I was looking for proofs using Calculus for the area of a circle and come across this one $$\int 2 \pi r \, dr = 2\pi \frac {r^2}{2} = \pi r^2$$ and it struck me as being particularly easy. The only other proof I've seen was by a teacher and it involved integrating $x = \sqrt{r^2 - y^2}$ from $-1$ to $1$, using trig substitutions and then doubling the area to get $\pi r^2$ but the above proof seemed much more straight forward.

Is it a valid proof, or is it based on circular logic or some other kind of fallacy?

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    $\begingroup$ Circle areas is twice the area under this curve. $\endgroup$ – user2468 Aug 28 '12 at 16:40
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    $\begingroup$ The one you came across with doesn't seem to make much sense without at least limits in the integral (to see what are we integrating) $\endgroup$ – DonAntonio Aug 28 '12 at 16:40
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    $\begingroup$ The same thing works for the volume of a sphere as integral of the surface area - and in higher dimensions too. See experimentX's diagram. $\endgroup$ – Mark Bennet Aug 28 '12 at 17:26
  • $\begingroup$ Related (duplicate?): math.stackexchange.com/questions/625 $\endgroup$ – BlueRaja - Danny Pflughoeft Aug 28 '12 at 18:17
  • $\begingroup$ @andreas.vitikan Note that your teacher's approach is what Jennifer Dylan describes above, and while the calculation of that integral is more difficult, the idea behind it is quite straightforward. On the other hand, the integral you have is quite easy to calculate but the background is not as intuitive, IMHO. Good question all around, and nice answers as well! $\endgroup$ – process91 Aug 28 '12 at 19:16

15 Answers 15


Possibly the proof that you found is what the Wikipedia article for the area of a disk calls "The Onion Proof".

Although I would probably use the following double integral instead:

$$ \text{Area of circle} = \iint_{x^2 + y^2 \leq R}1 \, dx\,dy $$

and then calculate the integral using polar coordinates to get

$$ \iint_{x^2 + y^2 \leq R}1 \, dxdy = \int_0^{2 \pi} \int_0^R r \, dr\,d\theta = \int_0^R 2\pi r \, dr = \pi R^2 $$

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    $\begingroup$ This seems more complicated than it needs to be. The circumference is $2\pi r$; the infinitely small width of the ring is $dr$; so the infinitely small area of the ring is $2\pi r\,dr$. Integrate that from $0$ to $R$ to get the whole area. $\endgroup$ – Michael Hardy Aug 28 '12 at 17:34
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    $\begingroup$ @MichaelHardy How do you know the circumference is $2\pi r$? Once you have that fact I agree that the onion is more straightforward, but the 'first-principles' approach of this proof is nice in its own right. $\endgroup$ – Steven Stadnicki Aug 28 '12 at 17:59
  • $\begingroup$ @StevenStadnicki : In the present day, that's usually taken to be the definition of $\pi$, i.e. it's the ratio of circumference to diameter. Of course, there's the more basic question: how do you know that that ratio is the same for all circles? Is that what you meant? $\endgroup$ – Michael Hardy Aug 28 '12 at 18:10
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    $\begingroup$ Following Michael's comment, somewhere you must define $\pi$. When I went to secondary school it was defined as the ratio of the circumference to the diameter of a circle. Nowadays I prefer $\pi = \int_{\mathbb{R}} \frac{dx}{1+x^2}$ :-). $\endgroup$ – copper.hat Aug 28 '12 at 18:30

The above integral seems geometrically as below figure.

enter image description here

\begin{align} \int_0^r 2\pi r\, dr& = 2\pi\int_0^r\ r\, dr \\ & = 2\pi\bigg\lvert_0^r\ \frac{r^2}{2} \\ & = \bigg(2\pi \frac{r^2}{2}\bigg)-\bigg(2\pi \frac{0^2}{2}\bigg) \\ & = \frac{2\pi r^2}{2} \\ & = \require{cancel} \frac{\cancel{2}\pi r^2}{\cancel{2}} \\ & = \color{red}{\pi r^2} \end{align}

