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I was looking for proofs using Calculus for the area of a circle and come across this one $$\int 2 \pi r \, dr = 2 \frac {r^2}{2} \pi = \pi r^2$$ 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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Circle areas is twice the area under this curve. –  user2468 Aug 28 '12 at 16:40
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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) –  DonAntonio Aug 28 '12 at 16:40
    
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. –  Mark Bennet Aug 28 '12 at 17:26
    
Related (duplicate?): math.stackexchange.com/questions/625 –  BlueRaja - Danny Pflughoeft Aug 28 '12 at 18:17
    
@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! –  process91 Aug 28 '12 at 19:16
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12 Answers 12

up vote 17 down vote accepted

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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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. –  Michael Hardy Aug 28 '12 at 17:34
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@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. –  Steven Stadnicki Aug 28 '12 at 17:59
    
@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? –  Michael Hardy Aug 28 '12 at 18:10
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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 $\pi = \int_{\mathbb{R}} \frac{dx}{1+x^2}$ :-). –  copper.hat Aug 28 '12 at 18:30
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$$ \int_0^a 2 \pi r \,dr $$ The above integral seems geometrically as below figure.

enter image description here

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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 $$\frac{1}{2}\int_0^{2\pi}r^2(\cos^2t+\sin^2t)\,dt=\frac{r^2}{2}\int_0^{2\pi}dt=\pi r^2$$

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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? –  DonAntonio Aug 28 '12 at 17:16
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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 $2πr$

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"..but we need to have infinite divided parts to get an exact rectangle shape". Can you prove that? :) –  Sawarnik May 13 at 17:36
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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.

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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.

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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 $0$.

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This is very nice –  Alyosha Jun 8 '13 at 12:31
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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.

Formally

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

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 $A(r)$.

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

Now,$$\int{dA}=\int_{0}^{2\pi}\frac{1}{2}(r^2){\delta\theta}$$

$$\int{dA}=\frac{1}{2}(r^2)\int_{0}^{2\pi}{\delta\theta}$$

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

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

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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.

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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. –  David H Sep 12 '13 at 8:19
    
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 –  Ibraheem Sep 12 '13 at 12:31
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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.

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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 $M$.

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