.e. that you cannot apply this definition of "area" to a sphere! – silph Apr 24 '20 at 07:04
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    The sphere is not a ruled surface meaning that you cannot apply a sheet of paper to it. If you take a definition that requires the surface to be ruled to have an area, then indeed there is no area. Or the definition is inappropriate. –  Apr 24 '20 at 07:09
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    I answered this question for myself like this: you need some amount of paint to cover the surface of sphere, right? So, the "surface area of a sphere" is real, not just some abstraction. It is possible to convert this answer into a strict definition of what is a surface area of arbitrary set of points (via the volume of the region close to our points) - but I am afraid it's possible to construct some weird surfaces such that this definition will be good for them. – lesnik Apr 24 '20 at 08:08
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    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? – Derek Allums Apr 24 '20 at 08:09
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    @DerekAllums actually, the perimeter of a circle was taught to me as "consider a piece of string, that you curl around the outside of a circle. what is the length of that piece of string?" and that satisfied me. it fit well into the precise idea of measuring a 1D line on a ruler. but sadly, i can't adapt this to a sphere; a piece of string can be wrapped along a circle; but no piece of flat paper (or plastic, or whatever) can be "wrapped around" a sphere! – silph Apr 24 '20 at 08:34
  • @silph But think about it further and it's the same issue: if you believe you can approximate a curved line by bending a straight one, why not approximate a curved surface by bending a flat one? Better explanations of this intuition are below in other users' answers. – Derek Allums Apr 24 '20 at 08:48
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    @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). – Barmar Apr 24 '20 at 16:11
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    Your backyard (if you have one) is on a curved surface (our globe). Does is strike you as problematic to talk about its area? – John Coleman Apr 24 '20 at 16:47
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    @JohnColeman this is interesting. i think that when i look at my backyard, it looks completely flat. but i were to zoom out far enough to see the entire globe, things feel problematic. i'm starting to realize that if i'm given a sphere i can hold in my hands, that i also want to use a square unit that's about one square cm big; and that using "one square millimetre" (ie, more the size of a backyard compared to a globe) is unsatisfying; my grade 3 brain has trouble visualizing any understanding of such small units of measurement. it wants to continue to think in the bigger square cm unit. – silph Apr 24 '20 at 17:04
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    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. – Jim Garrison Apr 24 '20 at 17:26
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    While I think for this case there are ways to "intuitively" visualize a sphere's surface area, realize that as you get to higher level math such an exercise becomes irrelevant and counterproductive. A 19 dimensional (hyper)sphere also has a surface area as well but good luck trying to come up a visual aid for it and why even bother with trying to conceptualize it in the first place. Personally, working with high dimensional objects got a lot easier once I realized trying to smash it down to the 3d world was pointless and just accepted it as an abstract concept with certain properties. – eps Apr 24 '20 at 17:49
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    @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. – Brondahl Apr 24 '20 at 17:58
  • You cannot apply a piece of paper on any curved surface, because the two faces would have different areas. This explains why a cylindre cannot have a surface area. –  Apr 24 '20 at 19:45
  • See gauss baynet theorem, and intrinsic curvature. A plane and cylinder are of the same genous but a sphere is not. They are not isometric and they require tearing or deformation to map one to other. Also can see poincare conjecture and ricci flow. – marshal craft Apr 25 '20 at 03:14
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    Lol just do $\int _{0}^{2\pi }\int _{0}^{\pi }r^{2}\sin \theta ,d\theta ,d\varphi =4\pi r^{2}.$ and that's the surface area :) – user600016 Apr 25 '20 at 11:15
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    Rather than try to explain this in terms of elementary-school mathematics, you should use this elementary question as motivation for calculus. More generally, calculus is what you naturally get when trying to generalize ideas from flat things to curved things (an important question here is "What is a curved thing?" -- the answer is "something that looks flat when you zoom in"). – Abhimanyu Pallavi Sudhir Apr 25 '20 at 14:42
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    @AbhimanyuPallaviSudhir admittedly, this /could/ be used as a motivation for calculus, for students who are ready/interested in the ides of calculus. calculus gives the precision required to describe what's happening in a very mathematically rigorous by-sets-of-axioms way. i know that i've always been interested in explanations, though, that give the intuitions/approximations of the insights, especially ways that very young students could understand. admittedly, the measure of how effective these intuitive explanations is far more subjective, but i think it's an interesting avenue of inquiry – silph Apr 25 '20 at 16:19
  • @silph Right, and I'm saying these insights/intuition should be equated to the discipline of calculus. – Abhimanyu Pallavi Sudhir Apr 25 '20 at 19:02
  • I've been somewhat annoyed that mathematically, a "sphere" is a 2D object (a circle is 1 dimensional). Technically, it's the 2D surface of a "ball" that they're talking about. Or just "the area of a sphere". This dichotomy of terms is annoying as hell when trying to figure out what physicists mean. For instance, a simplified universe model is commonly "a 4 dimensional surface of a 5 dimensional sphere". ......No, not really. What they meant was either a 4 dimensional sphere (which no one would understand) or a 4D surface of a 5D ball. OY! – tgm1024--Monica was mistreated Apr 26 '20 at 02:52
  • Perhaps an intuitive way to visualize area is to imagine covering each shape with paint to the same thickness. The area is then the volume of paint used divided by its thickness. Very spiky shapes become problematic due to discontinuities, but when the when the thickness is small compared to the shape (i.e. tends to zero), the ratio tends to the correct value. – Gnubie Apr 26 '20 at 18:25
  • Isn't this asking how to define the integral (in this case, two dimensional). – lcv Apr 26 '20 at 23:31
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    You might benefit from this (and, less directly other) 3Blue1Brown videos: But why is a sphere's surface area four times its shadow?. – Eric Towers Apr 27 '20 at 00:19
  • The question of what area means is to be understood in terms of "utility". For example, until area was "invented" people measured equivalence of plots of land in terms of their perimeter. It may seem bizarre to us, but intuitively it does conform to the idea that increasing the length or the width of a rectangle increases its size "linearly". So the answer must start with answering the question "What purpose does the area serve?" In that respect @YvesDaoust has given an answer which should be developed into a full answer. – Kapil Apr 27 '20 at 11:46

