Talk:Area of a circle/Archive 1

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

Zero area
Someone must erase this article because circles have not area --kiddo 02:40, 29 October 2006 (UTC) --kiddo 05:23, 2 November 2006 (UTC)
 * sorry I didn't mean to erase exactly

Humor
πr2---Pi R Squared

πr 2 ---Pi R Not Squared

πro---Pi R Round

cr2---Cornbread R Squared

— Preceding unsigned comment added by 68.151.35.220 (talk) 04:28, 10 December 2006 (UTC)

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.

Suppose the circle area, C, is greater than the triangle area, T = 1&frasl;2cr, by an amount &Delta;. 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 &Delta;, 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 &Delta;. 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&frasl;2nhs. But since h &lt; r and ns &lt; c, the polygon area must be less than the triangle area, 1&frasl;2cr, a contradiction. Therefore our supposition that C is greater than T must be wrong.

Suppose the circle area is less than the triangle area by an amount &delta;. 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 &delta;, slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the trapped area is less than &delta;. 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, 2&pi;r (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 &pi;, 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 &Delta; or &delta;. --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)

Some issues with the current presentation
There are several different things to prove about the area of a disk: I think it would be helpful to state more precisely which is being proved in each proof.
 * 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.

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. --C S (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:
 * Young students
 * Their teachers
 * Early calculus students
 * Curious adults
 * Fans of &pi;
 * 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.  --C S (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 &pi; 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 &pi;.
 * 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} $$
 * Here 2&pi;t is the circle at thickness t. In this approach we have no limits, no trigonometry, no change of variables. This approach also naturally draws us into higher dimensions. (Again Archimedes was there first; he showed that the area of a sphere equals the area of a circumscribed cylinder, (2&pi;r)(2r), which integrates to 43&pi;r3. He also found ways to show the volume of a sphere without calculus.)
 * (I like connections for at least four reasons. One is personal; I think that way. A second is pedagogical; research shows we remember connected facts better than isolated ones. A third is promotional; connections draw people into learning more mathematics. And a fourth is philosophical; all of mathematics is connected.)
 * Two alternative integrals are obvious possibilities, but each brings complications.
 * The first uses two semicircles.
 * $$ 2 r^2 \int_{-1}^{1} \sqrt{1-x^2} \, dx $$
 * Usually &pi; makes its appearance during a trigonometric substitution,
 * $$ 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) . $$
 * The first integral is a full period of a sinusoid, thus vanishes; the second is trivially &pi;. This gives us the desired &pi;r2, but the reader is forced to wade through a mess of calculus, algebra, and trigonometry to get there. (And we silently slipped in the trick of extracting r2 to work with a unit circle.) We get little insight, and little inspiration.
 * A second alternative sums radial wedges.
 * $$ \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)

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 (Originally Grundzüge der Mathematik, Vandenhoeck &amp; 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&radic;3. We have have luxury of decimal notation and our two equations, so we can quickly double seven times:
 * {| style="text-align:center" cellspacing="10"

A best rational approximation to the last average is 355&frasl;113, which is an excellent value for &pi;. But Snell proposes (and Huygens proves) a tighter bound than Archimedes.
 * 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
 * }
 * 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
 * }
 * 6 || 384 || 6.2831152 || 6.2833255 || 3.1416102
 * 7 || 768 || 6.2831678 || 6.2832204 || 3.1415970
 * }
 * }
 * $$ k \frac{3 \sin \frac{\pi}{k}}{2+\cos\frac{\pi}{k}} < \pi < k \frac{2 \sin \frac{\pi}{k} + \tan \frac{\pi}{k}}{3} $$

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

Derivation
Let one side of an inscribed regular n-gon have length sn and touch the circle at points A and B. Let A&prime; be the point opposite A on the circle, so that A&prime;A is a diameter, and A&prime;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&prime;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&prime; be the point opposite C on the circle. Thus the length of CA is s2n, the length of C&prime;A is c2n, and C&prime;CA is itself a right triangle on diameter C&prime;C. Because C bisects the arc from A to B, C&prime;C perpendicularly bisects the chord from A to B, say at P. Triangle C&prime;AP is thus a right triangle, and is similar to C&prime;CA since they share the angle at C&prime;. Thus all three corresponding sides are in the same proportion; in particular, we have C&prime;A : C&prime;C = C&prime;P : C&prime;A and AP : C&prime;A = CA : C&prime;C. The center of the circle, O, bisects A&prime;A, so we also have triangle OAP similar to A&prime;AB, with OP half the length of A&prime;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}$$ In the first equation C&prime;P is C&prime;O+OP, length r+1&frasl;2cn, and C&prime;C is the diameter, 2r. For a unit circle we have the famous doubling equation of Ludolph van Ceulen,
 * $$ c_{2n} = \sqrt{2+c_n} . \,\!$$

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

Ideal sources sought
If anyone would like to help, I thought it might be nice to find two more sources to cite. What I'm looking for is:


 * 1) A calculus text or MAA article or some such taking the "onion" approach. Lang is great in some ways, but not ideal for this. The typical calculus text integrates the area under (1−x2)1/2, which is not what we want. One likely context would be a discussion of multidimensional sphere areas and volumes, for by itself it is out of pedagogical sequence for a text and too trivial for an article.
 * 2) A Monte Carlo article or text that throws darts and that simply explains the variance issue. Buffon's needle is a popular example, but it's about computing &pi; rather than the area of a circle.

In both cases, a scholarly source that is freely available on-line would be extra nice. --KSmrqT 12:54, 29 May 2007 (UTC)

uh... I am about to go to bed, so I probably won't respond for a while, but check out my list of references (link on user page) and E-mail me if you think any of them would be helpful. Cheers --Cronholm144 13:14, 29 May 2007 (UTC)

Archimedes triangle
I have removed the image of circle and triangle because it is badly composed. This I know because I've long had a better image (in fact an elaborate SVG animation), but thought it not a benefit to the article, only a distraction. --KSmrqT 00:59, 2 June 2007 (UTC)

comments

 * Wow, a lot of hard work has been poured into this article. It certainly has no problems with breadth of coverage...
 * Does it need a history section? Many minds have apparently struggled with the question...
 * "Other approaches are also of interest." Of interest to whom?
 * (Archimedes 2002)? Egads. This is not an individual work that has been translated; it is a compilation and translation and editing etc. AFAIK, in cases such as this you're licensed to list Heath as the author e.g. (Heath 2002). Another method might be "... writing in the third century B.C., Archimedes showed that blah blah..." with a footnote which leads to  a Notes section. In the Notes you can say "For a modern English translation see Heath (2002)". Finally, in the References section you would have Heath, T. (2002). The Works of Archimedes. New York: Dover Publications.
 * "...great mathematician Archimedes". I know he was great. You know he was great. Everyone beyond say 5th grade or so knows he was great. But the language is unencyclopedic. Some would say this is POV. Maybe it is; that's arguable. But I think a stronger case could be made that it is unencyclopedic. Drag out your Encylopedia Brittanica and find an article which mentions Archimedes (not the actual Archimedes article). See whether it has such adjectives...
 * What does it mean to use regular polygons "in an essential way"?
 * I can come back and look again later; busy today... I'm supposed to be studying for prelims... But I made some edits to the article that show the gist of what I might do... spell everything out clearly ("the first argument that leads to a contradiction...") .. eliminate the word "we" (who is we?) etc. I'll look again another time... Ling.Nut 22:15, 5 June 2007 (UTC)


