#### (An archive question of the week)

While gathering answers to questions about volume and surface area formulas, I ran across this question that applies to all of them: Given all the approximations and assumptions we make in the derivations we show (without calculus), how can we claim that the resulting formula is exact? Or can we?

## How can approximations yield an exact formula?

Here is the question, from 2003, which starts by presenting a (reversed) version of the first derivation I showed last week relating the volume of a sphere, \(S = \frac{4}{3}\pi r^3\), to its surface area, \(S = 4\pi r^2\):

Sphere Surface Area Precision How can the formula 4*pi*r^2 be precise? The way that one arrives at this answer seems to have a flaw. Consider a sphere with a pyramid drawn inside it and the point of the pyramid at the center of the sphere and the base of the pyramid on the surface of the sphere. The volume of this pyramid would be r*(1/3)*(area of the base of the pyramid) Now we duplicate this pyramid to fill the entire sphere. This would mean that [r*(1/3)*(area of the base)]+[r*(1/3)*(area of the base)]+[r*(1/3)*(area of the base)]+... should be equal to the volume of the sphere. Now let's use the distributive property. r*(1/3)*(area of the base + area of the base...) This also means r*(1/3)*(surface area of the sphere) This should still equal the volume of the sphere, so let's say that the volume formula for the sphere is equal to it. (4/3)*pi*r^3 = r*(1/3)*(surface area) Now solve for the surface area 4*pi*r^2 = Surface Area of a SphereBut how can the area of the base of the pyramid be found with any precision?You can just do the w*h, but since the base of the pyramid is flat and the surface of a sphere is curved it will be a tiny bit off.You can draw many pyramids inside the sphere, making the answer more and more precise, but the answer will never be exact.Where does my logic break down, or am I right that one can't find the exact surface area of a sphere? I have thought about Zeno's Paradox having some relation to this. An infinite number of points can add up to a finite number in this geometric series (1/2+1/4+1/8+1/16+1/32...). Maybe if you put an infinite number of pyramids in the sphere the bases will add up to the exact surface area.

These are good things to be concerned about; the derivations we have been discussing have been only summaries, not careful proofs, and it’s important to see the difference.

Ultimately, this question lies at the foundation of calculus, though the reasoning was geometrical. Without calculus, or related ideas, our proofs are only rationalizations.

## The same problem is present in π

Doctor Rick replied:

Hi, Alex. The problem you are concerned about arises before you get to the sphere: in the calculation of the value of pi. This was first done by Archimedes, who proved that the ratio of the area of a circle to the square of its radius is between 3 10/71 and 3 1/7. He did this by computing the areas of inscribed and circumscribed 96-sided polygons; the area of the circle must lie between these two areas. In principle we could continue doubling the number of sides and narrowing down the value of pi as far as we wish. By other methods we have in fact gotten billions of digits of pi. Still, we have not found the _exact_ value of pi. Butwe know that it is a particular number, we just can't express it exactly.

This is a common source of questions, from people who ask, say, how can a circle have a circumference when π is an irrational number. A number we can’t write out is still a number; it has an exact value though we can’t say exactly what value it has.

The fact that it is a particular number has to do with the theory oflimits, which is part of the foundation of calculus. It is indeed related to Zeno's paradoxes. The following Web page discusses both matters in the history of Greek mathematics: Basic Ideas in Greek Mathematics (Michael Fowler) http://www.phys.virginia.edu/classes/109N/lectures/greek_math.htm See also this description of finding the volume of a sphere by a method different from yours: Surface Area and Volume of a Sphere http://mathforum.org/library/drmath/view/55230.html

The *Ask Dr. Math* link is to the question I started with last time, which links to two different methods. The other link is a nice overview of Greek ideas that were precursors of calculus, including how Archimedes approximated π and found the formula for the area of a circle, and culminating in his determination of the surface area of a sphere. It doesn’t quite deal explicitly with the question of exactness, but gives some valuable background.

Alex responded:

Are you saying that Archimedes trying to find pi issimilarto my problem (can't be found exactly) or are you saying that the formula isimprecisebecause the value of pi is irrational and we can't give a precise answer in decimal form?

Before I continue what Alex said, I’ll interject Doctor Rick’s response to this first paragraph:

Hi, Alex. I'm saying both. Since the volume of a sphere involves pi, it necessarily involves the same issue that Archimedes dealt with; but also the method Archimedes used has enough similarity to the one you describe that I think it's worth your study.This does not mean that either formula is not exact, however!

So “both” does not include the idea of the *formula* not being exact, but that it depends on π, and therefore the *value* you get is only as precise as the value one uses.

## The objection restated

Returning to what Alex wrote,

If it's the second answer, what I mean is that the surface area formula would be imprecise because when you put the pyramid inside the sphere and find the area ofthe base of that pyramid, it won't be the same area asthe surface of the sphere above it. A 2D-model/cross-section would look like a segment of a circle. The two lines that make up the central angle and extend to the edge of the circle would be the sides of the pyramid and the line that connects those two lines would be the base of the pyramid. The arc of the circle above it would be like the surface of the sphere above the surface of the flat base of the pyramid (the line connecting ends of central angle). The formula 4*pi*r^2 is like measuring the "base of the pyramid" and saying that measure is the same measure as the arc above it, which is false. I'll try to draw a picture: /|\ / | \ / | \ ./ | | \ | | \ | / \ | / \|/ The period is the center of the circle. The slash marks extending out form the central angle and the sides of the pyramid. The line connecting the ends of those slashes is the "base of the pyramid." The curved thing on the right is the arc of the circle.The base of the pyramid is of course not the same measure as the measure of the arc.