13 Answers13

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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.

trisct
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  • forgive me for asking: i actually don't have much more than a high school background of mathematics, so i don't understand the notation used in #1. what does the "S squared" and "Big U with subscript n" mean?

    At any rate, i'm thinking also about how to take this idea of "approximate a unit square onto the sphere" in a way that would make sense to a third grader (or even to my own brain!).

     – silph Apr 24 '20 at 07:12
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    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. – silph Apr 24 '20 at 07:15
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    @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. – J.G. Apr 24 '20 at 07:18
  • @silph Yes, I realized my notations can be disturbing if you don't have much background. I edited my answer to make the notations more friendly. – trisct Apr 24 '20 at 07:21
  • @silph To your second comment: not necessarily "slightly bigger", I only know it should be "slightly different". Whether this can make sense to third graders may depend on the individual. Some may not even realize this could be a problem. – trisct Apr 24 '20 at 07:28
  • @silph : These are variable symbols: you assign them on the spot a meaning and then they stand for that thereafter. So when he says, "let $S$ be the sphere", it means now that whenever you see $S$, that references that particular sphere (versus any other sphere). – The_Sympathizer Apr 24 '20 at 15:46
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    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.) – user3067860 Apr 24 '20 at 15:47
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    @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. – silph Apr 24 '20 at 15:57
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    @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. – trisct Apr 24 '20 at 16:13
  • @trisct Sorry, the question mark there was rhetorical. I just thought it would be an illuminating concept for the OP to see that this applies in lower dimensions, too. – user3067860 Apr 24 '20 at 16:19
  • This paragraph is contradictory

    ". 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 if as it is flat, and then add up the areas of the patches."

    You can't make patch small enough to make it flat then u say you can..

