Why is the area of a circle pi times the square of the radius? I was going through a lot of geometric representations but they all were approximations. Is there a solid, intuitive way to prove this?
Isn't the one above an approximation? Is there another intuitive way to think about this?
Its not an approximation,
Instead of looking at this as a regular circle, consider it as a function, $f$ on a $x,y$ cartesian plane.
Let $f(x, y):= x^2 + y^2 = 25$ or $g(x) = \pm \sqrt{25 - x^2}$
If you have done calculus, you would know the area under this function can be described by an integral:
$A = \int_{-5}^{5}{f(x, y)} = \int_{-5}^{5}{\sqrt{25 - x^2}}dx + |\int_{-5}^{5}-{\sqrt{25 - x^2}}dx| = 2\int_{-5}^{5}{\sqrt{25 - x^2}}dx = 25\pi$.
Another way to think about this, is substituting in what we know about trigonometry and the unit circle, where $x = r\sin(\theta), dx = r\cos(\theta)d\theta$
$2\int_{-5}^{5}{\sqrt{25 - x^2}}dx \to 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sqrt{25 - 5^2\sin(\theta)^2}}5\cos(\theta)d\theta \to 10\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sqrt{25(1 - \sin(\theta)^2)}}\cos(\theta)d\theta$
$\to 50\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cos(\theta)^2}d\theta = 25\pi = \pi r^2$
Another method would be to consider is to use the fundementals of calculus but in a slightly different way. Consider the image below:
Let the length of the sides of the polygon be $= a$, and the number of sides in the polygon $= n$. If we were to draw a line from the centre (lets call this the origin, $O$) to $2$ consecutive edges of the polygon, you would get a isosceles triangle, which the sides with of length $r$ (radius of the circle) and base length, $a$. Lets call this triangle, $ABO$.
Since the $n$ sides are of equal, $a$ length, $\angle AOB$ is $\frac{2\pi}{n}$
Also, the area of $\triangle = \frac{1}{2} * h * a$. However, this does not help us, we need to figure out the height and base in terms of $\theta$. If you draw line straight in the middle of the triangle to the base, cutting $\theta$ in half, and using trigonometry ... the base becomes $2r\sin{\frac{\pi}{n}}$. The height becomes $r\cos{\frac{\pi}{n}}$
Hence, area of $\triangle = \frac{1}{2} * r\cos{\frac{\pi}{n}} * 2r\sin{\frac{\pi}{n}} = r^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}} = \frac{r^2\sin{\frac{2\pi}{n}}}{2}$
The area of the polygon is the number of triangles that can fit in it, which if one thinks about it, corresponds to the number of sides. Therefore, area of polygon inside circle $= \frac{nr^2\sin{\frac{2\pi}{n}}}{2}$
Now, heres the calculus part. When $n$ starts becoming bigger and bigger the polygon starts approaching the circle (as seen in the diagram).
Hence, it can be stated, $$\lim_{n \to \infty}{\frac{nr^2\sin{\frac{2\pi}{n}}}{2}} \to \text{ area of circle }$$
Computing the limit as followed:
$$\to \lim_{n \to \infty}{\frac{nr^2\sin{\frac{2\pi}{n}}}{2}} \to \lim_{n \to \infty}{\frac{r^2\sin{\frac{2\pi}{n}}}{\frac{2}{n}}} \to \pi r^2$$
We shifted the $n$ to the denominator to use L'hopital's rule.
Hence, using both methods, the area is represented by $\pi r^2$.
Well... sure, the example you show isn't exact (you could call it an "approximation"). It's just meant to be an informal representation of the argument for the area of a circle which involves dividing the circle into concentric rings of circumference $2\pi x$ (where $x$ is the distance from the center of the circle) and infinitesimal width $dx$. If you've studied calculus before, you just perform a simple integration realizing that you're summing these concentric rings that each have area $2\pi x dx$ and taking the limit of that sum. That is all that your example wishes to show.
The reason why many of the approaches you have seen seem like approximations is because they try to demonstrate such calculus techniques without actually getting into any of the math. Of course, if you take a finite number of rings each with macroscopic width, you will always run into the problem of having to approximate a little bit. Calculus allows you to obtain an exact solution.