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    $\begingroup$ It's also worth noting that you can find the area of that triangle to get the area of a circle. Very nice. $\endgroup$ – adam Oct 24 '16 at 1:27

There's a particularly simple formula using line integrals: if $\,\gamma\,$ is a simple, closed and smooth (at least by parts) path (in the positive direction), the area of the inclosed region equals $$\frac{1}{2}\oint_\gamma x\,dy-y\,dx$$

In our case, we can take the path $\,\gamma(t)=(r\cos t\,,\,r\sin t)\,\,,\,t\in [0,2\pi)\,$ , and get $$\frac{1}{2}\int_0^{2\pi}r^2(\cos^2t+\sin^2t)\,dt=\frac{r^2}{2}\int_0^{2\pi}dt=\pi r^2$$

  • $\begingroup$ Do note that the above proof is the same that Adrian wrote after applying Green's theorem (either way, it doesn't matter)...nice, uh? $\endgroup$ – DonAntonio Aug 28 '12 at 17:16

I would like to show you another method how we can prove that the area of circle is $\pi r^2$ by using infinity part of a circle. I divided only 8 parts in my picture to demostrate how to apply that method but we need to have infinite divided parts to get an exact rectangle shape. After that We can write easily that

Area of circle = $\pi r .r =\pi r^2$

enter image description here Note: I assumed that we know the circumference is $2πr$


Just one remark. The so-called onion proof is a special case of the co-area formula. This formula is a rigorous justification of all those computations that we learned in the first course of general physics. It is a "curvilinear" generalization of Fubini's theorem: instead of slices, you integrate over hypersurfaces like a sphere. And also the fact that "differentiating the volume gives the area" is a consequence of the same theorem.


Here is another 'calculus proof'. (Basically integrating along $\theta$ rather than $r$.)

Approximate the circle from the inside using a regular n-sided polygon formed from the vertices $(r \cos k\theta_n, r \sin k\theta_n)$, where $\theta_n = \frac{2 \pi}{n}$, and $k = 0,...,n-1$. Draw lines between adjacent vertices and between the origin and each vertex. This splits the polygon into $n$ triangles with sides $r,r,2r \sin \frac{\theta_n}{2}$. The area of each polygon is given by $A_n = 2 r \sin \frac{\theta_n}{2} \frac{1}{2} r \sqrt{1 - (\sin \frac{\theta_n}{2})^2} = r^2 \sin \frac{\theta_n}{2} \cos \frac{\theta_n}{2}$.

Then the area of the circle is given by $\lim_{n \to \infty} n A_n = \lim_{n \to \infty} n r^2 \sin \frac{\theta_n}{2} \cos \frac{\theta_n}{2}$. Since $\lim_{x \to 0, x \neq 0} \frac{\sin x}{x} = 1$, we have $\lim_{n \to \infty} n \sin \frac{\theta_n}{2} = \lim_{n \to \infty} n \sin \frac{\pi}{n} = \lim_{n \to \infty} \pi \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}} = \pi$. Consequently we have $\lim_{n \to \infty} n A_n = \pi r^2$.

Note: To show that the area of the polygon converges to the area of the circle, note that the area between the polygon and the circle is bounded by $n r \sin \frac{\theta_n}{2} r ( 1 - \cos \frac{\theta_n}{2})$. A calculation along the above lines shows that this converges to $0$.

  • $\begingroup$ This is very nice $\endgroup$ – Meow Jun 8 '13 at 12:31
  • $\begingroup$ @copper.hat thanks for the great answer. Can you explain why the following is true? Since $\lim_{x \to 0, x \neq 0} \frac{\sin x}{x} = 1$, we have $\lim_{n \to \infty} n \sin \frac{\theta_n}{2} = \lim_{n \to \infty} n \sin \frac{\pi}{n}$ $\endgroup$ – Neo M Hacker Nov 27 '17 at 5:22
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    $\begingroup$ @NeoMHacker: Recall that ${\theta_n \over 2} = {\pi \over n}$. Then $n \sin {\pi \over n}= \pi {\sin {\pi \over n} \over {\pi \over n}}$. $\endgroup$ – copper.hat Nov 27 '17 at 5:24
  • $\begingroup$ Ah, I'm an idiot. Got it! $\endgroup$ – Neo M Hacker Nov 27 '17 at 5:26
  • $\begingroup$ @NeoMHacker: Everything is obvious in retrospect :-). $\endgroup$ – copper.hat Nov 27 '17 at 5:27

