Why is the area of the circle $πr^2$? 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.
 A: 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}
A: 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 ($A=\pi ab$).
A: 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 $O(\rho^4).$
A: How Archimedes viewed it:
 $%emptyuselesstextemptyuselesstextemptyuselesstext$
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 $\pi r^2$.
A: 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*}
A: 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:
$$\begin{eqnarray*}2\int_{-1}^{1}\sqrt{1-x^2}\,dx &=& 4\int_{0}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}x^{-1/2}(1-x)^{1/2}\,dx\\ &=& 2\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} = \color{red}{\Gamma(1/2)^2}.\end{eqnarray*}$$
