Archimedes and Early Infinitesimals: The Road to Calculus

What Were Archimedes And Early Infinitesimals?

Archimedes and early infinitesimals is the name for the chain of methods, from ancient Greece to the 1650s, that measured curved areas and volumes by cutting them into ever smaller or infinitely thin pieces. Archimedes (287 to 212 BC, Syracuse) used two such methods: the rigorous method of exhaustion for his proofs and a looser "mechanical method" of slices for discovery. Seventeenth-century mathematicians, including Kepler, Cavalieri, Fermat and Wallis, revived and extended those ideas, and their work fed directly into the calculus of Newton and Leibniz.

In modern notation, every one of these methods is computing a definite integral, the limit of a sum of thin pieces:

$$\text{Area} = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k),\Delta x = \int_{a}^{b} f(x),dx.$$

Here $f(x_k)$ is the height of the $k$-th thin strip, $\Delta x = \frac{b - a}{n}$ is its width, and $n$ is the number of strips. The condition is that $f$ is a well-behaved (for example, continuous) function on $[a, b]$. Archimedes never wrote this formula, but his answers match it exactly, which is why historians treat his work as the deepest root of integral calculus. The wider story, from Babylon to the 19th century, is on the history of calculus page; this article stays with the early infinitesimal methods themselves.

What Is The Method Of Exhaustion?

The method of exhaustion finds the area of a curved shape by fitting inside it a sequence of polygons whose areas get as close as you like to the true area, then proving that the area can be neither more nor less than a target value. The idea is credited to Antiphon in the late 5th century BC and was made rigorous by Eudoxus of Cnidus (about 408 to 355 BC), who used it to prove that a cone's volume is one third of the cylinder with the same base and height.

The classic exhaustion results run from Euclid to Archimedes:

• Euclid's Elements, Book XII: circles are to one another as the squares on their diameters, and the cone is one third of its cylinder.

• Archimedes' Measurement of a Circle: polygons with 96 sides pin $\pi$ between $3\frac{10}{71}$ and $3\frac{1}{7}$.

• Archimedes' Quadrature of the Parabola: a parabolic segment is $\frac{4}{3}$ of its inscribed triangle.

Every exhaustion proof ends with a double contradiction. Suppose the area is larger than the target; then a polygon can be built that breaks a known inequality. Suppose it is smaller; a different polygon breaks another. Only the target value survives, so the proof never needs an "infinitely small" piece or an actual infinite sum.

That last point is easy to miss. Exhaustion is closer to the modern limit than to the infinitesimal: it uses as many pieces as you like, always finitely many, which is the same move Weierstrass made 2,000 years later. The Greeks had good reason to avoid actual infinity, since Zeno's paradoxes had shown how quickly it produced contradictions.

How Did Archimedes Find The Area Of A Parabolic Segment?

Archimedes proved that a parabolic segment (the region between a parabola and a chord across it) has area exactly $\frac{4}{3}$ of the triangle inscribed in it. He did it by filling the segment with triangles whose areas form a geometric series.

The construction runs in stages:

1. First triangle. Join the chord's ends to the point of the parabola where the tangent is parallel to the chord. Call its area $T$.

2. Second stage. In each of the two leftover gaps, build a triangle the same way. Each has area $\frac{T}{8}$, so together they add $\frac{T}{4}$.

3. Every later stage. Each new stage adds $\frac{1}{4}$ of the area added by the stage before.

The total is the geometric series

$$T + \frac{T}{4} + \frac{T}{16} + \frac{T}{64} + \cdots = T \cdot \frac{1}{1 - \frac{1}{4}} = \frac{4}{3},T.$$

Archimedes did not "sum to infinity" the way this line does. He proved instead that after $n$ stages the triangles plus one third of the last stage's area equal exactly $\frac{4}{3}T$, then used the double contradiction to rule out every other value.

What Is The Method, And Why Did Its Rediscovery Matter?

The Method is a letter from Archimedes to Eratosthenes, librarian at Alexandria, explaining how he discovered results before proving them. According to the MacTutor biography of Archimedes, he wrote that certain things first became clear to him "by a mechanical method", although they had to be proved by geometry afterwards. The work was thought lost until J L Heiberg, a professor at the University of Copenhagen, identified it in the palimpsest in the summer of 1906.

