Talk:Area of a disk

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I know the formatting isn't very good, so if anyone could help with that, I would appreciate it. --Chuck 11:06, 12 April 2006 (UTC)


Someone must erase this article because circles have not area --kiddo 02:40, 29 October 2006 (UTC)

  • sorry I didn't mean to erase exactly

--kiddo 05:23, 2 November 2006 (UTC)

Contents

[edit] Humor

πr2---Pi R Squared
πr2---Pi R Not Squared
πro---Pi R Round
cr2---Cornbread R Squared

[edit] Archimedes

Maybe I'll get around to working on the article itself. Meanwhile, here's a description of the way Archimedes proved that the area of a circle must be exactly the same as the area of a right triangle whose base is the circumference and whose height is the radius. This can be found in T. L. Heath's translation of J. L.Heiberg's Greek version of "Measurement of a circle" in The Works of Archimedes, Dover, 2002 (originally Cambridge University Press, 1897), ISBN 978-0-486-42084-4.

The proof begins with the claim that if the area of the circle is not equal to that of the triangle, then it must be either greater or less. It then eliminates each of these by contradiction (using regular polygons), leaving equality as the only possibility. The proof uses what today is called the Axiom of Archimedes.

Circle with square and octagon inscribed, showing area gap
Circle with square and octagon inscribed, showing area gap

Suppose the circle area, C, is greater than the triangle area, T = 1⁄2cr, by an amount Δ. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those segments, S4 is greater than Δ, split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total area, S8. Continue splitting until the total segment area, Sn, is less than Δ. Now the area of the inscribed polygon, Pn = C−Sn, must be greater than that of the triangle.

\begin{align} \Delta &{}= C - T\\ &{}>S_n\\ P_n &{}= C - S_n\\ &{}> C - \Delta\\ P_n &{}> T \end{align}

But this forces a contradiction, as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides, ns, is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals 1⁄2nhs. But since h < r and ns < c, the polygon area must be less than the triangle area, 1⁄2cr, a contradiction. Therefore our supposition that C is greater than T must be wrong.

Circle with square and octagon circumscribed, showing area gap
Circle with square and octagon circumscribed, showing area gap

Suppose the circle area is less than the triangle area by an amount δ. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area trapped between the square and the circle, A4 is greater than δ, slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the trapped area is less than δ. The area of the polygon, Qn, must be less than T.

\begin{align} \delta &{}= T - C\\ &{}>A_n\\ Q_n &{}= C + A_n\\ &{}< C + \delta\\ Q_n &{}< T \end{align}

This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of length r. And since the total side length is greater than the circumference, the polygon consists of n identical triangles with total area greater than T. Again we have a contradiction, so our supposition that C might be less than T must be wrong as well.

Therefore it must be the case that the area of the circle is precisely the same as the area of the triangle.

The circumference of a circle with radius r is, of course, 2πr (a separate discussion). Therefore the area of the circle is

\begin{align} C &{} = T\\ &{}= \frac{1}{2} c r\\ &{}= \frac{1}{2} (2 \pi r) r\\ &{}= \pi r^2 . \end{align}

Archimedes then goes on to compute lower and upper bounds for π, again using inscribed and circumscribed polygons.

One appeal of this proof is that it uses no calculus, no integrals, no limits. It merely depends on the fact that polygon splitting will eventually force an area difference less than Δ or δ. --KSmrqT 23:07, 13 December 2006 (UTC)

this proof is of historical significance and should definitely be included in the article. seems to me one can make a pretty good case this is the very beginning of analysis. the idea of a limit is already present in the proof. approximating from inscribed and circumscribed polygons is essentially the same as the Riemann lower and upper sums. and this is how many centuries before Newton, et al.? pretty remarkable. Mct mht 01:25, 14 December 2006 (UTC)

[edit] Some issues with the current presentation

There are several different things to prove about the area of a disk:

  1. The area of a disk is some constant times the square of its radius. By convention, we can call that constant π.
  2. The constant π defined as the area of a unit circle is equal to some other non-circle-related definition of π.
  3. The area of a disk is equal to the half the circumference times the radius. Since we know (by some other proof) that the circumference is 2πr, the formula follows.