     – tryst with freedom Apr 24 '20 at 19:31
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    @DDD4C4U There's no contradiction, though it may be written somewhat unclearly. The answer is saying that it's not possible to get a sphere perfectly flat, but it is possible to get it flat enough. – Tanner Swett Apr 24 '20 at 23:21
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    @trisct I'm not sure if this contradicts your answer, but I proposed an edit with a visualization to help the grade 3 brain. – Leander Apr 25 '20 at 10:30
  • @Leander That is exactly what I've been looking for to put in my answer. – trisct Apr 25 '20 at 11:06
  • I think this won't work as it is written. You need to introduce some restrictions on how to choose those patches, otherwise one can come up with a 3D version of this (proof that pi is 4 by "extruding" parts of a square repeatedly). – Federico Poloni Apr 25 '20 at 16:03
  • @FedericoPoloni I've seen the same fake proof used to prove that the diagonal of a square has twice the side length. I stated in my answer that each patch should be flat enough. But if you use an "approximation" analogous to the fake proof you cited, you would notice that the patches you've chosen are never flat enough, and thus don't meet my requirements. – trisct Apr 25 '20 at 16:30
  • @FedericoPoloni Similarly, for the 2d version of the fake proof, you notice that you are using two line segments intersecting at a right angle to approximate an arc. As we go to infinitesimal, the arc would seem more straight", but the approximating line segments do not, as they always form a right angle. – trisct Apr 25 '20 at 16:34
  • All patches are squares, or line segments in the 1D version, so they are "flat enough" by themselves. (At least that's how I interpret that word.) You need to add a requirement on how they are joined one to the other. – Federico Poloni Apr 25 '20 at 16:36
  • @silph "in contrast, my grade 3 brain has trouble visually accepting a similar idea for a 1cm x 1cm piece of paper, applied to sphere" - perhaps it's better to think of it in a different way. Let's say that you're not applying peaces of paper to the sphere - suppose instead that you have a perfect sphere made of some material. Then you cut it up into small pieces - they are so small that they are almost flat, but they aren't really. But you can't tell the difference. 1/2 – Filip Milovanović Apr 25 '20 at 17:25
  • @silph You then place them on the floor, and then you measure each one's surface (as if flat), then add them up (maybe you have some sort of an overhead scanner that can do that automatically). You get a number. You then cut them up even further, repeat the measurement, and you get a number that's slightly higher. You keep going. The result ever so slightly increases, approaching some limit, but never goes over it. Mathematically, you don't do it literally that way, but there's a formal way of doing it that amounts to the same thing, and produces that limit as the result. 2/2 – Filip Milovanović Apr 25 '20 at 17:25
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    I forgot to mention that this method suffers from the Schwarz lantern problem. Even if you approximate the sphere using only triangles whose vertices lie on the sphere, no matter how small an upper bound you place on the size of the triangles, it's possible for the triangles to have an arbitrarily large surface area. So as much as "value of the approximation above should tend to a fixed number", it doesn't tend to a fixed number. (But the desired number is probably the limit inferior.) – Tanner Swett Apr 26 '20 at 03:55
  • @TannerSwett Thanks for mentioning this interesting example. But I don't think ithe desired number is the limit inferior, since the limit may exist, but fail to be the desired number. And when a limit exists, limit = limit inferior. – trisct Apr 26 '20 at 06:28
  • @trisct: I don't want to ruin the party in being pedant but your answer has a serious logical flaw. You argue correctly that no matter how small you take your patch $P_i$, it can never be flattened but then you say that you take $P_i$ to be small enough and compute $\textrm{Area}(P_i)$ "as if" each $P_i$ is flat. Since $P_i$ is never flat, you can't really compute its area as you run down into the same problem that you had with the original surface $S$ because you don't know in advance how to compute the surface area of something curved, no matter how small it is. – levap Apr 27 '20 at 02:08
  • This is a real non-trivial issue which is really different from the issue you encounter if you want to define the area of a planar shape. There, you can use approximation by squares (whose areas you really know how to compute) but here there aren’t any “pieces” you can actually use. – levap Apr 27 '20 at 02:08
  • This issue arises also in the case of curves in a simpler form. How can you define the length of something curved? You divide it into small pieces $P_i$ but then again you don’t know how to compute their length. So instead, you replace each piece with a line connecting the two endpoints and get a polygonal approximation whose length you can actually compute and sum everything. This is the meaning of “compute $\textrm{Length}(P_i)$ as if each $P_i$ is flat” part. – levap Apr 27 '20 at 02:08
  • You might expect that the same would work for surfaces (and it does) but the Schwartz lantern example shows that you really need to be careful on how exactly you do it. It is wrong to say that “the Schwarz lantern problem shouldn’t arise in method 1” because the pieces “were originally on the surface” because it really doesn’t help - you don’t know how to compute their surface area. – levap Apr 27 '20 at 02:08
  • Now, one can make your story precise by specifying how you compute the area of each $P_i$ as if it is flat. One way is to project $P_i$ onto the tangent plane at one of the point and then you get a planar shape which lies on the tangent plane (but not on the surface!) and you can compute its area (again, approximating using squares) and sum everything up but this not what the most common definition “does”. Instead, it “replaces” each piece completely with a small planar parallelogram and sums the areas. In any case, you must use “pieces” which are not originally on the surface. – levap Apr 27 '20 at 02:09
  • @levap You are absolutely right. Projecting onto the tangent plane is exactly how calculus deals with curved area. I didn't want to mention that because I was mainly focusing on the idea of cutting the surface into smaller pieces. As for how to compute the areas of the pieces, even though projecting onto a tangent plane is a viable solution, there's no guarantee that this is the only viable solution. If I put that in my answer I'd somehow have to introduce a criterion on what kind of approximation is viable and what is not.This criterion, if it exists, is what I meant by hard to make precise – trisct Apr 27 '20 at 02:45
  • @trisct: The point of my (overly lengthy) comment was to emphasize that in your description there are really two separate issues: the approximation part and the "as if" part and both are equally important. Since things stay curved, no matter how small they are it doesn't help to emphasize that the pieces were originally on the surface because, in the end, you need somehow to "take them off the surface and make them planar". Other than that, great answer. +1 :) – levap Apr 27 '20 at 02:54
  • @levap I've taken your suggestion and edited the answer. Actually I have thought about the issue before but decided to avoid it. But now with your comments I think it is better to mention it rather than ignoring it. Thanks! – trisct Apr 27 '20 at 03:08
  • The Schwarz problem is described and analyzed in RADO: "WHAT IS THE AREA OF A SURFACE?" https://doi.org/10.1080/00029890.1943.11991341 where he shows that the proper definition is the limit inferior of the surface area of the inscribed polyhedron. He credits Lebesgue for the definition. – hyportnex Apr 27 '20 at 13:34
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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.