The mechanical method treats a region as made of parallel slices and balances those slices on an imaginary lever:

• Slices as weights. A region becomes a stack of line segments (for areas) or flat cross-sections (for volumes), each with a weight equal to its length or area.

• The law of the lever. Two weights balance when weight times distance from the fulcrum is equal on both sides.

• Known centres of gravity. If one figure's balance point is known, the lever reveals the other figure's area or volume.

Archimedes did not count this as proof. Treating an area as a sum of infinitely many lines is exactly the step he could not justify, so he re-proved each result by exhaustion. The palimpsest's later history adds a twist: a colophon in the prayer book dates it to 13 April 1229, and the book was auctioned by Christie's in New York in 1998, after which modern imaging recovered much more of the hidden text.

Why did 1906 matter so much? Before then, historians saw only Archimedes' polished exhaustion proofs. The Method showed his working: he reasoned with indivisible slices, the same idea Cavalieri would rediscover independently in 1635, almost 1,900 years later.

How Did Archimedes Find The Volume Of A Sphere?

Archimedes showed that the volume of a sphere is two thirds of the cylinder that just encloses it, which gives the modern formula $V = \frac{4}{3}\pi r^3$. He valued this result so highly that he asked for a sphere inside a cylinder to be carved on his tomb, and the Roman writer Cicero reported finding that tomb in 75 BC.

The slicing argument from The Method, in modern notation, compares three solids of height $r$: a hemisphere of radius $r$, a cylinder of radius $r$, and a cone of radius $r$. Cut all three at height $z$ above the base:

$$\text{Hemisphere slice: } \pi\left(r^2 - z^2\right)$$

$$\text{Cylinder slice minus cone slice: } \pi r^2 - \pi z^2$$

The two expressions are equal at every height $z$. Equal slices at every level mean equal volumes, so

$$V_{\text{hemisphere}} = \pi r^3 - \frac{1}{3}\pi r^3 = \frac{2}{3}\pi r^3, \qquad V_{\text{sphere}} = \frac{4}{3}\pi r^3.$$

A full sphere sits inside a cylinder of radius $r$ and height $2r$, with volume $2\pi r^3$, and $\frac{4}{3}\pi r^3$ is exactly $\frac{2}{3}$ of that. Today the same calculation is a volume of solids of revolution problem, done with the integral $\int_{-r}^{r} \pi\left(r^2 - z^2\right),dz$.

What Did Archimedes Have, And What Did He Lack?

Archimedes came closer to integral calculus than anyone before the 1600s, yet he lacked the pieces that turned area-finding into a general subject. Sorting his work into what he had and what he lacked shows exactly where calculus begins.

What Archimedes had:

• The core idea of integration. Areas and volumes as limits of sums of thin pieces.

• Rigour. His exhaustion proofs meet modern standards of proof for the results they cover.

• Correct answers to hard problems. Parabolic segments, spheres, spirals, and centres of gravity.

What he lacked:

• A general algorithm. Each shape needed a new, clever construction. Nothing like "the integral of $x^n$ is $\frac{x^{n+1}}{n + 1}$" applied to everything at once.

• Notation. No symbols for functions, sums or integrals, so every argument was written out in geometric words.

• The derivative and the link to it. The fundamental theorem of calculus, which ties areas to rates of change, appears nowhere in his work.

That last gap is the decisive one. Calculus as a subject begins when Newton and Leibniz saw that finding areas is the reverse of finding tangents, so one set of rules handles both. Archimedes owned half the subject and none of the bridge.

How Did Kepler And Cavalieri Revive Infinitesimals?

Kepler and Cavalieri revived infinitesimal reasoning in Europe in the early 1600s, before The Method had been found, by treating curved shapes as made of infinitely many tiny pieces. Their work turned Archimedes' private discovery tool into a public technique.