 * Point by point:
 * 
 * A brief comment brought me here awhile back, I started editing, and the rest is history. It was fun as well as work. But thanks.
 * No, it does not need a history section, because it in many ways it already is a history, and to go further would essentially duplicate the history of numerical approximations of π.
 * The wording "are of interest" is not ideal, but another editor wanted to point out that the article went beyond Archimedes.
 * Yes, I know 2002 is an abomination. The problem is that the auto-linking of the Harvard citation mechanism has limitations. Ideally I would refer to Heath 2002, but Heath is a translator, not an editor. So the formal citation has Archimedes as the author and 2002 as the date of publication, and that is what the reference must use. We do not have a single definitive source by Archimedes himself; even the recent work with the palimpsest — which itself is a transcription — is limited. And, please, no #%#&*^!! footnotes, which are a worse abomination that should be banned from the Web.
 * The "great" serves a purpose, because Archimedes' proof did something radically new, using methods that would not be improved for centuries. There is zero controversy about his greatness, so I have little sympathy for knee-jerk claims of POV should they arise. In fact, scholarly works and histories regularly state that Archimedes is one of the greatest mathematicians of all time. The Encyclopædia Britannica] article begins "… the most famous mathematician and inventor of ancient Greece." Many lay readers will be more familiar with Napoleon than with Archimedes, so I don't share your view that almost everyone knows he was great.
 * If you do not understand how the use of regular polygons is essential to the proof, then you do not understand the proof. Sorry.
 * I removed every edit you made to the article, because each one (!) made the article worse.
 * Although a link to pi is a good idea, we have had repeated discussions elsewhere about linking mathematical symbols, and always conclude that a wikilink on the symbol itself should not be used.
 * You changed active voice to passive voice, changing two simple declarative sentences to a tortured combination.
 * The section headings make clear what you felt the need to state. If the reader is unable to grasp that, the whole proof is wasted on them. We don't need the clutter.
 * 
 * Unfortunately, I've seen many such examples of well-meaning editors who make things worse. Honestly, I appreciate the thought, just not its embodiment. It troubles me that so many writers — especially those with academic and technical backgrounds — have gaping blind spots in their sensitivity. For example, numerous readability studies show that active voice is easier to read, and our mathematics manual of style promotes it.
 * I'll find a way to add a link for pi, and look for a less awkward notice about other approaches. As for the rest, I think they're best left as they are. --KSmrqT 15:15, 6 June 2007 (UTC)


 * We all have blind spots KSmrq! ;) Anyway, I agree with you that the active voice is much more readable, and that a history section is unnecessary. Still, I thought Ling.Nut's comments were more helpful than your response suggests. In particular, "Archimedes 2002" would surely be better replaced by a reference to "Heath 1897", or possibly "Heath 1897, 2002", if that is compatible with auto-linking: I agree with Ling.Nut that this work is more than just a translation.
 * As you point out yourself, the sentence on regular polygons being "essential" adds nothing, either for the reader who understands the proof, or the reader who doesn't. Concerning the overall style, I know it can be quite a challenge to write engaging prose without using the first person plural, but I think it is worth the effort. Maybe I will give it a go, but I have rather a lot of blind spots of my own and this is already a very polished piece of work... Geometry guy 16:53, 6 June 2007 (UTC)


 * Much as I appreciate the citation template, it is still a work in progress, and the reference is as it must be for now. The true dates are a little awkward anyway, because we don't have Archimedes originals, only later transcriptions (as in the palimpsest); then we have especially Heiberg's work; and we have Heath; but the print version (cited) is a Dover Publications reprint of the original Cambridge University Press edition (linked). (Whew!) The way I would like to say it is "(Heath 2002)". Ah well, we do the best we can with what we have.
 * I certainly did not assert that the "essential" comment adds nothing, since that is the opposite of my view! It orients the reader to the nature of the method, which I strongly feel is helpful.
 * The use of "we" makes some editors uncomfortable. Apparently WP:MSM says to avoid it, which is dead wrong, and is contradicted by advice elsewhere. Specifically, the main Manual of Style says:
 * Nevertheless, it is sometimes appropriate to use we or one when referring to an experience that anyone, any reader, would be expected to have, such as general perceptual experiences. For example, although it might be best to write, “When most people open their eyes, they see something”, it is still legitimate to write, “When we open our eyes, we see something”, and it is certainly better than using the passive voice: “When the eyes are opened, something is seen.”
 * It is also acceptable to use we in mathematical derivations; for example: “To normalize the wavefunction, we need to find the value of the arbitrary constant A.”
 * Almost every effort to avoid "we" that I have seen is a disaster. It can be misused, but I try to be careful how I use "we". It is not meant to be personal, as the use of "you" or "I" must be. Instead, it refers to "we mathematicians" or "we humans" or the like. I get really annoyed at overuse of the epithet "unencyclopedic". When I was a youngster I spent long hours curled up with an encyclopedia, so I trust my judgment. Why do so many editors think precise writing must also be stilted? I've spent a lifetime fighting this battle.
 * Ah well, at least we still have pretty pictures to look at. Remember when geometry was about things we could draw? This article brings back some of the nostalgia. :-) --KSmrqT 21:05, 6 June 2007 (UTC)

Tarski's rearrangement theorem
If the proof uses the axiom of choice to construct the pieces, then I would say it does tell "how" to construct them, although the construction won't be completely explicit because of the choice required. The current wording suggests a proof by contradiction in which the nonexistence of a partition is shown to be impossible but no actual partition is constructed. CMummert · talk 11:59, 29 May 2007 (UTC)


