Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question

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

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

7  
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
11  
$\pi r^2$ is exact, as far as the human race knows. –  leonbloy Jan 19 '13 at 4:15
1  
what's wrong with $\pi r^2$? –  Foga Mar 13 at 19:19
add comment

3 Answers 3

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).

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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