  • i'm trying to think of how i could apply this idea when explaining it to my own third grade brain. the difficulty i'm experiencing is that "flat square pieces of paper" is tangible, and a precise definition. in contrast, "liquid covering a surface" doesn't allow for actual measurement, and it's hard for me to visualize how liquid actually lays itself down to "measure" a rectangle (or sphere) surface. -- but thanks for the idea, anyways. – silph Apr 24 '20 at 07:07
  • oh, i mean that the paint idea doesn't allow me to arrive at a number to say "the area is ___ square centimetres". i'd have to start saying something like "the area is 36 mL", but then again, i have trouble actually visualizing how one mL of paint lays itself down as it covers a 2d rectangle (or surface of a sphere). – silph Apr 24 '20 at 07:20
  • Yes you can arrive at a number. Paint 1 cm², taking 0.1 ml. Paint the sphere, taking 3 ml. The area is 30 cm². Think out of your box. –  Apr 24 '20 at 07:35
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    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. – Nuclear Hoagie Apr 24 '20 at 15:39
  • Assuming a paint thickness of 1 unit, this plot shows the ratio of the volume of paint needed to the area of the sphere. When the paint thickness is 1/100 the sphere radius, you're off by 2%, when it's 1/10 of the radius, you're off by a full 10%. The ratio approaches 1 for large spheres, but there's always some error: https://www.wolframalpha.com/input/?i=plot+%28%284%2F3+pi+%28x%2B1%29%5E3%29+-+%284%2F3+pi++x%5E3%29%29%2F%284+pi+x%5E2%29+from+x+%3D+0+to+100 – Nuclear Hoagie Apr 24 '20 at 16:06
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    @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. –  Apr 24 '20 at 16:35
  • @NuclearWang This is easy to resolve. Let $V(r,\varepsilon)$ be the volumn of paint needed to paint a sphere of radius $r$ if we paint with thickness $\varepsilon$. Assuming the sphere has area $S(r)$, the relation $V(r,\epsilon)=S(r)\varepsilon$ approximately holds when $\epsilon$ is small. Then $S(r)=\lim_{\epsilon\to0}V(r,\epsilon)/\epsilon$ is the area. – trisct Apr 24 '20 at 17:02
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    @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. – Nuclear Hoagie Apr 24 '20 at 18:36
  • @trisct Why don't you differentiate between $\epsilon$ and $\varepsilon$? – Allawonder Apr 24 '20 at 18:44
  • @Allawonder Sorry 'bout that...It was a typo. – trisct Apr 24 '20 at 18:50
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    See also How do I experimentally measure the surface area of a rock? – Brian Apr 24 '20 at 19:11
  • @NuclearWang If you paint both sides of the sphere, then the thickness makes you use extra paint on the outside but save paint on the inside. According to Weyl's tube formula those effects cancel out... almost. Well, not exactly for a sphere, but exactly for a torus :) (Yes it's terrible as an answer to "what is surface area". But since the comments section has already left that station, I'm just remarking that the problem of paint of finite thickness covering a curved surface has had surprisingly serious study.) – echinodermata Apr 24 '20 at 19:43
  • @FakeMod: the amount of paint is proportional to the area. Sorry, you are clearly wrong. –  Apr 24 '20 at 20:27
  • @FakeMod Isn't it the other way around? Gabriel's Horn has infinite surface area but finite volume. – user76284 Apr 24 '20 at 22:30
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    @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." – user76284 Apr 24 '20 at 22:34
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    @YvesDaoust Did you think about FakeMod's example at all (when you wrote he/she was clearly wrong)? It has finite area, but for any constant thickness of a layer of paint you need an infinite amount of paint. – doetoe Apr 25 '20 at 00:01
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    @doetoe: this is ridiculous. Gabriel' Horn has infinite area, and will OF COURSE require an infinite amount of paint. –  Apr 25 '20 at 09:09
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    @YvesDaoust Sorry, I didn't read it well and had the following in mind (should have looked up what actually is Gabriel's Horn), my bad. I was thinking of a cusp like one that could be obtained as a surface of revolution of the graph of $e^{-x}$ for positive $x$. This has finite area and will require an infinite amount of ink. – doetoe Apr 25 '20 at 11:33
  • @doetoe: errare humanum est, perseverare diabolicum. You are drowning. –  Apr 25 '20 at 12:56
  • @doetoe if one fills the horn with a finite amount of paint, and then empties the finite amount (less a little left on the surface) out again, one has painted the entirety of the surface. This works for any object with a finite volume. – Tim Apr 26 '20 at 15:25
  • @Tim That's true (nice argument btw), but the amount of paint will not be proportional to the area. For that you will need a layer of a fixed thickness (and then you would need to take the limit for the thickness going to 0). But such a horn (not of the shape given by the $1/x$ of Gabriel's horn (which has infinite area), but $e^{-x}$ or $1/x^2$, etc, whose area also is finite, will require an infinite amount of paint for any layer thickness. – doetoe Apr 26 '20 at 15:53
  • @yves for the skeptics you can spell out the grade-school math to calculate the area: if a can of paint can cover $1m^2$, how many cans (or what fraction of a can) does it take to cover the sphere? That's how you measure the area. – alexis Apr 26 '20 at 22:00
  • @alexis: the OP doesn't seem to agree. –  Apr 27 '20 at 05:45
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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.)