 * I added explicit language about the axiom of choice to clarify. Perhaps it would help if you skimmed the two relevant references (one of which is the paper itself). Your wording conveys the wrong meaning to me, and mine to you, eh?
 * The original estimate was that 1050 pieces would be required, so that alone would make a figure difficult. :-D
 * But setting that frivolous obstacle aside, wording poses a dilemma. At the level this article is otherwise written, the axiom of choice certainly does not say how. Nor is this a non-constructive proof in the sense that a constructive proof might be possible, as your wording would suggest. The fact that we must use the axiom of choice means we throw up our hands and say we can so choose, and we can never say how. --KSmrqT 12:23, 29 May 2007 (UTC)


 * I scanned through the paper this morning. The proof requires the existence of two vectors independent over the rationals with some special property, and proceeds by showing that almost every pair of vectors will do.  This isn't the axiom of choice per se.  The only possible place I see that AC might be used is in the measure-theoretic facts used throughout the paper.  So I rephrased the last sentence to remove the claim that AC is crucial to the proof. CMummert · talk 14:16, 1 June 2007 (UTC)


 * Without AC it is legitimate to assume all sets are measurable, so the fact that the construction involves non-measurable sets implies that some form of choice is crucial, doesn't it? Geometry guy 22:00, 6 June 2007 (UTC)

send it thru FAC then..
Eh, sorry, but "between you and me there is a great gulf affixed." I see the article as seriously in need of  help. To me it reads exactly like a Mathematics textbook... and one of the older/moldier/dustier/dead-er ones, at that. :-) KSmrq sees it as  needing only trivial tweaks; G-guy said it is "highly polished." ... No. Not even close. No no no.

The chief problem is that the article makes no attempt whatsoever to consider its audience. Witness:


 * "If the reader is unable to grasp that, the whole proof is wasted on them. We don't need the clutter."
 * "If you do not understand how the use of regular polygons is essential to the proof, then you do not understand the proof."

That attitude is clearly reflected in the text of the article itself! I dunno if you can see that, but I can... The article is written for the author(s), not the reader(s).

Finally, I dunno if you noticed, but one of my degrees is in *teaching English* ...the perception regarding the evils of the passive voice is also demonstrably incorrect. Umm.. I could try to track down the literature... genre analysis of academic literature (particularly *hard sciences* literature!) has shown that the passive voice has a wide range of important functions and uses in academic text.. I'm not saying my edits were perfect; they weren't. It was a quick run-through. I *am* saying that they were a step in the correct direction... that is.. a reader-centered rather than writer-centered one. Sorry if you don't see that. :-)--Ling.Nut 00:08, 7 June 2007 (UTC)


 * Indeed, our views seem to differ. Perhaps teaching mathematics is more demanding than teaching English; it won't work with a passive audience. As for passive voice, certainly it has its uses; and I have been amused and enlightened by the rogue perspective of Richard Lanham's Style: An Anti-Textbook (ISBN 978-0-300-01720-5). But I dare say I've read (and reviewed) far more technical articles than you, and had my fill of excess passivity. I believe Strunk &amp; White said "Use the active voice" for good reason.
 * I do appreciate your generous impulse to improve the article. Perhaps we can meet again and collaborate on something less mathematical, where our impulses are more closely aligned. --KSmrqT 16:02, 7 June 2007 (UTC)

Over explained
You ever think that wikipedia might go too far in explaining concepts? —Preceding unsigned comment added by 67.166.28.81 (talk) 04:52, 15 November 2007 (UTC)

Why then do we do it this way
Why do they teach us in school that the area of a circle is Pi times the squared radius, when this can't give an exact answer but Area=.5 radius times diameter gives us an exact answer? 79.180.238.41 14:42, 17 October 2007 (UTC)


 * Questions like this belong on the Reference Desk; this page is about editing the article. Besides, the question is based on two false premises. (1) The area is exactly &pi; times the squared radius. (2) The article clearly states that the area is also half the radius times the circumference (not the diameter). --KSmrqT 05:21, 15 November 2007 (UTC)

Monte Carlo
The illustration to accompany the Monte Carlo estimation (darts) technique is confusing, as nowhere is it stated that the square has a sidelength of two! Here I was thinking it was a sidelength of unity... —DIV (128.250.80.15 (talk) 08:04, 14 January 2008 (UTC))
 * I have stated the length. –Pomte 08:13, 14 January 2008 (UTC)

Semicircles proof: Circular dependency?
Personally, one of my favorite methods of finding the area of a disk is the integration of 2 semicircles. However, it is obviously heavily dependent on the derivatives of the trigonometric functions, which rely on the limit as x tends to 0 of sinx/x, which relies on enequalities involving the unit circle that use the fact that the unit circle has an area of pi in order to prove that sinx < x. Maybe there is another way to find the derivatives of the trigonometric functions, but it seems that the semicircle integral relies on the area of a disk itself. —Preceding unsigned comment added by 69.151.149.254 (talk) 03:40, 17 July 2008 (UTC)

Un-Encyclopaedic style
This article is very informative, and I congratulate the primary writer, however I can find major stylistic faults running throughout. Openings words such as "Let" and "We" are not passive, and therefore are rather un-encyclopaedic. These should be fixed in order for this to pass the GA review process. MasterOfHisOwnDomain (talk) 22:27, 7 November 2009 (UTC)

Correction to Snell-Huygens Formulas
Snell-Huygens did NOT use the circumscribed polygons, like Archimedes did.

When he talks about the Snell-Huygens bound, there was an error in the formula for the UPPER bound, which I have corrected. I derived the correct formula from the original diagrams that Snell used. It turns out that his approximations are exact with the 1st and 3rd order terms of the resulting power series, and a very small error with the 5th order term.

21:21, 28 January 2011 (UTC)Bill Sinclair (billsincl@aol.com) — Preceding unsigned comment added by Billsincl (talk • contribs)

onion proof
Since the circumference and the area are functions of the radius, shouldn't the integral be with respect to "r"? — Preceding unsigned comment added by 90.222.147.151 (talk) 02:05, 20 January 2012 (UTC)

Rearrangement proof
The "Rearrangement proof" seems to have a error in the drawing label with "Circle area by rearrangement". The drawing shows part of the circle segment below the x-axis and uncounted (area which is yellow) with the next segment is measured and counted as the full radii. The segments should be completely measured. John W. Nicholson (talk) 11:37, 26 December 2012 (UTC)

First Sentence
came up from reference desk, I think the first sentence is erroneous, area of a disk doesn't mean area of a circle....a disk is a vague object that is sorta circular, circles could be disks, but most disks are not circles....should be "area of a circle's interior" if anything....68.48.241.158 (talk) 13:45, 14 April 2016 (UTC)


 * See Disk (mathematics) for the precise definition of "disk" as it is used mathematically. "In geometry, a disk (also spelled disc) is the region in a plane bounded by a circle. A disk is said to be closed or open according to whether it contains the circle that constitutes its boundary." -- ToE 14:33, 14 April 2016 (UTC)

Move history
-- ToE 02:16, 15 April 2016 (UTC)
 * 02:42, 29 October 2006 (UTC) Juan Marquez (talk | contribs) moved Area of a circle to Area of a disk: circle is a 1-dimensional objet so it hasn't area at all...
 * 03:44, 12 April 2016 (UTC) GeoffreyT2000 (talk | contribs) moved page Area of a disk to Area of a circle over redirect: Per WP:COMMONNAME.


