Why is the area of the circle $πr^2$? [duplicate]

Question

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.

Answer 1

How Archimedes viewed it:

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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 $\pi r^2$.

Answer 2

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: $$\begin{eqnarray*}2\int_{-1}^{1}\sqrt{1-x^2}\,dx &=& 4\int_{0}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}x^{-1/2}(1-x)^{1/2}\,dx\\ &=& 2\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} = \color{red}{\Gamma(1/2)^2}.\end{eqnarray*}$$

Answer 3

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}

Answer 4

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 $O(\rho^4).$

Answer 5

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 $x=r$.

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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 ($A=\pi ab$).

Answer 6

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*}