He is right: In Archimedes’ work with π, the chords forming the edges of the polygons are not the same length as the arcs of the circle that they approximate; and the bases of the pyramids in the sphere derivation are not parts of the sphere itself.

Another way to visualize it would be like forming a round object out of a lot of pyramids. It would be nearly round but not quite. When you found the surface area of that nearly round object it would becloseto the surface area of a corresponding sphere (if you drew a tight fitting sphere around the nearly round object) butit would not be exactly correct. The math is like this (=~ means approximately equal to): r*(1/3)*(area of the base of the pyramid + area of the base of the pyramid...) = r*(1/3)*(surface area of the nearly round object) = (4/3)*pi*r^3 = r*(1/3)*(surface area of the nearly round object) = 4*pi*r^2 = Surface Area of the nearly round object =~ surface area of the sphere The formula 4*pi*r^2 is finding the surface area of the nearly round object and not the sphere around it, so it must be a little bit off. Is there a fault in my logic? Also, even if pi is irrational couldn't you still write an exact answer (assuming that I’m wrong and the formula is precise). For example, if the radius was 2 cm you could write the surface area as: 16*pi cm^2

This last paragraph touches on an important distinction: The formula claims that the area is exact, when expressed in terms of the symbol π, even though we can’t write an exact numerical value.

## No perfection until the limit

Here is Doctor Rick’s answer to this part:

Do I understand correctly that you are referring to the derivation of the volume of the sphere assuming that we have already derived its surface area? This is the approach taken in the Archive page to which I referred you: first derive the surface area, then the ratio of volume to surface area. Do you have no problem with the derivation of surface area? I see the same sort of difficulty in both cases!

Alex’s version is actually the reverse of mine, assuming the volume has already been derived and using it to find the surface area; but the idea is the same.

You're correct thatthe surface area of a compound of pyramids is not exactly equal to the surface area of the sphere. But thevolumeof the compound of pyramids is not exactly r/3 times the surface area, either. Note that you used r as the height of each pyramid, but this is not exactly true; r is the slant height of the pyramid, which is slightly greater than the height, as your figure shows. Thus any compound of pyramids with a finite number of faces has a volume that is slightly less than r/3 times the surface area of the sphere. The point is thatwe imagine continuing the process of adding faces to the polygon without limit. The more faces it has, the closer the ratio of volume to surface area comes to r/3, because the slant height comes closer to the height.

This is the key idea: for any *finite* number of pyramids (together constituting a polyhedron inscribed in the sphere), the volume and surface area will both be a little less that those of the sphere; but both get closer and closer, while retaining the same relationship. It is only “in the limit” that the formula is exactly correct, not at any point on the way there.

To make a solid proof, we need to do the same thing Archimedes did, and calculate another ratio that we can prove to begreaterthan the ratio for the sphere. Then we need to show thatthe two ratios both approach r/3, getting as close as you wish as long as you choose a number of faces that is great enough. When we have done this, we have established that the ratio of volume to surface area of a sphere is _exactly_ r/3. Technically, we say that r/3 isthe _limit_ of the ratio of volume to surface areaas the polygon approaches a sphere.The limit is an exact value, not an approximation.

For π, Archimedes found that it is *between* a lower value and an upper value. For a full proof of the volume (or surface area) of a sphere, we can similarly show that it lies *between* two polyhedra, whose volumes both approach the same number, which must be the volume of the sphere.

Limits are a concept from calculus, and this is where we make the leap from mere approximations to exact values.

### Similar problems with limits

See the following answer in the Dr. Math archives for some further discussion of limits that may help you understand why a limit is exact. The sort of difficulty you are facing is frequently raised in the context of the fact that 0.9999... (repeated without end) equals exactly 1. The Infamous .999... = 1 http://mathforum.org/library/drmath/view/55748.html

This is discussed in the post Frequently Questioned Answers: 0.999… = 1, which mentions the same link, though it uses others like it.

.

Here is a more in-depth explanation of limits, in the context of calculus where they are dealt with formally. If it's too much for you, don't worry; I just want you to see thatthe issue is an important one to which mathematicians do pay careful attention. Understanding the Need for Limits http://mathforum.org/library/drmath/view/53754.html

This page is the basis for the post What’s the Point of Limits?

Alex almost had the idea, responding,

So you're saying that becausethe nearly round object's surface areais so incredibly close to the surface area of the corresponding sphere that it might as well be the same using the concept of limits?

Close. Doctor Rick replied:

Hi, Alex. Something like that, yes.The focus isn't on any one "nearly round object,"but on a *sequence* of polyhedra with increasing numbers of sides. The surface area of each is greater than that of the one before it in the sequence, but the surface area never exceeds that of the sphere. Thus the difference between the surface area of the polyhedron and the surface area of the spherekeeps getting smalleras you go farther in the sequence.