 * It appears to me that both moves were made in good faith and both had valid reasons. We could go round and round on this (as with so many other Wikipedia discussions) but ultimately we will have to decide based on what is the best title for the expected readership. The technically correct "disk" would certainly be preferred by the mathematically knowledgeable, but the traditional and elementary "area of a circle" would be the name that people seeking more information would most likely use. If you are sophisticated enough to know that a circle has no area, the area of the region enclosed by a circle will be no mystery to you. The article should indicate the correct terminology, but the title should remain as it now is. Bill Cherowitzo (talk) 04:14, 15 April 2016 (UTC)


 * I strongly concur. --JBL (talk) 14:20, 15 April 2016 (UTC)


 * Concur as well.68.48.241.158 (talk) 15:03, 15 April 2016 (UTC)


 * Comment. I have moved it back.  Up until now, the article has played it very fast and loose with the word "circle".  Sometimes in this very article, "circle" meant the boundary, and sometimes it meant the disk enclosed by the circle.  This is genuinely confusing, and adopting an article title that uses the confusing, incorrect, term goes against WP:COMMONNAME: "Ambiguous or inaccurate names for the article subject, as determined in reliable sources, are often avoided even though they may be more frequently used by reliable sources."  Furthermore, it seems inappropriate to use an article title if we are to insist on using correct terminology in the article (which I think we must, since otherwise there are parts of the article that are genuinely confusing and go against established mathematical practices).   S ławomir  Biały  18:23, 25 April 2016 (UTC)


 * I think the title of the article should be "area of a circle"...as what is described in the article is what "area of a circle" means to literally everyone. I googled "area of a circle is zero" and "area of a circle is technically zero" and basically nothing comes up...(not that I'm disputing that people can choose to define things such a way in certain contexts)...I also googled "area of a circle" and this article doesn't appear anywhere...I'd bet it would be on top of Google if it were properly titled....the consensus above agrees and I suspect if you queried others they'd predominately agree....(of course what is technically meant by "area of a circle" can be explained right away in opening paragraph...ie area enclosed by a circle, area of a circle's disk etc)....68.48.241.158 (talk) 22:53, 2 May 2016 (UTC)


 * The problem, as I mentioned but will reiterate, is that the article would then be nonsensical. In the article, by "circle" we mean the circle, not the disk.  By "disk", we mean the disk.  We cannot have a title that confuses these two things that must remain distinguished in the actual text of the article.   S ławomir  Biały  23:09, 2 May 2016 (UTC)
 * I don't think it's a problem as I see no confusion from titling the article "area of a circle" for the sake of the general population finding the article/knowing they're in the right place etc and then going on to clarify what that means very technically and even using the phrase "area of a disk" throughout...it could even be put in a parenthetical note that "area of a disk" will be the predominately used phrase throughout the article...but the consensus seems to be for "area of a circle"...perhaps a request for comment would be good as this could go back and forth forever...and it's a pretty important/high profile topic...68.48.241.158 (talk) 23:24, 2 May 2016 (UTC)


 * "Area of a disk" is technically correct, and "Area of a circle" redirects here. I have no doubt that "circle" is what almost every Wikipedia reader expects. But my understanding of WP:COMMONNAME is that the uncommon, correct name with a redirect is the way to go. Mgnbar (talk) 00:12, 3 May 2016 (UTC)
 * at a glance I find a lot more to support "area of a circle" in that policy page (just look at the four bullets at the top)...particularly since "area of a circle" isn't technically incorrect in any real sense but for special and rare contexts...68.48.241.158 (talk) 00:26, 3 May 2016 (UTC)


 * It might help you to know that I occasionally clash with Slawomir Bialy over issues of accessibility. Indeed, I was sympathetic to your argument, until I read WP:COMMONNAME closely.
 * Of the bullet point goals that you mention, the first two goals suggest that "circle" is to be preferred. The third and fifth goals suggest that "disk" is to be preferred. The fourth goal treats "circle" and "disk" equally. So those bullet points are not a clear win for either side.
 * Practically speaking, any reader who searches for "area of a circle" will be redirected here and will immediately understand that "area of a disk" means the same thing. So this whole argument seems quite theoretical to me. It's about which title best conforms to Wikipedia policy, not about whether Wikipedia's readers are finding what they need.
 * I disagree with your assessment of "special and rare contexts". The distinction between a circle and a disk is important in topology/geometry, which is the area of mathematics that governs this article.
 * All that said, let me reiterate my opinion that "area of a circle" is what almost every reader will be looking for. Respectfully, Mgnbar (talk) 00:57, 3 May 2016 (UTC)