What you have here:

$$ \int 2 \pi r \, dr = 2 \frac {r^2}{2} \pi = \pi r^2 $$

does not represent an area because the integration is not bounded (also, a constant is missing on the RHS). An area should be for something with bounds (limits). However, the formula you mentioned is used in what is known as Onion proof for area of the circle (please do a find on 'onion'). This proof divides the circle into rings as explained in the link.


enter image description here Consider a circle of radius $r$ with $O$ as its center.

Now, consider an arc $XY$of the circle which subtends an angle of $\theta$ at the center.

let, $dA$ be the area of the segment $XOY$

if we take infinitesimally small angle $\delta\theta$ then we have, $dA=\frac{1}{2}(r^2)(\sin{\delta\theta})$

we know that $\lim_{\delta\theta ->0} \frac{\sin\delta\theta}{\delta\theta} =1$

using, the above fact, we have, $dA=\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$

Now, $\lim_{\delta\theta ->0} dA = \lim_{\delta\theta ->0}\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$

we, have, $dA=\frac{1}{2}(r^2){\delta\theta}$



$$A=\frac{1}{2}(r^2)(2\pi)$$ $$A={\pi}r^2$$

we, now have the area of a circle of radius $r$, and it is ${\pi}r^2$


You can actually convince yourself geometrically that

$$A'(r)=2 \pi r (*)$$

Intuitively, the rate of change of the area of the circle is the circumference.


$$A'(r) = \lim_{\Delta r \to 0} \frac{A(r+\Delta r) -A(r)}{\Delta r}$$

Now, geometrically it is pretty clear (but not really easy to prove mathematically) that the area of a corona between circles satisfies

$$2 \pi r_\text{in} (r_\text{out}-r_\text{in}) < \text{Area} < 2 \pi r_\text{out} (r_{out}-r_{in})$$

Using these inequalities it is easy to calculate the above limit which leads to $(*)$.

The formula you provided solves $(*)$ for $A(r)$.

  • $\begingroup$ If anyone is notified of this comment, could you please link to a resource that will prove the provided inequality 2𝜋𝑟in(𝑟out−𝑟in)<Area<2𝜋𝑟out(𝑟𝑜𝑢𝑡−𝑟𝑖𝑛) ? I would be very interested if such a proof exists $\endgroup$ – Math Enthusiast Aug 1 at 16:45

I just want to point out that your proof (as formalized by some of the answers above) is a special case of a more general fact. Let $\gamma(s)$ be a smooth, closed curve, and $N(s)$ the normal along the curve. Consider "inflating" the curve by moving every point a distance $\epsilon$ along the normal, to get a new closed curve $\gamma(s)+\epsilon N(s)$. Then the area enclosed by the inflated curve is

$$A(\gamma+\epsilon N) = A(\gamma) + \epsilon L(\gamma) + \frac{1}{2} \epsilon^2 \int_\gamma \kappa(s)\,ds$$ where $A(\gamma)$ is the area enclosed by $\gamma$, $L(\gamma)$ is the total arc length of $\gamma$, and $\kappa(s)$ is the signed curvature of the curve at $s$ (for simple curves, the total curvature is always equal to $2\pi$, by the Whitney-Graustein theorem, and the last term is just $\pi \epsilon^2$).

In particular, this formula tells us that perimeter of a region in the plane is the first derivative of area, with respect to inflation along the boundary normal. For the disk, inflation along the normal just amounts to scaling the radius.