J.G.
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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.

CCTO
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    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!

     – silph Apr 24 '20 at 17:12
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    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] – TonyK Apr 24 '20 at 19:32
  • ...Your idea is not without merit, by the way; it's just not quite as straightforward as it seems. – TonyK Apr 24 '20 at 19:34
  • 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! – MrWonderful Apr 24 '20 at 21:47
  • @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. – mike rodent Apr 26 '20 at 13:58
  • @TonyK The error will be in the same order of magnitude as the one introduced by the fact that you can emphatically not tile any part of the Earth with flat, rectangular tiles. In other words, the error will be just as negligible if the tiles are small enough. Your particular error can also easily be avoided by making the width of the tiles the proper fraction of the respective latitude length -- the concept here does not really depend on all tiles being evenly sized. – Peter - Reinstate Monica Apr 27 '20 at 13:24
  • @Peter: I agree entirely. See my second comment. – TonyK Apr 27 '20 at 13:53
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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)

ti7
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    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. – jawheele Apr 25 '20 at 20:26
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    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 $\pi=4$: https://math.stackexchange.com/questions/12906/the-staircase-paradox-or-why-pi-ne4 (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. – jawheele Apr 25 '20 at 20:37
  • 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. – jawheele Apr 27 '20 at 19:19
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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.

Allawonder
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  • 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. – mike rodent Apr 26 '20 at 14:24
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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.

TonyK
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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.

Daniel R Hicks
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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 $1\text{cm}^2$.

5xum
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    And, moreover, the tinier (and thus flatter) you make the pieces, the closer the area would be to $1$ square centimeter. – PrincessEev Apr 24 '20 at 06:51
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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.

Martin Argerami
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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. :-)

Ed999
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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.

Stephen Montgomery-Smith
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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

TomF
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