Circular Argument
There is a problem with one of these proofs of the area of the circle: it is circular. I am referring to the calculation of the integral $$ \int_{-r}^{r}\sqrt{r^2-x^2}dx$$. One cannot calculate the area in this way. It requires a substitution of $$ x=r\sin \theta$$, and this substitution requires us to know that the derivative of $$ \sin\theta $$ is $$ \cos\theta $$ so that we can replace $$ dx$$ with $$ r\cos\theta d\theta$$. This derivative requires the limit $$ \lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1$$. The standard proof of this limit (e.g. https://proofwiki.org/wiki/Limit_of_Sine_of_X_over_X ) requires a knowledge of the formula for the area of a sector. And this in turn requires knowledge of the area of a circle. Thus the proof is circular. — Preceding unsigned comment added by 78.100.47.243 (talk) 05:48, 22 April 2016 (UTC)
 * If the definition of sine and cosine is understood to be the non-geometric one in terms of power series, we can easily avoid this problem. Additionally, it too is justified by multivariable calculus because the proof of the multivariate substitution rule makes no reference to the definition of sine and cosine, and the change of variables formula does not involve them either. In symbols:
 * $$A = 2\int_{-r}^{r}\sqrt{r^2-x^2}dx = \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} dy dx = \int_0^r \int_0^{2\pi} r dr d\theta = 2\pi \int_0^r r dr$$.
 * I'm also sure that in any case, a geometric proof of that fact does not have to rely on knowing the area of the sector of a circle.--Jasper Deng (talk) 15:40, 22 April 2016 (UTC)
 * I don't think the problem is easily solved by using the analytical definitions of sine and cosine in terms of their power series. This is because we are asking a geometric question (the area of a circle) and so if we are to use the analytical definitions, we must show the equivalence of the power series definition and the geometric definition. To do this we must calculate the power series of sine (defined geometrically), and to do this we will need to differentiate sine, need to calculate $$ \lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta} $$, and the problem remains. If we were to ignore this, and use the analytical definitions without proving the equivalence to the geometric definitions, we would get a result in terms of the analytical definition of $$ \pi $$ -- half the period of the power series of sine -- which would take us well away from the geometric definition of $$ \pi $$ as being the ratio between the circumference and diameter of a circle.
 * The multivariable argument you present is correct and is not circular and is essentially the "Onion Proof" which I have no problems with.
 * To be more concise about the problem as I see it: The area of the circle is indeed $$ 2\int^r_{-r}\sqrt{r^2-x^2} dx$$. The issue arises in the methods we are allowed to use to solve this integral. We may not use any method that implicitly uses the formula for the area of a circle. I believe any method (e.g. trigonometric substitution) that relies on the fact that $$ \frac{d}{dx} \sin x = \cos x $$ (defined geometrically) does implicitly use this fact. I may be wrong in this, but all the standard arguments I have seen calculating this derivative implicitly use the formula for the  area of a circle.  — Preceding unsigned comment added by 78.100.47.243 (talk) 20:19, 22 April 2016 (UTC)
 * Well, the analytic definitions suffice for the substitution as long as it is known ahead of time that $$\sin(\pi) = 0$$. I could just as well define $$f(x) = \sum_{i = 0}^\infty \frac{(-1)^i x^{2i + 1}}{(2i + 1)!}, g(x) = \sum_{i = 0}^\infty \frac{(-1)^i x^{2i}}{(2i)!}$$, show that $$f(x)^2 + g(x)^2 = 1$$ for all x (this can be done easily using Euler's formula or Cauchy products), and that $$f'(x) = g(x)$$. The only issue this leaves is that it's not immediately clear how $$\pi$$enters into the calculation. I am sure, however, that one can prove that e.g. $$\pi$$ is a root of $$g(x)$$, without resorting to geometric principles. After all, the geometric proof of the limit relation does not alone even imply that sine and cosine are analytic - for that, one has to demonstrate that the power series are absolutely convergent for all x (which they are).
 * As for my multivariable argument, the point of that was to show that one can evaluate that particular definite integral without resorting to the trigonometric substitution.--Jasper Deng (talk) 21:28, 22 April 2016 (UTC)
 * I would also take issue with the multivariable calculus formulation (contrary to what I said above when I said it was not circular). I do not believe you can obtain a formula for the area of a circle C of radius $$ r $$ by calculating $$ \int\int_C 1 r dr d\theta $$ any more than you can obtain the formula for the area of a rectangle $$ R $$ by calculating $$ \int\int_R 1 dx dy $$. The reason for the latter is one cannot know the area of an infinitesimal rectangle of width $$ dx $$ and height $$ dy $$ is $$ dx dy $$ if one doesn't  already know the formula for the area of a rectangle. Similarly, how does one calculate the infinitesimal region in polar coordinates? It is generally calculated as the difference between two sectors - and the areas of sectors is unknown if we don't know the area of a circle. Or if you want to convert coordinate systems more formally using the Jacobian determinate, you need to be able to differentiate the sine and cosine functions. (see  https://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_.28area.29) This leads to the same issues as before. I am rapidly reaching the conclusion that elementary calculus - single or multivariable -  is actually useless in trying to calculate the area of a circle.
 * Your idea about using power series substitutions is workable as long as you calculate both the area and the circumference using the analytical substitutions you give. The circumference would be calculated using the standard formula for arc length. Then you should be able to show that the area is the circumference multiplied by $$ r/2 $$. But this sort of argument is probably not appropriate for the target audience. At least it would take a lot of work to make them readable and accurate.
 * Reading back on the talk page I see that similar points have been made before by David Epstein and CMummert. I think it is appropriate to completely remove the calculus arguments from the page. They obscure rather that enlighten and seem to be little more than meaningless symbolic manipulations. — Preceding unsigned comment added by 78.100.47.243 (talk) 14:09, 23 April 2016 (UTC)
 * Err, no. The only thing you need for the change of coordinates is the Pythagorean identity, which as I said before, can be shown to hold without relying on the area of the circle, so you can calculate the Jacobian determinant completely analytically; the power series definition also gets around the issue of differentiation. Also, the area of a rectangle is one of the basic axioms defining area so it can be taken for granted. That article also says that Riemann integration is explicitly a way to find area, so the symbolic manipulations are completely workable.
 * Even Archimedes' proof relies on the notion of the limit, or at the least, the supremum and infimum, of areas of polygons. A similar argument also justifies the "triangle proof" which is done completely independent of trigonometric functions, using only the geometric definition $$C = 2\pi r$$.
 * No-one needs to know the geometric properties of sine and cosine, other than their values at $$\pi$$ and other important values, to do this, so an analytic definition suffices. Besides, many readers are beginning calculus students for which this is an elementary exercise, so it is definitely appropriate for our target audience. --Jasper Deng (talk) 18:17, 23 April 2016 (UTC)
 * The multivariable calculus argument seems fine to me. Define a function $$\rho:\mathbb R^2\to\mathbb R$$ by $$\rho(x,y)=\sqrt{x^2+y^2}$$.  Also, let D be the disc $$r<1$$ in $$\mathbb R^2$$.  Note that $$|\nabla\rho|=1$$. Let $$\mathcal L^2$$ be the two-dimensional Lebesgue measure in $$\mathbb R^2$$ and $$\mathcal H^1$$ the one-dimensional Hausdorff measure.  Assume that we know that the one-dimensional Hausdorff measure of a circle of radius r is $$2\pi r$$ (this appears to be the IP's preferred "definition" of &pi;, but see below.  In any case, we can prove it as a theorem if necessary.)  By the coarea formula, $$\mathcal L^2(D) = \int_D |\nabla \rho|\,d\mathcal{L}^2 = \int_{\mathbb R} \mathcal H^1(\rho^{-1}(r)\cap D)\,dr = \int_0^1\mathcal H^1(\rho^{-1}(r)\cap D)\,dr = \int_0^1 2\pi r\, dr = \pi.$$  S ławomir  Biały  15:06, 24 April 2016 (UTC)

The anonymous editor's complaint seems valid to me. In particular, it's hard to tell which facts are being assumed as axioms at the start of this "proof". Nonetheless, the example seems valuable for Wikipedia's audience.

