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How did mathematicians prior to the coming of calculus derive the area of the circle from scratch, without the use of calculus?

The area, $A$, of a circle is $\pi r^2$. Given radius $r$, diameter $d$ and circumference $c$, by definition, $\pi := \frac cd$.

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    $\begingroup$ 'how did they" — I have no idea how, but this malicious group of people surely was not doing math, because you cannot even define area without some bits of calculus. $\endgroup$ – xsnl Mar 31 '18 at 7:34
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    $\begingroup$ You can't rigorously prove it without calculus, but you can give pretty convincing heuristic arguments. See, for instance, this wikipedia article for some examples. $\endgroup$ – Arthur Mar 31 '18 at 7:37
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    $\begingroup$ It is perfectly possible to define the area of plane figures bounded by straight lines without calculus - break them up into triangles. It is also quite easy (by using polygonal approximations to the circle) to show that if it is assumed that the area exists, it scales as $r^2$. You should note that it is at least as difficult to decide the arc length of a curve (and hence $C$) as it is to define Area enclosed by a curve ($A$). The radius and diameter are straight lines. $\endgroup$ – Mark Bennet Mar 31 '18 at 7:47
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    $\begingroup$ proofwiki.org/wiki/Area_of_Circle/Proof_3, this claims to prove the formula using some simple geometry and proof by contradiction (no calculus involved). $\endgroup$ – Sil Mar 31 '18 at 10:07
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    $\begingroup$ @xsnl that seems backwards to me. At least from the Riemann formulatiom of of integrals, don't we need to know the area of a rectangle before we can use a sum of many rectangles to compute an area of a function? $\endgroup$ – Tyberius Apr 1 '18 at 0:06

11 Answers 11

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There's an interesting method using which you can approximately find the area. Split up the circle into many small sectors, and arrange them as a parallelogram as shown in the image (from wikipedia)

enter image description here

The higher the number of sectors you take, the more it tends to a parallelogram, with one of it’s height the radius $r$, and the other side half the circumference $\pi r$. Thus, its area tends to $\pi r \cdot r = \pi r^2$

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    $\begingroup$ Is this not just integration in disguise? $\endgroup$ – HBeel Mar 31 '18 at 9:15
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    $\begingroup$ @HBeel well, you do split it up into very-small approximate triangles which we then sum up, so yeah it can be interpreted as a form of integration. Nonetheless, you can see through the machinery and understand what is happening here more intuitively. $\endgroup$ – Prathyush Poduval Mar 31 '18 at 9:18
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    $\begingroup$ Upvoted because this is the same "intuitive" explanation taught to me as a schoolkid. The most tricky bit is that takes a fair bit of intuition (or at least hand-waving) to see that the "sort-of parallelogram" becomes a rectangle at the limit. $\endgroup$ – Deepak Mar 31 '18 at 11:52
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    $\begingroup$ @HBeel Any way of doing this will be integration in disguise. $\endgroup$ – lisyarus Mar 31 '18 at 13:25
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    $\begingroup$ @HBeel It depends on what do we call integration. The technique of taking the limit of approximating sums was known long before the rigorous study of integration began. $\endgroup$ – lisyarus Mar 31 '18 at 14:56
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Calculus depends on the concept of a limit, and does apply that to the problem of determining the area of a curved object. But the concept of a limit and the ability to reason about the area of a curved object using that concept existed before Calculus.

As far as recorded history is concerned, Archimedes was the first to derive $A = \pi r^2$. Though he didn't call it $\pi$, I think we can still say this is the answer to your question. His proof depends on the concept of a limit. He showed that, given a circle, with radius $r$ and circumference $c$, the area of that circle can't be more than that of a triangle with height $r$ and base $c$, and that it can't be less than the area of that triangle either. He did this by examining the area of polygons with an increasing number of sides, both inside the circle and outside the circle. This involved the concept of a limit and calculating area, which FEELS like calculus, but isn't.