This is important in any discussion of limits, as it was when we looked at 0.999…: **the limit is not any one of the numbers we reach on the way, but the destination that we are approaching**.

The limit concept is like a game: On your move, you get to pick any positive number, as small as you like. On my move, I have to find a polyhedron in the sequence such that the difference between its surface area and that of the sphere is less than the number you picked. If I can always do that, then the limit of the sequence of surface areas isexactly the sameas that of the sphere. With care, we could show that this is always possible. In other words, the difference between the areas of a polyhedron in the sequence and the sphere can be made smaller than any number you can name. The only number that is smaller than any positive number is zero. Therefore we say thatthe surface areas of the polyhedra approach a LIMIT that is EXACTLY the surface area of the sphere. Here's another way to see that the limit of the polyhedron surface areas isexactly4*pi*r^2. We know that the polyhedron surface area is never greater than 4*pi*r^2. Thus the limit cannot be greater than 4*pi*r^2, either. If the limit is not *exactly* 4*pi*r^2, then it must be *less* than this. But if so, then we could show that at some number of faces, the polyhedron area was greater than 4*pi*r^2. Beyond that point, the polyhedron surface areas would get *farther* from the limit, which contradicts the definition of a limit.

Another way to express this is that for each of the polyhedra we use, we have found the relationship between the volume and the area; that relationship, true for all of them, is also true (exactly) in the limit, so we know that the volume is equal to that of a pyramid whose base is the surface area of the sphere, and whose height is the radius: $$V = \frac{1}{3}Sh = \frac{1}{3}4\pi r^2\cdot r = \frac{4}{3}4\pi r^3$$

Alex concluded:

Thanks a lot, Dr. Rick! I understand how that formula can be precise now. I first stumbled upon this site when I was looking for a formula to calculate Pi, which I them implemented into a program. Ever since then I've loved this site. It's a great service.

That’s what we like to hear.

Dr. Ibrahim OnaranHi,

As I understand that here you are relating the surface area of a sphere with its volume using limit and pyramidal shapes. As in the pyramid if the height is linearly proportional to the surface of the intersection at that height, the volume is 1/3 times the height and the base area like also in cones and other shapes with different base shapes. The surface of the sphere and the radius exactly fits in this definition, therefore we have such relation.

However, using limit we can directly compute the surface area of the sphere. Assume that we devide the angle phi on x-y plane into M sections and angle theta on x-z plane into N sections. They would be the angles of the spherical coordinates, or they would be like earth’s latitude and longitude (See https://cdn1.byjus.com/wp-content/uploads/2018/11/maths/2017/08/07072953/Sphere-Diagram-300×300.png for the setup I am describing). Lets label each section of phi as dp = 2*pi/M and theta as dt = pi/N. We can express the area of a rectangle that lives between phi and theta lines as,

A(m,n) = (R*dt)*(R*sin(n*dt)*dp) for mth phi and nth theta line,

and surface area as,

S = Sum(m = 0 to M-1) Sum(n=0 to N-1) A(m,n)

Where R is the radius of the sphere.

Since R, dp, dt are constants we can move them out of the summation.

S = R^2*M*dp*dt*sum(n=0 to N-1) sin(n*dt), here M*dp = 2*pi

S=R^2*(2*pi)*dt*(1+cos(dt))/sin(dt)

If we let dt goes to zero, the term dt*(1+cos(dt))/sin(dt) goes to 2, hence S=4*pi*R^2.

Note:

Sum(n=0 to N-1) sin(n*dt) = IM(sum(n=0 to N-1) exp(j*dt)^N) , if you use geometric sum formula and a bit of math to simplify the expression you would obtain (1+cos(dt))/sin(dt).

Dave PetersonThis post is not about how to actually calculate surface area, but about the validity of limits for such a calculation. You apparently missed the link to the previous post, Volume and Surface Area of a Sphere – Without Calculus, which derives the formulas. Yours is similar to the first derivation there.

Jeffrey R TimmisMathematics can not be an exact science since some of the most important numbers are irrational. Any results using irrational numbers will always be a close approximation. It may be that there is a chapter of mathematics yet to be discovered.

Jeffrey R Timmis

Dave PetersonHi, Jeffrey.

Actually, I think you have it backward.

If “exact science” means that all numbers involved must be known exactly, then mathematics is the

onlyexact science! (If you count it as a science at all.)All experimental sciences rely on

measurements, which can never be exact, and oninferencesfrom data, which can never be entirely certain.In mathematics, on the contrary, we can give a name to an irrational number like pi, and talk about it exactly. Moreover, we can state

axiomsthat define the subject we are studying, and make logical inferences that arecertain, given those axioms.What is inaccurate is the

applicationof mathematics to the real world, which depends on the accuracy of the data as well as on the accuracy of approximations to mathematical constants (which are far, far more precise than any measurement in the universe).The post Significant Digits: Measurements and Exact Numbers has some relevance to this. You may also find this definition of “exact science” worth pondering.