Johannes Kepler (1571 to 1630) published Nova stereometria doliorum in Linz in 1615, a study of the volumes of wine barrels and other solids of revolution. He treated a circle as infinitely many thin triangles meeting at the centre, each with height $r$, so their bases add up to the circumference $2\pi r$:

$$\text{Area} = \frac{1}{2} \cdot r \cdot 2\pi r = \pi r^2.$$

Bonaventura Cavalieri (1598 to 1647) set out a full theory of indivisibles in Geometria indivisibilibus continuorum (1635). Its central rule, now called Cavalieri's principle, says: if two solids lie between the same two parallel planes, and every plane parallel to them cuts both in cross-sections of equal area, then the two solids have equal volumes. That is the same comparison Archimedes used for the sphere.

Cavalieri's indivisibles were lines (for areas) and planes (for volumes), objects one dimension lower than the shape they built. Kepler's infinitesimals were tiny pieces of the same dimension, such as thin triangles. The difference matters, and it is exactly what Cavalieri's critics attacked.

How Did Fermat And Wallis Push The Ideas Toward Calculus?

Fermat and Wallis turned slicing arguments into something close to algebra, which is the step that made a general calculus possible. Both worked in the generation just before Newton.

Pierre de Fermat (1601 to 1665) did two things:

• Maxima, minima and tangents. His method of adequality, in a treatise circulated around 1636, compares $f(x + e)$ with $f(x)$, divides by $e$, and then sets $e$ aside. It is the derivative in all but name.

• Areas under $y = x^n$. He split the interval into pieces of unequal length that shrink in a geometric progression, which turns the area into a geometric series he could sum.

John Wallis (1616 to 1703) published Arithmetica infinitorum in 1656. He computed areas by ratios of sums, extended the area rule for $x^n$ to fractional and negative powers, and found the infinite product for $\pi$ now named after him. He had introduced the symbol $\infty$ a year earlier, in a 1655 tract on conic sections. MacTutor's Newton biography records that the young Newton read Wallis's indivisible method for areas under parabolas and hyperbolas, so the line from Archimedes runs straight into Newton's own notebooks.

Examples Of Archimedes And Early Infinitesimals

Each example reproduces a historical computation in modern notation, then checks it against the integral that calculus would use.

Example 1

Use Archimedes' series to find the area of the parabolic segment cut from $y = x^2$ by the line $y = 1$.

The inscribed triangle has vertices $(-1, 1)$, $(1, 1)$ and $(0, 0)$:

$$T = \frac{1}{2} \cdot 2 \cdot 1 = 1$$

$$\text{Segment} = \frac{4}{3},T = \frac{4}{3}$$

Check with an integral:

$$\int_{-1}^{1} \left(1 - x^2\right),dx = 2 - \frac{2}{3} = \frac{4}{3}$$

Final answer: the segment's area is $\frac{4}{3}$.

Example 2

Find the area of the same parabolic segment by reasoning about its shape.

A tempting first move is to say the segment is "a triangle with a bent side", so its area should be close to the triangle's, $T = 1$. Test that guess. The construction in the parabola section places two more triangles in the gaps beside $T$, each with area $\frac{1}{8}$, and they lie entirely inside the segment. So the segment holds at least

$$1 + \frac{1}{8} + \frac{1}{8} = \frac{5}{4},$$

which is already more than $1$. The guess fails because it ignores how much area hides in the curved gaps. The rescue is to keep filling: each stage adds a quarter of the stage before.

$$1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots$$

$$= \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}$$

The partial sums $1$, $1.25$, $1.3125$, $1.328125$ climb toward $1.3333$ and never pass it.

Final answer: $\frac{4}{3}$, which is larger than the triangle by exactly one third of it.

Example 3

Show by slicing that a cone has one third of the volume of its cylinder.

Take a cone of radius $r$ and height $h$, with its tip at height $0$. Cut it into $n$ discs of thickness $\frac{h}{n}$; disc $k$ has radius $\frac{k}{n}r$:

$$V_n = \sum_{k=1}^{n} \pi\left(\frac{k}{n}r\right)^2 \cdot \frac{h}{n}$$

$$= \pi r^2 h \cdot \frac{(n + 1)(2n + 1)}{6n^2}$$

Let $n \to \infty$:

$$\lim_{n \to \infty} \frac{(n + 1)(2n + 1)}{6n^2} = \frac{1}{3}$$

With $n = 10$ the factor is $0.385$, and with $n = 100$ it is $0.33835$. This is Eudoxus' result, here found by the disc stacking of the disk method.