Polygons with increasing number of sides inside: enter image description here Polygons with increasing number of sides outside: enter image description here

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    $\begingroup$ Welcome to MSE. Nice first answer! Please use MathJax. $\endgroup$ – José Carlos Santos Mar 31 '18 at 12:09
  • $\begingroup$ So how do we define Calculus? I would say that any infinitesimal argument is the use of Calculus. $\endgroup$ – Michael Hoppe Mar 31 '18 at 12:20
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    $\begingroup$ @MichaelHoppe If you define calculus as using limits to calculate an area, then Archimedes invented it instead of Newton or Leibniz. Calculus is much more than that, and it became its own field once we discovered that the derivative of the area function was the original function. $\endgroup$ – De Novo Mar 31 '18 at 12:27
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    $\begingroup$ This understates the subtlety of Archimedes' proof. It not only involves using limits to calculate areas (even Euclid did that!), but also requires a convexity argument which — if fully justified for curved figures — will end up using some notion of tangent line. I don't know that delineating whether something is or is not calculus is particularly productive, but Archimedes is a lot closer to doing calculus than you're giving him credit for. $\endgroup$ – Micah Mar 31 '18 at 16:32
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    $\begingroup$ @MichaelHoppe If the term calculus is ambiguous, you might make a distinction between infinitesimals and analysis or standard analysis. $\endgroup$ – Davislor Mar 31 '18 at 23:14
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Euclid initiates this is Book XII, Prop. 2., showing indirectly that there is a constant of proportionality between the area of a circle and its diameter. He does this by showing that the area of two circles goes as the square of their diameters, which forces this constant of proportionality between, in the Question's notation, $A$ and $r^2$. Lets call that constant $p$: $A = p r^2$. So Euclid shows that $p$ exists, but does not show that $p$ is related to the ratio of the length of the circumference to that of the diameter.

The preceding proposition in that book, that the areas of similar polygons inscribed in circles are in the same ratio as the squares of the diameters of the circles, gives us a method to approximate $p$ from below, by using sequences of polygons that cover more and more of the circle. Archimedes uses this method (as well as the result that a circumscribed polygon contains all the area of the circle) to bound $p$ in his Measurement of a Circle. But first, he uses these facts about inscribed and circumscribed polygons to show $p = \pi$.

His first proposition is the result you ask about: The area of a circle is equal to the area of a right triangle with one leg having the length of the radius and the other leg having the length of the circumference. That is, using the definition you give that the circumference is $\pi$ times the diameter and also that the diameter is twice the radius, $$ A = \frac{1}{2} \cdot r \cdot C = \frac{1}{2} \cdot r \cdot (\pi \cdot 2r) = \pi r^2 \text{,} $$ so $p = \pi$.

Archimedes proves this result by the method of exhaustion. It is common to claim that this method is elementary calculus, but that isn't entirely correct. As it is used here (and in many other places), it is an application of trichotomy to the completeness of the reals and is a method more commonly used in advanced calculus and real analysis. Archimedes observes that if the area of the triangle is not the area of the circle, it must be either greater or less than the area of the circle. He shows it cannot be greater, then that it cannot be less. Therefore, the area of the triangle is the same as the area of the circle.

(Using completeness of the reals, one would phrase this as: for all $\varepsilon >0$, the area of the triangle differs from the area of the circle by less than $\varepsilon$, so is the same as the area of the circle. If we imagine a shrinking sequence of $\varepsilon$s, this gives the nested sequence of intervals for the completeness link above.)

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Another way to derive the formula for the area of a circle is by decomposing it into several concentric rings (annuli) to form a "triangle" as seen in the below diagram. As the width of the concentric rings approach zero, it forms a triangle of base circumference ($C$ = $\pi$$d$) and height $r$. decomposing a circle into concentric rings diagram

Since we know the area of a triangle is $\frac{bh}{2}$, by substitution $$\frac{\pi d\cdot r}{2}$$ Since $d$ = 2$r$, $$A = \frac{\pi \cdot 2r\cdot r}{2} \rightarrow A = \frac{\pi \cdot 2r^2}{2} \rightarrow A = \pi \cdot r^2$$.

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    $\begingroup$ +1, this is what I was trying to say in my badly worded answer... $\endgroup$ – IanF1 Apr 1 '18 at 12:55
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It is easy to go from the area of a triangle to the area of a regular polygon (one whose sides all have the same length and whose angles all have the same measure) by breaking the polygon into triangles and summing the areas of the triangles. But we need to first review the formula for the area of a polygon for reference. It can be shown that every regular polygon can be inscribed in a circle. If we draw a perpendicular line from the center of the circle to any side of the inscribed polygon, that line is called an apothem. It is a fact that the area of a regular polygon is $1/2ap$ where $a$ is the apothem and $p$ is the perimeter of the polygon. It is also a fact that as the number of sides of the inscribed regular polygons increases, the lengths of the apothems of the polygons approach the radius of the circle. enter image description here

enter image description here

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The area is very much a concept belonging to Calculus. In this regard to find the area of something without using Calculus is almost an oxymoron, but we can have an intuitive (and not too general) notion of area based on the following points:

