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Can the area of a circle be derived without calculus or Archimedes approach? The area is given as$\pi R^2$, where $\pi$ is defined by circumference=$2\pi R$. It is easy to derive it as an integral or by using the limit as a sequence of polygons (Archimedes). Is there a more elementary geometry derivation?

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    $\begingroup$ What geometry exists to calculate area at all? For curves.... no, nothing exist that isn't essentially calculus. $\endgroup$ – fleablood Apr 11 '18 at 1:05
  • $\begingroup$ Here is an intuitive approach to motivate the conclusion that the area of a circle equals half the product of the radius and circumference: quora.com/… $\endgroup$ – John Wayland Bales Apr 11 '18 at 1:09
  • $\begingroup$ This question has a number of proofs including my own. math.stackexchange.com/questions/2593324/… $\endgroup$ – Rene Schipperus Apr 11 '18 at 1:10
  • $\begingroup$ I'm also the type of person who likes to know more of what is going on in a problem so I upvoted this question. $\endgroup$ – Timothy Aug 29 at 20:43
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An elementary proof cannot exist because $\pi$ is a transcendental number. It cannot be expressed as a fraction, nor even as the root of a polynomial with integer coefficients. It cannot be computed geometrically, with ruler and compass. To be expressed from integers using algebraic functions only, it requires an infinite process.

This problem is known as the impossible squaring of the circle.


Note that if the formula for the circumference is taken for granted, the proof for the area is easy. But that just displaces the problem to that of the circumference.

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I don't think such a simple proof exists. I guess you're the type of person who doesn't want to accept a proof unless you know enough about what's going on in the problem. For example, you might be like the person who asked the question Infinity and Hilbert's hotel paradox and then accepted this answer to it which explained more of what was going on with the problem, if that really is the reason he or she accepted the answer.

If you want to do research in another area of math such as the Peano arithmetic, you might want to not accept a proof of any statement until you figured out a proof in Peano arithmetic. Then you could completely avoid using the theorem that the area of a circle is $\pi r^2$ with no problem. You may also want to keep trying until you figure out how to show that when when ever you can prove a statement describable in Peano arithmetic in an extension of Peano arithmemetic that allows you to use the theorem that the area of a circle is $\pi r^2$, you can also prove the same statement just in the formal system of Peano arithmetic.

If you do want a proof that the area of a circle is $\pi r^2$. I think you will like this method the best. You might want to read and understand the proof in this answer that an integral is an antiderivative which I never checked because I felt like it would take too much time to figure out how to understand and I had better things to do. After that, you might want to read this proof by integration that the area of a circle is $\pi r^2$.

$\forall r \in \mathbb{R} - \mathbb{R}^-\int_{-r}^r 2\sqrt{r^2 - x^2} dx = \int_{r \sin \frac{-\pi}{2}}^{r \sin \frac{\pi}{2}} 2\sqrt{r^2 - x^2} dx = \int_{r \sin \frac{-\pi}{2}}^{r \sin \frac{\pi}{2}}2r\cos(\sin^{-1}(\frac{x}{r})) dx = \int_{\frac{-\pi}{2}}^\frac{\pi}{2}2r\cos(x)\frac{d}{dx}(r \sin(x)) dx = \int_{\frac{-\pi}{2}}^\frac{\pi}{2}2r^2\cos^2(x) dx = \int_{\frac{-\pi}{2}}^\frac{\pi}{2}2r^2\frac{\cos(2x) + 1}{2} dx = \pi r^2$

This proof might also help you figure out how to show that when ever you can prove a statement describable in Peano arithmetic in that extension of Peano arithmetic, you can also prove that statement in Peano arithmetic itself.

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  • $\begingroup$ The OP already knows that. $\endgroup$ – Yves Daoust Aug 30 at 5:52

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