# Is there an exact expression for the area of a circle, other than the approximate “$\pi r^2$”, known to the human race? [closed]

Let, area A, of a circle of radius R = ${\alpha}R^{\beta}+{\gamma}$, where, $\alpha$, $\beta$, $\gamma$ are constants

When, $R=0$, the area A is $0$, which gives $\gamma = 0$

the, Area A = ${\alpha}R^{\beta}$, where, $\alpha, \beta \gt 0$

Is there a way to get these constants $\alpha, \beta$

Is the assumption flawed ? If "YES", give reasons ?

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## closed as not constructive by BenjaLim, Ittay Weiss, Erick Wong, Fabian, AangJan 19 '13 at 6:32

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Yes, there is a way: $\alpha=$ area of the unit circle, and $\beta=2$. –  Giuseppe Negro Jan 19 '13 at 3:19
NOTE: Even when R tends to $\infty$ the area of a circle of radius R will be less than the area of a square of side R. –  Rajesh K Singh Jan 19 '13 at 4:08
$\pi r^2$ is exact, as far as the human race knows. –  leonbloy Jan 19 '13 at 4:15
what's wrong with $\pi r^2$? –  Foga Mar 13 at 19:19

There are some good reasons based on Euclid's axioms of geometry that we consider the area of a circle to be $\pi R^2$ (though changing the axioms can certainly change things). The approach you are taking should lead to the same conclusion, if we examine the geometry of the situation.

To find another constraint on your equation, we need a "test" circle (beyond the trivial case you already mentioned, of radius zero). The assumptions made in defining the circle will in turn define how the circle operates (for example, simply deciding that the circumference of a circle is equal to $2\pi R$, and that calculus is allowed, means you can use the Onion Proof, adding up concentric rings).

A common geometrical proof first used by Archimedes involved circumscribing a regular polygon around a circle, and inscribing another within the circle, and increasing the number of sides of the regular polygons towards infinity. The area of the circle would lie somewhere between the area of the polygons (which could be composed in turn, by triangular regions).

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It can be shown that the area of a circle of radius $r$ is exactly $\pi r^2$.

If $C$ is the circumference of the circle, then the area of the circle is exactly $\dfrac{C^2}{4\pi}$. That's an "exact expression for the area of the circle other than $\pi r^2$". But just what you intended in your question is not altogether clear.

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Let, $A_2, A_3$ be the area of circles with radius 2,3 respectively.
$A_2/A_3=(\frac{2}{3})^\beta$
Logically, $A_2 \lt A_3$, which implies, $\beta \gt 0$
$\alpha \gt 0$ for a positive area.