  1. Any rectangle deserves an area (i.e. it is a measurable set) and the area of a rectangle $R$ with side lengths $a,b,a,b$ is $\mu(R)=ab$;
  2. Isometric measurable sets have the same area and if a line splits some measurable set $R$ into two components $R_1,R_2$ then $R_1,R_2$ are measurable and $\mu(R_1)+\mu(R_2)=\mu(R)$. This fixes the area of parallelograms, triangles, bounded polygons;
  3. Not every compact subset of $\mathbb{R}^2$ is a polygon, so we want to extend the notion of measurability in the following way: given $K$, a compact and connected subset of $\mathbb{R}^2$, we define $\mathcal{I}$ as the family of polygons contained in $K$, $\mathcal{E}$ as the family of polygons containing $K$. If $$ \sup_{P\in\mathcal{I}} \mu(P) = \inf_{P\in\mathcal{E}}\mu(P) = L $$ we say that $K$ is measurable and define its area as $\mu(K)=L$.

Notice that 1.,2.,3. do not allow to measure unbounded sets like $\{(x,y)\in\mathbb{R}^2: x\geq 0, 0\leq y\leq\frac{1}{(x+1)^2}\}$ or subsets of $\mathbb{R}^2$ with countable connected components, like $\bigcup_{n\in\mathbb{N}}\{(x,y): (x-n)^2+y^2\leq \frac{1}{(n+1)^3}\}$.
Still we may use this naive notion of area to compute the area of the unit circle. By considering the regular polygons in $\mathcal{I}$ and $\mathcal{E}$ we have $$ \mu\left(\{(x,y):x^2+y^2\leq 1\right) = \sup_{n\geq 3}n\cdot\sin\frac{\pi}{n} = \inf_{n\geq 3}n\cdot\tan\frac{\pi}{n}=\pi $$ hence the area of the unit circle is just half the length of its boundary and $\pi\in(3,4)$ follows from the fact that

If $A,B$ are two convex, bounded sets in $\mathbb{R}^2$ and $A\subsetneq B$, the length of $\partial A$ is less than the length of $\partial B$. (see also this question)

An Archimedean-like geometric approach for the numerical approximation of $\pi$ is the following: a circle with radius $1$ can be decomposed as the union of an octagon with side length $\sqrt{2-\sqrt{2}}$ and eight circular segments. Such segments can be approximated by parabolic segments, whose area is simply $\frac{2}{3}\text{base}\cdot\text{height}$. The parabolic segments are slightly smaller than the corresponding circle segments, hence the following construction

enter image description here

leads to the lower bound $$\pi > \frac{16}{3}\sqrt{2-\sqrt{2}}-\frac{2}{3}\sqrt{2}= 3.13914757\ldots$$ whose accuracy is comparable with the actual Archimedean approximation $\pi\leq\frac{22}{7}$. The parabolic method applied to the regular dodecagon leads to the nice bound $$ \pi > 4\sqrt{6}-4\sqrt{2}-1 = 3.1411\ldots $$ which also explains the proximity between $\pi$ and $\sqrt{2}+\sqrt{3}$.

A broader notion of area comes from defining the Riemann/Lebesgue integral and defining the area of $K\subset\mathbb{R}^2$ as $$ \mu(K) = \iint_{\mathbb{R}^2}\mathbb{1}_K(x,y)\,dx\,dy $$ provided that the RHS makes sense. 1,2,3 are fulfilled, and this also allows us to measure the previous troubling sets. Additionally, this gives us new ways for approximating $$\pi=4\int_{0}^{1}\sqrt{1-x^2}\,dx = 6\arcsin\frac{1}{2} = 3\sum_{n\geq 0}\frac{\binom{2n}{n}}{16^n(2n+1)}=3+\frac{1}{8}+\frac{9}{640}+\left(0\leq\varepsilon\leq 3\cdot 10^{-3}\right) $$ also through $$ \frac{1}{5\cdot 4^8}\approx\int_{0}^{1}\frac{x^8(1-x)^8}{1+x^2}\,dx=4\pi-\frac{188684}{15015}.$$


Improving the answer: why is it enough to study the unit circle? Because a circle is highly symmetric, os by applying a dilation with factor $\lambda>0$ the area is multiplied by $\lambda^2$ and the length of the boundary is multiplied by $\lambda$ (the fundamental property of affine maps is that they preserve the ratio of areas). In particular if the area of the unit circle is $\beth$, the area of a circle with radius $R$ is $\beth R^2$.

