16
$\begingroup$

This question already has an answer here:

I searched many times about the cause of the circle area formula but I did not know anything so ...

Why is the area of the circle $\pi r^2$?

Thanks for all here.

$\endgroup$

marked as duplicate by Hans Lundmark, Aditya Hase, Joonas Ilmavirta, daw, Joel Reyes Noche Dec 1 '14 at 8:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ That the area depends on the radius-squared is somewhat of an observation from considering inscribed and circumscribed regular polygons. The occurrence of $\pi$ is strictly for keeping track of exactly what the proportionality constant is between $r^2$ and the area of the circle. $\endgroup$ – Cameron Williams Nov 30 '14 at 22:32
  • 2
    $\begingroup$ Has this really never been asked here before? $\endgroup$ – HDE 226868 Nov 30 '14 at 22:53
  • 1
    $\begingroup$ what is $\pi$ in your formula? $\endgroup$ – abel Nov 30 '14 at 22:53
  • 1
    $\begingroup$ this is my favorite explanation/animated gif ... people.wku.edu/tom.richmond/Pir2b.html $\endgroup$ – John Joy Dec 1 '14 at 16:27
24
$\begingroup$

How Archimedes viewed it:

enter image description here $%emptyuselesstextemptyuselesstextemptyuselesstext$

As the width of the slices approaches $0$, the object on the right-hand side approaches a rectangle of width $2\pi r /2 = \pi r$ and height $r$, hence area $\pi r^2$.

$\endgroup$
  • 3
    $\begingroup$ (+1) Excellent answer, at the right level for someone who needs to ask this question. $\endgroup$ – John Bentin Nov 30 '14 at 23:07
  • $\begingroup$ Actually, Archimedes used a method of exhaustion: en.wikipedia.org/wiki/Area_of_a_disk#Archimedes.27s_proof $\endgroup$ – Alan C Dec 1 '14 at 2:26
  • $\begingroup$ is "exhaustion" Greek for "calculus"? $\endgroup$ – John Joy Dec 1 '14 at 16:28
10
$\begingroup$

The area of a circle with radius $r$ is just $r^2$ times the area of the unit circle, by homothety. So the area of the circle is the square of the radius times a universal constant, given by: $$\begin{eqnarray*}2\int_{-1}^{1}\sqrt{1-x^2}\,dx &=& 4\int_{0}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}x^{-1/2}(1-x)^{1/2}\,dx\\ &=& 2\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} = \color{red}{\Gamma(1/2)^2}.\end{eqnarray*}$$

$\endgroup$
  • 26
    $\begingroup$ Seriously? Beta and Gamma functions? $\endgroup$ – user2345215 Nov 30 '14 at 22:39
  • 5
    $\begingroup$ I like Jack's answer. It may not help OP, but I wouldn't have thought to do that -- so it helps me out. (+1) $\endgroup$ – user137731 Nov 30 '14 at 22:41
  • 3
    $\begingroup$ I think that in mathematics we should not regard the $\Gamma$ function as an alien. It is not so different from the sine or cosine function, speaking about its Weierstrass product. $\endgroup$ – Jack D'Aurizio Nov 30 '14 at 22:41
  • 4
    $\begingroup$ That's definitely an overkill. You can evaluate the integral by $x\mapsto \sin x$ $\endgroup$ – UserX Nov 30 '14 at 22:43
  • 2
    $\begingroup$ So $\pi$ is half the length of the unit circle, or the fist positive root of the sine function, but we are all stating the same thing. The only difference is that, through the Bohr-Mollerup theorem, the $\Gamma$ function exists even without knowing any circle or trigonometric function. In a way, it looks more natural to me to define $\pi$ as $m!^2$ with $m=-\frac{1}{2}$. $\endgroup$ – Jack D'Aurizio Nov 30 '14 at 22:44
5
$\begingroup$

A number of different ways of showing this exist. First notice that the area has to be $(\text{constant}\cdot r^2)$ because the area of a region of any shape in a plane must be proportional to the square of the distances. E.g. if you multiply all distances by $3$, then the area is multiplied by $9$. And "constant" in this case means it's the same number regardless of what $r$ is. So now the question is: Why must the "constant" be the same as the ratio of circumference to diameter?