What about this solution: Instead of calling it a "proof", say that it "verifies", "corroborates", or "reproduces" the well-known expression for the area of a circle. Mgnbar (talk) 20:30, 23 April 2016 (UTC)


 * I don't agree that the proof is circular. The sine and cosine can be defined as certain convergent power series (e.g., in Rudin) or as solutions of a differential equation (e.g., in Ahlfors).  Facts like $$\frac{d}{d\theta}\sin\theta=\cos\theta$$ then do not require any special facts about circles.  Further, it has been argued (for reasons that are not very convincing) that analytical tools like standard facts about sine and cosine are off-limits because "we must show the equivalence of the power series definition and the geometric definition".  This is not true.  One of the standard definitions of the number &pi; is twice the least positive zero of the cosine, where the cosine is defined analytically; see pi or the text of Ahlfors for details.  So all we need to do is show that the area of a unit semicircle is equal to the smallest positive zero of the cosine.  That is, that the integral $$\int_{-1}^1\sqrt{1-x^2}\,dx=\pi/2$$.  This is easily done with a trigonometric substitution, as discussed in the article.   S ławomir  Biały  22:03, 23 April 2016 (UTC)


 * I think the relevant question is what is taken as an assumption and what is not. If we adopt the standpoint of the ancient Greeks, where everything is defined in terms of geometry (not unreasonable, since you can see geometry all around you, but where have you ever seen a number?) then the area of the circle, the sine, and the cosine all have a priori meanings, and everything must be justified in terms of those.  Whereas if we adopt set-theoretic foundations, then we have to pick definitions of area and of trigonometric functions, and we are free to adopt analytic definitions in terms of Lebesgue measure and power series if we so please.
 * If we adopt geometric foundations, then I think it's no surprise that an analytic proof ends up being circular. Conversely, if we take analytic foundations, I suspect that any purely geometric proof of the area formula will be circular.  Myself, I find analytic foundations more convenient but geometric foundations more intuitive.
 * For the proof under discussion, I do not see any circularity. For the purposes of this proof, we want to compute the measure of the set $$\{ (x, y) : x^2 + y^2 \le r^2 \}$$.  We assign a geometric meaning to this set, but really it's just some set given by some formula, and if we make the right substitution, a number comes out.  Which is fine.  Ozob (talk) 02:12, 24 April 2016 (UTC)


 * I think Jasper Deng puts a very fine point on the issue. The problem is to show $$\int_{-1}^1\sqrt{1-x^2}\,dx=\pi/2$$. But let us be clear what this means. The functions and constants in this discussion seem to have multiple definitions, and this I believe is causing misunderstanding. So let's be explicit. Let's refer to $$ f(x) $$ and  g(x)  when referring to the analytical "definitions" of $$ \sin x $$ and $$ \cos x $$. The terms $$ \sin x $$ and $$ \cos x $$ will refer to the $$ y $$ and $$ x $$ coordinates of a point on the unit circle. The symbol $$ \pi will refer to the constant that is the ratio of the circumference of a circle to its diameter. Let  the symbol p represent twice the least positive root of g(x).


 * Although we have talked of analytical and geometrical definitions of sine and cosine, this is speaking rather loosely. To be exact, mathematical objects cannot have more that one definition. Hence the scare quotes in the previous paragraph. Now suppose we want to calculate the integral . Normally we would use the substitution and, but as was argued above this requires us to know that , and this assumes we know the area of the circle. Alternatively, we could use the substitution x = f(u). Now we can calculate analytically that , convert the bounds of the integration (using the fact that p is  twice the least positive root of g(u)) and complete the integral analytically. This will give us the result that  (or by a simple generalisation that the area of a circle of radius r is pr^2)  .In other words it will tell us that the area of the unit circle is 2 times the least positive root of g(u). That is an interesting theorem but it is not the one we want. As I understand it, we want to show that the area of a circle of radius r is  where C is the circumference of the circle. We could get this desired result by setting up an integral to calculate the arc-length of a circle and solving it analytically to get the answer C= 2pr. However this analytical proof is now nothing at all like the proof presented on the page.  — Preceding unsigned comment added by 185.37.110.40 (talk) 05:33, 24 April 2016 (UTC)
 * But showing that latter fact is straightforward: You agree that one half of the circle is given by for |x| \leq R, right? Then we can compute . Using the (analytic) trigonometric substitution shows this quantity to be precisely 2Rp as expected.
 * But I assume you know this fact. I would support adding this as a footnote or other elaboration, but no more than just that, since I believe that this formality is a distraction for beginning calculus students, which are a substantial audience here.
 * Instead, I think this should be discussed in the article on the trigonometric functions themselves, for curious readers.--Jasper Deng (talk) 07:06, 24 April 2016 (UTC)
 * To add to what Jasper Deng has said, there is an error here in assuming that \pi=C/D is a rigorous definition of &pi; that can be presented independently of an analytical treatment. But it is not, because circumference is not something we can calculate using just pre-calculus geometry.  The length of a rectifiable plane curve is defined as its one-dimensional Hausdorff measure.  To compute this ultimately requires us to calculate an integral such as .   S ławomir  Biały  12:43, 24 April 2016 (UTC)


 * But the problem has not been addressed! The problem is that the argument as presented on this page is circular. Specifically, when the trigonometric substitution is used in the argument in the text of the article, the derivative  is used - and this fact relies on the fact that the area of the circle is \pi r^2. All the talk about analytic definitions is just a distraction from this basic point. The only way to demonstrate that this argument isn't circular is to produce a proof that  without using the formula for the area of a circle. If this can be done I will concede the point. If not, then my objection remains.

Sure. The series converges uniformly on \mathbb R, by the ratio test. So.  S ławomir Biały  12:30, 24 April 2016 (UTC)


 * And how do you know that ? If you are not calculating the Taylor series of \sin x, then you must simply be defining \sin x $$ to be $$ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$. In which case after the integration you get $$ A=pr^2 $$, where $$ p $$ is the least positive root of  $$ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$.  How do you know $$ p=\pi $$? The definition of $$ \pi $$ is given at the top of the page as the ratio between the circumference of a circle and its diameter .  — Preceding unsigned comment added by 78.100.47.243 (talk) 13:09, 24 April 2016 (UTC)