Final answer: $V = \frac{1}{3}\pi r^2 h$.

Example 4

Use 96-sided polygons to bound $\pi$, as in Archimedes' Measurement of a Circle.

For a circle of diameter $1$, the inscribed regular 96-gon has perimeter $96\sin\left(\frac{\pi}{96}\right)$ and the circumscribed one has perimeter $96\tan\left(\frac{\pi}{96}\right)$:

$$96\sin\left(\frac{\pi}{96}\right) = 3.1410$$

$$96\tan\left(\frac{\pi}{96}\right) = 3.1427$$

Archimedes worked without trigonometry, using hand-computed square roots, and rounded outward to fractions that are easier to state:

$$3\frac{10}{71} = 3.1408 < \pi < 3\frac{1}{7} = 3.1429$$

His bounds are slightly wider than the polygon values, so they remain correct. Today's value of pi to 4 decimal places is $3.1416$.

Final answer: $3\frac{10}{71} < \pi < 3\frac{1}{7}$.

Example 5

Use Fermat's adequality to divide a line of length $b$ into two parts whose product is as large as possible.

Let one part be $x$, so the product is $f(x) = bx - x^2$. Compare $f(x + e)$ with $f(x)$:

$$f(x + e) - f(x) = be - 2ex - e^2$$

Divide by $e$:

$$b - 2x - e$$

Set $e$ aside and make the expression adequal to zero:

$$b - 2x = 0$$

$$x = \frac{b}{2}$$

In calculus terms, Fermat solved $f'(x) = b - 2x = 0$, the first-derivative test for maxima and minima.

Final answer: cut the line in half; the largest product is $\frac{b^2}{4}$.

Example 6

Use Fermat's geometric partition to find the area under $y = x^2$ from $0$ to $b$.

Mark points $b, br, br^2, br^3, \ldots$ with $0 < r < 1$. The strip between $br^{k+1}$ and $br^k$ has width $br^k(1 - r)$ and height $\left(br^k\right)^2$:

$$S = \sum_{k=0}^{\infty} \left(br^k\right)^2 \cdot br^k(1 - r)$$

$$= b^3(1 - r)\sum_{k=0}^{\infty} r^{3k}$$

$$= \frac{b^3(1 - r)}{1 - r^3} = \frac{b^3}{1 + r + r^2}$$

As $r \to 1$, the strips become infinitely thin and

$$S \to \frac{b^3}{3}$$

With $b = 1$, $r = 0.9$ gives $0.3690$ and $r = 0.99$ gives $0.3367$. The answer matches $\int_{0}^{b} x^2,dx = \frac{b^3}{3}$.

Final answer: the area is $\frac{b^3}{3}$.

Why Did Mathematicians Need Infinitesimals Before Calculus?

"Curved shapes refuse to be counted in squares" is the short version of why these methods exist. Every area rule before Eudoxus worked by cutting a shape into rectangles and triangles, which works for straight edges and fails for anything curved.

Three pressures kept the problem alive for 2,000 years:

• Practical measurement. Merchants, builders and astronomers needed volumes of barrels, domes and orbits. Kepler's barrel study began at his own wedding, where he watched merchants estimate a barrel's volume with a rod pushed diagonally through the bung-hole and wondered why that worked.

• Incommensurable lengths. The Greeks knew the diagonal of a square cannot be a whole-number ratio of its side, so area could not always be computed by counting.

• Fear of the infinite. Zeno's paradoxes made "infinitely many pieces" sound like nonsense, which is why Archimedes proved by exhaustion even after discovering by slices.

The pay-off came in the 1660s and 1670s. Once Newton and Leibniz showed that areas undo tangents, the slicing ideas became the integral, now used every time an engineer computes the area under a curve or a scientist totals a changing quantity. A wider account of what Archimedes' slicing became is in the Stanford Encyclopedia of Philosophy entry on continuity and infinitesimals.

Common Confusions About Archimedes And Early Infinitesimals

The mistakes below come from treating "infinitely small", "as small as you like" and "one dimension lower" as the same idea. They are not.