Incidentally, all of the above carries over beautifully to surfaces $M$ enclosing volumes in $\mathbb{R}^3$:

$$V(M + \epsilon N) = V(M) + \epsilon A(M) + \epsilon^2 \int_M H\,dA + \frac{\epsilon^3}{3} \int_M K\,dA$$ Where $V$ is enclosed volume, $A(M)$ is the surface area of $M$, and $H$ and $K$ are mean and Gaussian curvature. By the Gauss-Bonnet theorem, the integral in the last term is always a multiple of $4\pi$ and depends only on the topology of $M$.


Start off with a first principle proof that $\lim_{x\to 0}\frac{\sin x}{x}=1$ is true
(we only need to know that the derivative of $sin(x)$ at $0$ is equal to 1).

Using the notation from there, you divide the circle into a $2^n$-gon and approximate the area with $2^n$ times the area of the small isosceles triangle wedges, each with a vertex angle of $\frac{2 \pi}{2^n}$ radians:

$\tag 1 2^n .5 \; a \sqrt{r^2 - (.25) a^2}$

Since $a(\theta)=r\, 2\sin\frac{\theta}{2}$, you can rewrite (1) as follows
$2^n (.5)\, r\; 2sin(\frac{\pi}{2^n}) \sqrt{r^2 - .25\, {(r\; 2sin(\frac{\pi}{2^n}))}^2}$
$2^n \, r^2\; sin(\frac{\pi}{2^n})\; cos(\frac{\pi}{2^n}) $
$\pi \, r^2\; \frac{sin(\frac{\pi}{2^n})}{\frac{\pi}{2^n}}\; cos(\frac{ \pi}{2^n}) $

and as $n$ goes to infinity you get the result,

AREA OF CIRCLE = $\pi r^2$

Note: Using properties of the sagitta you can show that this increasing sequence of area approximations gives you the only 'area definition' that makes sense.


The proof that depends on $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to calculate the area of circle, is not complete as the proof of that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ depends on area of circle equation and limit squeeze theorem.

  • $\begingroup$ This is false. The formula for the area of a circle is doesn't figure into the proof. See www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X/Geometric_Proof. The point is moot though, since there are a number of non-geometric alternative ways of defining the trig functions and their properties, such as power series or differential equations. $\endgroup$ – David H Sep 12 '13 at 8:19
  • $\begingroup$ Thank you for you reply. in the link you sent "From Area of Sector, the sector formed by arc AB subtending O is θ/2 ." that is using the circle are formula $\endgroup$ – Ibraheem Sep 12 '13 at 12:31

the integral of r(d(theta)) from 0 to 2π is 2πr (the circumference of a circle with radius r), now integrate 2πr(dr) from 0 to r and the answer is πr^2 (the area of a circle of radius r). This is my derivation.


If you have a regular polygon with n sides and r being the distance from the center to a vertex, and take the area of that, you get: A = nr^2sin(180(n - 2)/2n)cos(180(n - 2)/2n). As n goes to infinity, the polygon turns into a circle. The sine term goes to 1 so you can take that out and r^2 is a constant so we can put that back in at the end. We're left with ncos(180(n - 2)/2n)as n goes to infinity. The cosine term goes to 0 and n goes to infinity but the product turns out to be pi so we have pi*r^2.

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    $\begingroup$ please format your answer using the Mathjax tutorial $\endgroup$ – Surb May 18 '15 at 14:55
  • $\begingroup$ Wlecome to math.SE! First things first, you might find useful some tips about typesetting. Second, I think your argument relies more on geometry than it does on calculus. For instance: while the geometrical intuition behind "the polygon turns into a circle" stands clear, I think a proper explanation on the reason why the area of the $n$-sided polygons tends to the area of the circle would be needed. Also, the definition of the sine of a sexagesimal angle is not that natural, coefficient-wise. $\endgroup$ – user228113 May 18 '15 at 14:55

Consider the standard equation of circle $$x^2 + y^2 = a^2$$ where $a$ is the radius of the circle.


  • $\begingroup$ Can you please edit your post and add some new information that explains the proof. It will be particularly useful if you add information not included in the other posts. And while you're at it, please use LaTeX / MathJax. $\endgroup$ – Ertxiem May 11 at 18:33

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