 * This follows from the definition of the sine. The sine function is defined as the power series $$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ (or it is defined as a solution of a certain differential equation, in which case the property you are after is similarly trivial).  I thought you said you would drop your objection now?  Anyway, as I already pointed out, the "definition" of &pi; as the ratio of circumference to diameter is not a proper definition to begin with, because "circumference" is not something that can be computed without reliance on an integral.  So the modern way to do this is to define all of the trigonometric functions first, and define &pi; as a period.  Then we prove things about circumference and area as theorems.  These are not hard theorems to prove, and it is left as an exercise to prove these on your own using what has already been said on this discussion page.  There is no circularity, period.
 * As I've already said, but will repeat at greater length here, the apparent circularity comes from the inadequacy of the "definition" of $$\pi=C/D$$. The difficulty is that the circumference is not a primitive geometrical concept, and requires some form of calculus to compute.  Archimedes used exhaustion, together with a certain pair of axioms (only one of which now bears his name), which was one of the earliest uses of a limit in mathematics.  This is not used in modern treatments, because Archimedes' second axiom is not a part of the standard foundations of mathematics.  Instead, the circumference of the circle is defined as a certain one-dimensional Hausdorff measure.  A little work shows that the one-dimensional Hausdorff measure of the unit semicircle is the integral $$\int_{-1}^1\sqrt{1+\frac{x^2}{1-x^2}}\,dx$$, an integral which is equal to a half-period of the sine.  (Indeed, one can show that the inverse trigonometric functions are given as integrals of this form, and the periods of the trigonometric functions are contour integrals around a simple pole of the integrand.)  To arrive at this integral, we exponentiate the normal bundle of the circle into a collar neighborhood, compute the area of the collar neighborhood as a Lebesgue integral, and then pass to the limit using dominated convergence and Fubini's theorem.   S ławomir  Biały  14:21, 24 April 2016 (UTC)
 * "How do you know $$ p=\pi $$? The definition of $$ \pi $$ is given at the top of the page as the ratio between the circumference of a circle and its diameter ." Then you clearly aren't listening because my comment above and Sławomir's comments clearly show why.--Jasper Deng (talk) 16:32, 24 April 2016 (UTC)
 * In case one is not convinced about the fact that sine and cosine indeed parametrize the unit circle in terms of the angle, note that for any complex number $$\theta$$, the Pythagorean identity holds. In other words, for real $$\theta$$, the point $$(\cos(\theta), \sin(\theta))$$ lies on the unit circle. Now note that when $$0 \leq \theta \leq p/2 = \pi/2$$, the restrictions of sine and cosine to that interval are bijective. Because sine and cosine are continuous on the compact interval $$[0, \pi/2]$$, they map it to the compact interval $$[0, 1]$$. Now define $$ f : \mathbb [0, \pi/2] \to \{(x, y) : x \geq 0, y \geq 0, x^2 + y^2 = 1\} $$ by $$ \theta \to (\cos(\theta), \sin(\theta))$$. As stated before, for any $$\theta$$, $$f(\theta)$$ lies on the unit circle. It is not hard to show that f is bijective, as follows: Let $$g : \{(x, y) : x \geq 0, y \geq 0, x^2 + y^2 = 1\} \to [0, \pi/2]$$ be defined by $$(x, y) \to \arcsin(y) = \int_0^y \frac{dt}{\sqrt{1 - t^2}} $$. We can write $$\arcsin(y) = \int_0^y \frac{dt}{\sqrt{1 - t^2}}  $$ because, with sine bijective on this interval, the derivative of arcsine is $$\frac{1}{\sqrt{1 - t^2}}$$ (by the properties of bijective differentiable functions, and the use of the Pythagorean identity (which, again, can be proven without geometric means)), and with the fact that $$\sin(0) = 0$$, we can use the fundamental theorem of calculus to recover arcsine from its derivative. Note also that for any $$x_1, x_2$$ in $$[0, 1]$$, $$x_1 < x_2 \implies y_1 = \sqrt{1 - x_1^2} > y_2 = \sqrt{1 - y_1^2}$$, since $$\sqrt{1 - t^2}$$ is also bijective on this interval, and the postimage of f contains only points where $$x_1^2 + y_1^2 = 1, x_2^2 + y_2^2 = 1$$. Therefore, g maps only one point on the unit circle back to only a single point in the interval from 0 to $$\pi/2$$. Since it is an inverse relation of f, and is a function, g is clearly the inverse of f. A similar argument can be used for the other quadrants. Note too that the integral $$\arcsin(y) = \int_0^y \frac{dt}{\sqrt{1 - t^2}}$$ is precisely the length of the arc between the x-axis and the point $$(\sqrt{1 - y^2}, y)$$, and hence equal to the angle subtended by that arc, by definition of the radian. In other words, an angle can be well-defined for every point on the unit circle in a manner consistent with the so-called geometric definitions of sine and cosine, and no matter what "definition" you use for sine and cosine, the proof is sound (this implicitly resolves the question of how to prove the limit relation, because I just showed that the "geometric definition" is completely in line with the analytic ones which were chosen to be particular power series).--Jasper Deng (talk) 19:50, 24 April 2016 (UTC)

The new proof, using Green's theorem, does not even use trigonometric functions. In fact, it doesn't even mention &pi;! All that is shown, in a simple and direct way, is that the area enclosed by a circle is equal to R/2 times the circumference of the circle. We can then show that this is equal to $$2\int_{-R}^R\sqrt{R^2-x^2}\,dx$$ by the good old fashioned Fubini theorem. No trigonometry required.  S ławomir Biały  23:26, 24 April 2016 (UTC)


 * It seems this conversation is drawing to an end. I feel I have made my case and won't be continuing after these comments. It seems we have wandered quite a distance from the text of the article. This text presents itself as a simple trigonometric substitution to calculate the area of a circle. A method that will be familiar to any second semester calculus student. However it is being argued that the argument is anything but this. The definition of the sine and cosine functions are not the simple geometrical ones that are familiar to the calculus students, but instead are those presented in introductory Analysis courses; the symbol $$ \pi $$ does not represent the ratio of the circumference of a circle to its diameter, as used in the introduction to the article and in every other argument on the page up till then, but instead refers to the period of a certain power series - the existence of such a number being generally demonstrated in graduate analysis courses.


 * If these are the intended meaning of the symbols, then this should be stated in the proof of the theorem, or it is, as you say "apparently circular". If one definition of terms makes the argument circular and another doesn't, then it is essential to state which definitions you are using. Different definitions will affect the development the argument, and of course you still end up with two distinct conclusions depending of what definition of $$ \pi $$ you use. Any proof that the two definitions are equivalent also presupposes the conclusion.