How is this related with the length of the boundary? Through convexity. If we consider the annulus between two concentric circles with radii $R$ and $R+\varepsilon$, the area of the annulus divided by $\varepsilon$ tends to the length of the boundary of the inner circle as $\varepsilon\to 0^+$. In particular if $\beth R^2$ is the area of the circle with radius $R$, the length of its boundary is given by $$ \frac{d}{dR}\beth R^2 = \color{red}{2\beth}R.$$

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  • $\begingroup$ Euclid is already comparing areas in Book I, Prop. 34. (By congruence, so, as per normal, shows two things are the same without measuring either of them.) Prop. 35 compares (barely) different objects, by subtracting a common subobject from two congruent objects to yield two non-congruent objects with the same "area" (whatever that is). So Euclid has your items 1 and 2 in Book I (with your 1 generalized to parallelograms). $\endgroup$ – Eric Towers Apr 1 '18 at 22:04
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    $\begingroup$ When you switch to approximating $\pi$, you skip right past OP's question: why is the constant of proportionality of areas the same as the constant of proportionality of circumferences? $\endgroup$ – Eric Towers Apr 1 '18 at 22:05
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    $\begingroup$ I agree that there is a constant of proportionality, which is what you and Euclid show. Why is it the same constant, which is what the OP asks? $\endgroup$ – Eric Towers Apr 1 '18 at 23:28
  • $\begingroup$ @EricTowers: I have updated my answer by including clearer explanations at the end. $\endgroup$ – Jack D'Aurizio Apr 1 '18 at 23:51
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    $\begingroup$ The clearer explanation at the end is not responsive to OP's "How did mathematicians prior to the coming of calculus derive the area of the circle from scratch, without the use of calculus?" $\endgroup$ – Eric Towers Apr 2 '18 at 12:50
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Monte Carlo method.

Pick a random x/y by throwing lots of dice. Ensure the points fall randomly on a square of width equal to circle's diameter. Repeat many times, or until satisfied, very bored, or until Happy Hour and Martinis are on the Casino (hence 'Monte Carlo' method. It is just about April 1st).

Take the ratio of the points that fall in the circle to those that fall outside. Note how as the number of random points increases, the ratio converges on a certain amount.

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    $\begingroup$ Good answer, but I doubt ancient mathematicians used Montecarlo method to derive the area $\endgroup$ – user Mar 31 '18 at 17:15
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    $\begingroup$ no, that is nonsense. there is no evidence that "mathematicians prior to the coming of calculus derive the area of the circle" n this way. $\endgroup$ – miracle173 Mar 31 '18 at 21:36
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    $\begingroup$ The question specifically asks how mathematicians worked out the area of a circle before the introduction of calculus. This derivation came later, and is therefore not an answer to the question. $\endgroup$ – Xander Henderson Apr 1 '18 at 2:37
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    $\begingroup$ The question in the title asks how to derive the area of a circle without calculus, which this question does. I took the the specification "before calculus" in the body to mean "without calculus", which is closer to OP's intent. Upvoted! $\endgroup$ – user1717828 Apr 1 '18 at 11:02
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    $\begingroup$ The title is not the question, it is a marker/pointer to help find it more easily. If it doesn't match the body, it should be edited, not used to justify posts NAA. $\endgroup$ – Nij Apr 2 '18 at 21:20
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Since two circles are similar, their areas are proportional to the squares of their radii. Now the area of a circle with radius $1$ is the proportional factor, just call it $\pi$.

Addendum. If you insist in the usual definition of $\pi$, i. e., the ratio of circumference to diameter, I doubt there will be no connection to my approach without an infinitesimal argument, that is calculus.

One imho nice approach would be to define a discrete $\pi_n$ for every regular $n$-gon, namely $\pi_n:=n\cdot\tan(\pi/n)$, which is the ratio of the $n$-gon’s circumference to its diameter. (Here the radius $r_n$ is defined by the distance of the $n$-gon’s center to one of its sides.)

In that case the area $A_n=\pi_n\cdot r_n^2$ and the circumference is $2\pi_n\cdot r_n$.