Mistake 1: Calling Exhaustion An Infinitesimal Method

Where it slips in: Summaries that say Archimedes "used infinitesimals" to prove the parabola result.

Don't do this: Describe the exhaustion proof as adding infinitely many triangles.

The correct way: Say the proof uses finitely many triangles at every stage plus a double contradiction; only The Method used infinitely thin slices, and Archimedes did not count that as proof.

The memorizer who learned "Archimedes did calculus" as a fact tends to make this one. A student who can explain the double contradiction sees at once why exhaustion is closer to the modern limit.

Mistake 2: Treating Indivisibles As Thin Rectangles

Where it slips in: Cavalieri-style arguments, where an area is described as "made of lines".

Don't do this: Add up the lengths of infinitely many lines and call the total an area without comparing it to another figure.

The correct way: Use indivisibles only to compare two figures slice by slice, as Cavalieri's principle does; for an actual number, switch to thin rectangles of width $\Delta x$ and take a limit.

Mistake 3: Stopping The Series Too Early

Where it slips in: Parabola and $\pi$ calculations, where the first few stages look close enough.

Don't do this: Report $1.3125$ (three stages) as the segment's area.

The correct way: Sum the whole geometric series or bound the remainder; three stages fall short by $\frac{1}{48}$, exactly one third of the last stage added.

Practice Problems On Archimedes And Early Infinitesimals

1. Find the area of the parabolic segment cut from $y = x^2$ by the line $y = 4$, first with Archimedes' $\frac{4}{3}$ rule, then with an integral.

2. A cone has radius $3$ and height $5$. Use the one-third rule of Eudoxus to find its volume.

3. Find the volume of a sphere of radius $3$ and check that it is two thirds of its enclosing cylinder.

4. Use adequality to find the value of $x$ that maximises $f(x) = 12x - 3x^2$.

5. Fermat's geometric partition for $y = x^2$ on $[0, 2]$ gives $S = \frac{8}{1 + r + r^2}$. Find $S$ for $r = 0.5$, then the limit as $r \to 1$.

6. Wallis compared $0^2 + 1^2 + \cdots + n^2$ with $(n + 1)n^2$. Compute the ratio for $n = 3$ and state its limit.

Answer to Question 1: the inscribed triangle has area $8$, so the segment is $\frac{32}{3}$; the integral $\int_{-2}^{2} \left(4 - x^2\right),dx$ also gives $\frac{32}{3}$.

Answer to Question 2: $\frac{1}{3}\pi \cdot 9 \cdot 5 = 15\pi$.

Answer to Question 3: the sphere is $36\pi$ and the cylinder is $54\pi$, and $\frac{36\pi}{54\pi} = \frac{2}{3}$.

Answer to Question 4: $f(x + e) - f(x) = 12e - 6ex - 3e^2$, so $12 - 6x = 0$ and $x = 2$.

Answer to Question 5: $S = \frac{8}{1.75} = 4.5714$ to 4 decimal places, with limit $\frac{8}{3}$.

Answer to Question 6: $\frac{0 + 1 + 4 + 9}{4 \cdot 9} = \frac{14}{36} = \frac{7}{18}$, and the ratio tends to $\frac{1}{3}$.

Key Takeaways

• Archimedes and early infinitesimals measured curved areas and volumes long before calculus, using exhaustion for proof and slicing for discovery.

• A parabolic segment is exactly $\frac{4}{3}$ of its inscribed triangle, found with a geometric series of triangles.

• The Method, rediscovered in 1906, showed Archimedes balancing infinitely thin slices on a lever.

• Kepler (1615), Cavalieri (1635), Fermat (about 1636) and Wallis (1656) turned slicing into techniques Newton and Leibniz could build on.

• Indivisible lines compare figures; strips with width and a limit produce actual areas.

Where To Go From Here

Work through practice problems 1 to 3 first, then 4 to 6; if the series in problem 1 feels shaky, return to Example 2 and the section on the parabolic segment. For a live session that connects these historical methods to modern integration, book a free demo with a Bhanzu trainer.

Read More:

• Definite integrals

• Fundamental theorem of calculus

• Infinite series

• Applications of integration

• Circle area formulas