I think it would be helpful to state more precisely which is being proved in each proof.

The proof by calculus appears to be attempting to prove the first statement, that the area scales as the square of the radius. But this statement isn't really about circles: the area of a figure scaled by a factor of r is r2 times the area of the figure, regardless of whether the figure is a circle or not. A proof of this needs to depend more carefully on how we define area, but is a straightforward consequence of the definition e.g. for Lebesgue measure. For that matter the generalization to ellipses doesn't really depend on ellipses, it's a more general fact about affine transformation of a circle. So I don't really see what the point of all the calculus is here.

I like the general approach of the proof using limits, which I view as being about the relationship between area, perimeter, and radius, but I think it might be more clear with more words and less trig formulas:

Consider a sequence of regular 2k-gons, inscribed in the circle. Subdivide each polygon into triangles by connecting the vertices to the center of the circle. As k increases, each step reduces the area not covered by triangles to less than half its previous value, so the total area of the triangles converges to the area of the disk. At each step, the total area of triangles is (by the triangle area formula) half the perimeter of the polygon times the triangle height. The perimeter converges to that of the circle, while the height converges to the radius of the circle, so half their product converges to the circle's perimeter times half its radius.

What more is needed than that? —David Eppstein 00:51, 14 December 2006 (UTC)

I don't believe anybody has mentioned the nicest reason I've seen for the formula. If you chop up the disc into pie wedges and reassemble it into a parallelogram (it gets closer to a rectangle the thinner the wedges), you notice that area of the parallelogram is approximately the radius multiplied by half the circumference. This clearly is a better approximation the thinner the wedges get. The nice thing about this method is that you can draw a very nice, understandable picture. --Chan-Ho (Talk) 01:54, 14 December 2006 (UTC)

Let's take a step back and consider what we want to say and to whom we may be saying it. I see a few obvious audiences:
  1. Young students
  2. Their teachers
  3. Early calculus students
  4. Curious adults
  5. Fans of π
What do we want to say to them, being mindful of the overlap with our pi article? For the young students and their teachers, we prefer visual methods with few formulas and no trigonometry or calculus. For the calculus students we need an explicit integration. Perhaps nothing additional is needed for adults. For the fans, we could mention a mind-bender (and our article):
As the Archimedes proof makes clear, we need no trigonometry and no limits to connect the area to the circumference. A modern approach to the connection is found in Serge Lang's little book, Math! Encounters With High School Students, Springer, 1985, ISBN 978-0-387-96129-3. He does use limits, but in a simple way that requires no explicit calculus.
A while back, in connection with pi, I created an animation using SVG that depicts a rearrangement method attributed to Leonardo da Vinci. Although it is really much the same mathematics as the inscribed polygon limit, visually it is different. Perhaps I'll polish its rough edges and make it available. --KSmrqT 16:43, 14 December 2006 (UTC)
Re "For the calculus students we need an explicit integration" — I agree it's helpful for calculus students to be shown the relation between area and integration, but what does the explicit integration actually prove? And how can we tell that it isn't just circular reasoning, unless we state more clearly what we're proving and how those π's got into the integration formulas we're using? I mean, going through the actual integration proof again, I can see that the π comes into it as part of a change of variable from Cartesian to polar, so the π in the formula is the π of the 2πr circumference formula rather than just some constant of convenience that we're calling π, but I think the formula-heavy presentation of the proof obscures that fact. I do think your visual approach could help with the limit based proof. —David Eppstein 18:44, 14 December 2006 (UTC)
The other potential problem with the integration proof is that it uses the antiderivative formula for cosine. Working with trig functions just invites circular definitions, so a lot of care must be taken. CMummert 03:49, 15 December 2006 (UTC)
Nice pun. Nothing about the definition of sine or cosine or the antiderivation formula under discussion requires the formula for the area of a disk. --Chan-Ho (Talk) 04:38, 15 December 2006 (UTC)
(ARGH! A web site crashed my browser, as I was almost finished composing a large post. I shall try to recreate what I can.)
I agree with David that the appearance of π should be explained. Ultimately, it comes from the connection shown by Archimedes. Previously it was known that the ratio of circumference to diameter was a constant independent of the size of the circle. Archimedes proved that the area involved the same number, and then used area to approximate the number π.
For calculus students, the easiest integration builds in the area–circumference–radius relationship, using the "onion" approach.
\begin{align} \mathrm{Area}(r) &{}= \int_0^{r} 2 \pi t \, dt \\ &{}= \left[ (2\pi) \frac{t^2}{2} \right]_{t=0}^{r}\\ &{}= \pi r^2 \end{align}
Here 2πt is the circle at thickness t. In this approach we have no limits, no trigonometry, no change of variables. This approach also naturally draws us into higher dimensions. (Again Archimedes was there first; he showed that the area of a sphere equals the area of a circumscribed cylinder, (2πr)(2r), which integrates to 43πr3. He also found ways to show the volume of a sphere without calculus.)