 * Of course you are free to use Green's Theorem, complex analysis, or any other method ("we exponentiate the normal bundle of the circle into a collar neighborhood, compute the area of the collar neighborhood as a Lebesgue integral, and then pass to the limit using dominated convergence and Fubini's theorem." ) to prove the theorem. But if this is your intent, then you should present those arguments in the text and not on the talk page. If it is essential for a reader to know the power series $$ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ is periodic in order to make sense of the proof and to remove its "apparent circularity", then this fact must be established. Otherwise readers will see the proof and assume that it is exactly what it looks like: a simple trigonometric substitution of the geometrically defined sine function computing the area of a circle in terms of the straightforward definition of $$ \pi $$. Nothing in this deceptive argument is  what it appears to be. And any attempt to clarify it seems always  to descend into a morass of analytical obscurantism.78.100.47.243 (talk) 09:46, 25 April 2016 (UTC)

It's not "analytical obscurantism". The fact is that the "definition" $$\pi=\frac{C}{D}$$ is not really a definition at all, because one doesn't know what the quantity "C" is. Making proper sense of this formula requires the use of mathematical analysis (e.g., either using ideas of rectifiability or Hausdorff measure). You've offered no alternative in any of your arguments here. Perhaps if you could be bothered to say how you would define the circumference in a way that is suitable for doing some rigorous mathematics, then it would be much easier to respond in a way that you find satisfactory (and doesn't rely on "analytical obscurantism"). But instead, it seems like you don't really know what circumference means, and yet continue to defend $$\pi=C/D$$ as the gold standard of mathematical definitions. You want a rigorous calculus proof that apparently uses no analysis (is such a thing even possible?) In order to appease you, I have added a section of the article on the definition of &pi;. I have also added a paragraph explaining how to do the onion proof without relying on any trigonometry whatsoever (although of course what was written was already correct), and a section on Green's theorem that doesn't even mention either of your two "versions" of &pi;. These are things that you complained about. I expect thanks, not a dismissive rant.  S ławomir Biały  11:33, 25 April 2016 (UTC)

Okay, now you're plain out refusing to listen and it's becoming tiring. I and Sławomir demonstrated multiple times why the "geometric definition" of trigonometric functions and pi are completely in agreement with the analytical ones, yet you still seem to think that they are somehow distinct or otherwise incompatible. That's just not true. It is also plainly false that the power series definitions are not those given in introductory calculus courses. Quite contrarily, introductory calculus courses often do emphasize that the modern definition of the trigonometric functions is the power series one, especially when they get to Taylor series. The "geometric proof" the limit relationship you seem to hold steadfast to (but as we have shown, is not needed when there are other proofs) is only presented in a handwavy fashion in these classes, in order to suggest intuition, but at no time do these classes claim to rigorously define sine and cosine using just a unit-circle-based definition.--Jasper Deng (talk) 15:27, 25 April 2016 (UTC)


 * Is this a fair summary of the dispute? The area can be computed using calculus in a non-circular fashion, as Slawomir Bialy and Jasper Deng have argued. However, the article (at least at the time that this section was created) did not really carry out that proof. Rather, it started with unstated and unclear definitions/assumptions, did some computations, and got the right answer. So yes the proof can be done (as multiple paragraphs of argumentation on this talk page show), but it was not done in the article. Is that fair?
 * I would be wary of making assumptions about what is taught around the world. My sense of the USA is that teachers typically define the trig functions using right triangles or points on rays, and make heavy use of these functions long before they introduce power series. My guess about the typical USA reader of this article is that they have never heard of power series and have no appreciation of how &pi; is rigorously defined. So, while it is important to make the arguments here technically correct, please do not forget the audience. Mgnbar (talk) 17:11, 25 April 2016 (UTC)
 * The problem is that those definitions are not suitable for analytic purposes. They do not, on their own, establish that sine and cosine are the entire functions we know them to be: how do we know that the functions are continuously differentiable and hence suitable for calculus? In particular, the right triangle definition cannot capture the full periodicity of the functions. Both definitions also suffer from the fact that they do not admit a concrete way to compute the functions for an arbitrary real number: most students who learn the functions before calculus are merely given calculators to compute the values, without any insight into how those values are actually obtained. I think, however, that Sławomir's additions are more than sufficient to clear this up.--Jasper Deng (talk) 17:40, 25 April 2016 (UTC)
 * The article currently goes to great lengths to clarify the sense in which these proofs can be made properly rigorous. These methods were dismissed by the OP as "analytical obscurantism".  Apparently, to give a rigorous proof did not fit his narrative that there is no proof that $$A=\pi r^2$$ that did not rely in some way on circular reasoning.  Although it is true that US college freshman are not taught rigorous mathematical methods, that is hardly surprising.  But it also does not change the fact that these are rigorous proofs, when the terms involved have their usual mathematical meanings (as opposed to the lies that we tell college freshman).  For example, that the sine and cosine functions actually have formal definitions, which do not correspond exactly to what appears in a pre-calculus textbook, but which are easily proven to be equivalent.  Is there anything further to discuss?   S ławomir  Biały  18:09, 25 April 2016 (UTC)


 * Let me say it a different way. The version of April 25 is much better than the version of April 21, because the April 25 version summarizes the theoretical underpinnings needed to make sense of the rigorous proofs. These underpinnings were not obvious to the original poster, who was thinking about this material far harder than the typical reader. In this way, the original poster has done us a great service. Although this talk page section has been contentious, both he and the article editors deserve our thanks. Mgnbar (talk) 20:02, 25 April 2016 (UTC)


 * I would've contended that those assumptions were meant to be already understood, but I will concede that thanks to a lack of rigor in many introductory calculus courses, the clarifications were necessary.--Jasper Deng (talk) 01:20, 26 April 2016 (UTC)

Hey, I don't think that the proof is circular. In geometric (not power series) sense, differentiate sin x and we get cos x requires the limit sin x/x, when x tends to zero, and x is measured in radians. It comes out that we should know area of sector = (1/2) r^2 theta, and somebody thinks that it must comes from the area of circle is pi r^2. But in my opinion that is not the case. Area of sector can be proved by getting A=(1/2)rs, where s is arc length. How can I prove A=(1/2)rs? It's easy. Just cut the arc length in many infinitesimals, and let it be delta s. As we all know that area of triangle is (1/2) base times height, we can deduce that $$\sum \frac{1}{2} r \Delta s = \frac{1}{2} r \sum \Delta s$$. When I take the limit, the result become accurate. So the result is (1/2) rs. Besides, we can prove that s= r theta. You can refer to this page. So finally we get that A=(1/2)r^2 theta, WITHOUT assuming the area of circle is pi r^2. That is my opinion. --Raycheng200 (talk) 03:03, 26 June 2016 (UTC)

By the way how can you prove that integrate sin^2 theta from x=0 to x=pi/2 equals integrate cos^2 theta from x=0 to x=pi/2 ? Use integration by parts? --Raycheng200 (talk) 03:20, 26 June 2016 (UTC)


 * $$\sin\theta = \cos(\pi/2-\theta)$$.  Sławomir Biały  (talk) 11:02, 27 June 2016 (UTC)