As $r$ increases, we have \begin{align} \text{rate of growth of area} & = \text{size of boundary} \times\text{rate of motion of boundary} \\ & = \text{circumference} \times \text{rate at which $r$ is changing} \\ & = 2\pi r \times \text{rate at which $r$ is changing}. \end{align} From calculus, recall that $$ \text{rate of change of $r^2$} = 2 r \times\text{rate of change of $r$}. $$ So we have $$ \text{rate of growth of area} = \pi\times \text{rate of change of $r^2$}. $$ So the area grows at the same rate at which $\pi r^2$ grows. That makes them always equal if they're equal when $r=0$. And it's easy to see that they're equal when $r=0$.

That's only one way to do it; there are others.

PS: Supposing we don't know calculus; how would we know that $$ \text{rate of change of $r^2$} = 2 r \times\text{rate of change of $r$}\text{ ?} $$ We could do that as follows. A square whose side has length $r$ is growing because its north side is moving northward and its east side is moving eastward. Then \begin{align} & \text{rate of growth of $r^2$} \\ = {} & \text{rate of growth of square's area} \\ = {} & \text{size of moving boundary}\times\text{rate of motion of boundary} \\ = {} & 2r \times \text{rate of change of $r$}. \end{align}

$\endgroup$
  • 1
    $\begingroup$ I really like this approach. It makes a lot of intuitive sense. $\endgroup$ – Chantry Cargill Nov 30 '14 at 22:54
  • $\begingroup$ @ChantryCargill : Thank you; I'm glad you like it. $\endgroup$ – Michael Hardy Nov 30 '14 at 23:09
3
$\begingroup$

Some variety, the circle of geodesic radius $\rho$ on the unit sphere has area $$ 2 \pi (1 - \cos \rho).$$ i will look that up in a minute...Yep, compared with THIS, take their $r=1$ and $h=1-\cos \rho.$

Less familiar, the circle of geodesic radius $\rho$ in the non-Euclidean plane of constant curvature $(-1)$ is $$ 2 \pi (\cosh \rho - 1). $$

Go Figure.

In both cases, the limit as $\rho \rightarrow 0$ agrees with $\pi \rho^2,$ the error is of size $O(\rho^4).$

$\endgroup$
3
$\begingroup$

It's the evauluation of the definite integral $$2\int_{-r}^{r}y\,dx$$ where $y=\sqrt{r^2-x^2}$, so it's really $$2\int_{-r}^r\sqrt{r^2-x^2}\,dx$$ In other words, it's twice the area under the curve of $y$, from $x=-r$ to $x=r$.


There's no way to evaluate this integral with a simple formula, like would be if, say, $y=x^3$. The most common method used is a form of trig substitution, where $x \to \sin u$. From here, it's easy to show that $$dx=\cos u \, du$$ and so the previous integral becomes $$\int_{-r}^{r} \cos u \sqrt{r^2-\sin^2u}\,du$$ If $r=1$, we can use $$\cos^2u=1-\sin^2u$$ to make this easier: $$\int_{-r}^{r} \cos^2u\,du$$ Otherwise, it takes some playing around to get the answer.

By the way, this method is particularly valuable with ellipses ($A=\pi ab$).

$\endgroup$
  • 1
    $\begingroup$ It wouldn't hurt to mention the $x\mapsto r\sin x$ substitution, it may not be obvious how to solve this to a beginner. $\endgroup$ – user2345215 Nov 30 '14 at 22:47
  • $\begingroup$ @user2345215 True. And that should explain the $\pi$. $\endgroup$ – HDE 226868 Nov 30 '14 at 22:47
  • $\begingroup$ @MichaelHardy Thank you for the edits. $\endgroup$ – HDE 226868 Nov 30 '14 at 22:54
  • 2
    $\begingroup$ @HDE226868 : Actually it's the value of, not the "solution to", the integral. One doesn't "solve" integrals; one evaluates integrals. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 2 '14 at 16:39
2
$\begingroup$

For another integration method, let $C$ be the interior of the circle $x^2+y^2=r^2$ with radius $r$. Then the area, with the aid of polar coordinates, is given by

\begin{align*} \iint_C dA &= \int_0^{2\pi}\int_0^r\rho\,\mathrm{d}\rho\,\mathrm{d}\theta\\ &=\frac{1}{2}r^2\int_0^{2\pi}\,\mathrm{d}\theta\\ &=\pi r^2. \end{align*}

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.