(I like connections for at least four reasons. One is personal; I think that way. A second is pedagogical; research shows we remember connected facts better than isolated ones. A third is promotional; connections draw people into learning more mathematics. And a fourth is philosophical; all of mathematics is connected.)

Two alternative integrals are obvious possibilities, but each brings complications.
The first uses two semicircles.
2 r^2 \int_{-1}^{1} \sqrt{1-x^2} \, dx
Usually π makes its appearance during a trigonometric substitution,
x = \sin \theta , \qquad dx = \cos \theta \, d\theta,
via the limits of integration,
-1 = \sin \left( -\frac{\pi}{2} \right) , \qquad 1 = \sin \left( \frac{\pi}{2} \right) .
The integral becomes
2 r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta .
Now we invoke a double-angle trigonometric identity,
\cos 2\theta = 2\cos^2 \theta - 1 , \,\!
to produce
r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2 \theta + 1 \, d\theta ,
then split and change variables to get a sum of known integrals.
r^2 \left( \int_{-\pi}^{\pi} \frac{\cos \phi}{2} \, d\phi + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \, d\theta \right) .
The first integral is a full period of a sinusoid, thus vanishes; the second is trivially π. This gives us the desired πr2, but the reader is forced to wade through a mess of calculus, algebra, and trigonometry to get there. (And we silently slipped in the trick of extracting r2 to work with a unit circle.) We get little insight, and little inspiration.
A second alternative sums radial wedges.
\int_0^{2\pi} r \cos \left( \frac{d\theta}{2} \right) r \sin \left( \frac{d\theta}{2} \right) .
Although formally correct, this appears completely unfamiliar to the beginning calculus student. We must show
\cos d\phi = 1 , \qquad \sin d\phi = d\phi \,\!
and use a half-angle substitution to get a familiar form,
r^2 \int_0^{2\pi} \frac{1}{2} \, d\theta ,
which we can then trivially integrate. The "we must show" part is an exercise in limits.
But if we're going to do that, perhaps we should instead directly compute the limit of the area of a circumscribed regular polygon. This has the side benefit of bringing us back to Archimedes. --KSmrqT 20:31, 15 December 2006 (UTC)

[edit] Numeric computation

Here's another bit for the article. This uses ideas of Willebrord Snell (Cyclometricus, Lugduni Batavorum: Elzevir, 1621) followed up by Christiaan Huygens (De Circuli Magnitudine Inventa, 1654), described in

  • Gerretsen, J.; Verdenduin, P. (1983). "Chapter 8: Polygons and Polyhedra", in H. Behnke, F. Bachmann, K. Fladt, H. Kunle (eds.): Fundamentals of Mathematics, Volume II: Geometry, S. H. Gould (trans.), MIT Press, pp. 243–250. ISBN 0-262-52094-X (ISBN 978-0-262-52094-2). 
    (Originally Grundzüge der Mathematik, Vandenhoeck & Ruprecht, Göttingen, 1971.)