From here we see that if the proportional factor for the area is $\pi_n$, the factor for the circumference $2\pi_n$. That’s true for $n$-gons from above.

But how to show that without calculus for circles?

I know that there are some curves which lengths may be calculated purely algebraic, such as Neil’s parabola $t\mapsto(t^2,t^3)$. But I don’t know an algebraic way for the circle. Since a purely algebraic rectification was thought impossible since Aristotle, I presume there is none for the circle.

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    $\begingroup$ That is not how $\pi$ is usually defined. $\endgroup$ – Rory Daulton Mar 31 '18 at 11:28
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    $\begingroup$ But how to prove that this is the same pi as that defined in the OP as c/d? $\endgroup$ – IanF1 Mar 31 '18 at 11:31
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    $\begingroup$ @RoryDaulton I agree, but the OP wants a derivation of the circle’s area without calculus, and I gave one. $\endgroup$ – Michael Hoppe Mar 31 '18 at 11:38
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    $\begingroup$ This is not a very satisfactory derivation, mainly because that's not how $\pi$ is commonly defined based on circular geometry. It is imperative that one go from the $\pi = \frac cd$ definition to $A = \pi r^2$, in my opinion.' $\endgroup$ – Deepak Mar 31 '18 at 11:51
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    $\begingroup$ In my course Calculus 1, $\pi$ was defined as the smallest positive zero of $\sin$, which was defined as a power series. Now what do you call “commmonly defined”? See also my addendum, please. $\endgroup$ – Michael Hoppe Mar 31 '18 at 12:04
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  1. Get two large sheets of papyrus with uniform thickness
  2. Cut a circle of unit radius from one, and a unit square from the other
  3. Measure the weight of each shape
  4. From their ratio you get the area of the circle
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    $\begingroup$ BTW, this method is very similar to the one posted by DRPR. $\endgroup$ – GregT Apr 3 '18 at 8:38
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By trial and error and by numerical approximations. The ancient mathematicians (Babylonians, 1800 BC) tried to square a circle (approximating the area of circle with a square, constructing a square with the same area as a circle, proved impossible in 1882 AD), to calculate $\sqrt{2}$, $\pi$, etc.

The precise calculation of the area of circle requires an infinitesimal analysis (calculus).

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I am going to use a non conventional method here:-

If you stack circles on top of circles , it forms a cylinder of considerable height.

Now the volume of this cylinder is equal to $Vol\, (cylinder) =A*H \quad\\where\, A=\text{Area of the circle (still unknown)}\\ \text{and H = height of the cylinder that you have created} $

Now the volume of the cylinder is equal to the increase of liquid level when the e object is immersed in the liquid (usually water).(Application of Archimedes principle) .This is how you get the volume of the cylinder.

Divide the volume by $H(\text{measurable quantity})$ and you get the Area.

To get the formula of the area of the circle you may have to use numerical methods.

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  • $\begingroup$ This isn't just non-conventional, it's nonsensical. A circle has zero volume. Make an infinite stack of them and it's still a height of zero. $\endgroup$ – Nij Apr 1 '18 at 7:54
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    $\begingroup$ @Nij Take 10 coins and stack them up on top of each other . i am sure you will get certain height .Then divide the volume you get with that height . If you dont get the area of the circle/base of the cylinder , then what do you get? $\endgroup$ – DRPR Apr 1 '18 at 12:59
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    $\begingroup$ Coins are already a cylinder, so you should have just measured one of them already. They're not circles which have zero volume. $\endgroup$ – Nij Apr 1 '18 at 19:56
  • $\begingroup$ Do you think that just because Archimedes took a bath one time in his life, he used water to solve math problems? He was a troublemaker who was properly disposed of by the Roman army. What is interesting is how you can tilt the cylinder to get it exactly half full. $\endgroup$ – richard1941 Apr 5 '18 at 1:46
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    $\begingroup$ The original question was "How did mathematicians prior to the coming of calculus ...". Well, we are talking about time before Arkhimedes then. The profession mathematician didn't exist, all its practitioners were philosophers, engineers etc. And their concepts were much less abstract than ours. So it's plausible to say that they didn't differentiate the concept of area from the material whose area they were measuring. So from a historical point of view, it's a perfectly correct answer IMO. How would a modern mathematician approach this problem, is a different question entirely. $\endgroup$ – GregT Apr 5 '18 at 13:37

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