Given a circle, let un be the perimeter length of an inscribed regular n-gon, and let Un be the perimeter length of a circumscribed regular n-gon. Then we have the following doubling formulae.

\begin{align} U_{2n} &{}= \frac{2 U_{n} u_{n}}{ U_{n} + u_{n}} , &\qquad& \text{harmonic mean} \\ u_{2n} &{}= \sqrt{U_{2n} u_{n}} , &\qquad& \text{geometric mean} \end{align}

Archimedes doubled a hexagon four times to get a 96-gon. For a unit circle, an inscribed hexagon has u6 = 6, and a circumscribed hexagon has U6 = 4√3. We have have luxury of decimal notation and our two equations, so we can quickly double seven times:

n k uk Uk (uk+Uk)/4   k=6×2n
0 6 6.0000000 6.9282032 3.2320508
1 12 6.2116571 6.4307806 3.1606094
2 24 6.2652572 6.3193199 3.1461443
3 48 6.2787004 6.2921724 3.1427182
4 96 6.2820639 6.2854292 3.1418733
5 192 6.2829049 6.2837461 3.1416628
6 384 6.2831152 6.2833255 3.1416102
7 768 6.2831678 6.2832204 3.1415970

A best rational approximation to the last average is 355⁄113, which is an excellent value for π. But Snell proposes (and Huygens proves) a tighter bound than Archimedes.

k \frac{3 \sin \frac{\pi}{k}}{2+\cos\frac{\pi}{k}} < \pi < k \frac{2 \sin \frac{\pi}{k} + \tan \frac{\pi}{k}}{3}

Thus we could get the same approximation, with decimal value 3.14159292…, from a 48-gon. --KSmrqT 22:32, 16 December 2006 (UTC)

[edit] Derivation

Circle with similar triangles: circumscribed side, inscribed side and complement, inscribed split side and complement
Circle with similar triangles: circumscribed side, inscribed side and complement, inscribed split side and complement

Let one side of an inscribed regular n-gon have length sn and touch the circle at points A and B. Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem, this is a right triangle with right angle at B. Let the length of A′B be cn, which we call the complement of sn; thus cn2+sn2 = (2r)2. Let C bisect the arc from A to B, and let C′ be the point opposite C on the circle. Thus the length of CA is s2n, the length of C′A is c2n, and C′CA is itself a right triangle on diameter C′C. Because C bisects the arc from A to B, C′C perpendicularly bisects the chord from A to B, say at P. Triangle C′AP is thus a right triangle, and is similar to C′CA since they share the angle at C′. Thus all three corresponding sides are in the same proportion; in particular, we have C′A : C′C = C′P : C′A and AP : C′A = CA : C′C. The center of the circle, O, bisects A′A, so we also have triangle OAP similar to A′AB, with OP half the length of A′B. In terms of side lengths, this gives us

\begin{align} c_{2n}^2 &{}= \left( r + \frac{1}{2} c_n \right) 2r \\ c_{2n} &{}= \frac{s_n}{s_{2n}} . \end{align}

In the first equation C′P is C′O+OP, length r+1⁄2cn, and C′C is the diameter, 2r. For a unit circle we have the famous doubling equation of Ludolph van Ceulen,

c_{2n} = \sqrt{2+c_n} . \,\!

If we now circumscribe a regular n-gon, with side A″B″ parallel to AB, then OAB and OA″B″ are similar triangles, with A″B″ : AB = OC : OP. Call the circumscribed side Sn; then this is Sn : sn = 1 : 1⁄2cn. (We have again used that OP is half the length of A′B.) Thus we obtain

c_n = 2\frac{s_n}{S_n} . \,\!

Call the inscribed perimeter un = nsn, and the circumscribed perimenter Un = nSn. Then combining equations, we have

c_{2n} = \frac{s_n}{s_{2n}} = 2 \frac{s_{2n}}{S_{2n}} ,

so that

u_{2n}^2 = u_n U_{2n} . \,\!

This gives a geometric mean equation. We can also deduce

2 \frac{s_{2n}}{S_{2n}} \frac{s_n}{s_{2n}} = 2 + 2 \frac{s_n}{S_n} ,

or

\frac{2}{U_{2n}} = \frac{1}{u_n} + \frac{1}{U_n} .

This gives a harmonic mean equation.

Illustration to come. --KSmrqT 07:32, 20 December 2006 (UTC)

I gladly see how this page had evoluzioned (and revoluzionated some thinkings) to see it for a perfect nomination to an excelent page of the moment :) kid--148.202.11.52 16:32, 12 March 2007 (UTC)

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