The above integral seems geometrically as below figure.

\begin{align} \int_0^r 2\pi r\, dr& = 2\pi\int_0^r\ r\, dr \\ & = 2\pi\bigg\lvert_0^r\ \frac{r^2}{2} \\ & = \bigg(2\pi \frac{r^2}{2}\bigg)-\bigg(2\pi \frac{0^2}{2}\bigg) \\ & = \frac{2\pi r^2}{2} \\ & = \require{cancel} \frac{\cancel{2}\pi r^2}{\cancel{2}} \\ & = \color{red}{\pi r^2} \end{align}

Possibly the proof that you found is what the Wikipedia article for the area of a disk calls "The Onion Proof" ^[1].

Although I would probably use the following double integral instead:

$$ \text{Area of circle} = \iint_{x^2 + y^2 \leq R}1 \, dx\,dy $$

and then calculate the integral using polar coordinates to get

$$ \iint_{x^2 + y^2 \leq R}1 \, dxdy = \int_0^{2 \pi} \int_0^R r \, dr\,d\theta = \int_0^R 2\pi r \, dr = \pi R^2 $$

There's a particularly simple formula using line integrals: if $\,\gamma\,$ is a simple, closed and smooth (at least by parts) path (in the positive direction), the area of the inclosed region equals $$\frac{1}{2}\oint_\gamma x\,dy-y\,dx$$

In our case, we can take the path $\,\gamma(t)=(r\cos t\,,\,r\sin t)\,\,,\,t\in [0,2\pi)\,$ , and get $$\frac{1}{2}\int_0^{2\pi}r^2(\cos^2t+\sin^2t)\,dt=\frac{r^2}{2}\int_0^{2\pi}dt=\pi r^2$$

I would like to show you another method how we can prove that the area of circle is $\pi r^2$ by using infinity part of a circle. I divided only 8 parts in my picture to demostrate how to apply that method but we need to have infinite divided parts to get an exact rectangle shape. After that We can write easily that

Area of circle = $\pi r .r =\pi r^2$

Note: I assumed that we know the circumference is $2πr$

Consider a circle of radius $r$ with $O$ as its center.

Now, consider an arc $XY$of the circle which subtends an angle of $\theta$ at the center.

let, $dA$ be the area of the segment $XOY$

if we take infinitesimally small angle $\delta\theta$ then we have, $dA=\frac{1}{2}(r^2)(\sin{\delta\theta})$

we know that $\lim_{\delta\theta ->0} \frac{\sin\delta\theta}{\delta\theta} =1$

using, the above fact, we have, $dA=\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$

Now, $\lim_{\delta\theta ->0} dA = \lim_{\delta\theta ->0}\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$

we, have, $dA=\frac{1}{2}(r^2){\delta\theta}$

Now,$$\int{dA}=\int_{0}^{2\pi}\frac{1}{2}(r^2){\delta\theta}$$

$$\int{dA}=\frac{1}{2}(r^2)\int_{0}^{2\pi}{\delta\theta}$$

$$A=\frac{1}{2}(r^2)(2\pi)$$ $$A={\pi}r^2$$

we, now have the area of a circle of radius $r$, and it is ${\pi}r^2$

Just one remark. The so-called onion proof is a special case of the co-area formula ^[1]. This formula is a rigorous justification of all those computations that we learned in the first course of general physics. It is a "curvilinear" generalization of Fubini's theorem: instead of slices, you integrate over hypersurfaces like a sphere. And also the fact that "differentiating the volume gives the area" is a consequence of the same theorem.

Here is another 'calculus proof'. (Basically integrating along $\theta$ rather than $r$.)

Approximate the circle from the inside using a regular n-sided polygon formed from the vertices $(r \cos k\theta_n, r \sin k\theta_n)$, where $\theta_n = \frac{2 \pi}{n}$, and $k = 0,...,n-1$. Draw lines between adjacent vertices and between the origin and each vertex. This splits the polygon into $n$ triangles with sides $r,r,2r \sin \frac{\theta_n}{2}$. The area of each polygon is given by $A_n = 2 r \sin \frac{\theta_n}{2} \frac{1}{2} r \sqrt{1 - (\sin \frac{\theta_n}{2})^2} = r^2 \sin \frac{\theta_n}{2} \cos \frac{\theta_n}{2}$.

Then the area of the circle is given by $\lim_{n \to \infty} n A_n = \lim_{n \to \infty} n r^2 \sin \frac{\theta_n}{2} \cos \frac{\theta_n}{2}$. Since $\lim_{x \to 0, x \neq 0} \frac{\sin x}{x} = 1$, we have $\lim_{n \to \infty} n \sin \frac{\theta_n}{2} = \lim_{n \to \infty} n \sin \frac{\pi}{n} = \lim_{n \to \infty} \pi \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}} = \pi$. Consequently we have $\lim_{n \to \infty} n A_n = \pi r^2$.

Note: To show that the area of the polygon converges to the area of the circle, note that the area between the polygon and the circle is bounded by $n r \sin \frac{\theta_n}{2} r ( 1 - \cos \frac{\theta_n}{2})$. A calculation along the above lines shows that this converges to $0$.

What you have here:

$$ \int 2 \pi r \, dr = 2 \frac {r^2}{2} \pi = \pi r^2 $$

does not represent an area because the integration is not bounded (also, a constant is missing on the RHS). An area should be for something with bounds (limits). However, the formula you mentioned is used in what is known as Onion proof for area of the circle ^[1] (please do a find on 'onion'). This proof divides the circle into rings as explained in the link.

You can actually convince yourself geometrically that

$$A'(r)=2 \pi r (*)$$

Intuitively, the rate of change of the area of the circle is the circumference.

Formally

$$A'(r) = \lim_{\Delta r \to 0} \frac{A(r+\Delta r) -A(r)}{\Delta r}$$

Now, geometrically it is pretty clear (but not really easy to prove mathematically) that the area of a corona between circles satisfies

$$2 \pi r_\text{in} (r_\text{out}-r_\text{in}) < \text{Area} < 2 \pi r_\text{out} (r_{out}-r_{in})$$

Using these inequalities it is easy to calculate the above limit which leads to $(*)$.

The formula you provided solves $(*)$ for $A(r)$.

The proof that depends on $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to calculate the area of circle, is not complete as the proof of that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ depends on area of circle equation and limit squeeze theorem.

I just want to point out that your proof (as formalized by some of the answers above) is a special case of a more general fact. Let $\gamma(s)$ be a smooth, closed curve, and $N(s)$ the normal along the curve. Consider "inflating" the curve by moving every point a distance $\epsilon$ along the normal, to get a new closed curve $\gamma(s)+\epsilon N(s)$. Then the area enclosed by the inflated curve is

$$A(\gamma+\epsilon N) = A(\gamma) + \epsilon L(\gamma) + \frac{1}{2} \epsilon^2 \int_\gamma \kappa(s)\,ds$$ where $A(\gamma)$ is the area enclosed by $\gamma$, $L(\gamma)$ is the total arc length of $\gamma$, and $\kappa(s)$ is the signed curvature of the curve at $s$ (for simple curves, the total curvature is always equal to $2\pi$, by the Whitney-Graustein theorem, and the last term is just $\pi \epsilon^2$).

In particular, this formula tells us that perimeter of a region in the plane is the first derivative of area, with respect to inflation along the boundary normal. For the disk, inflation along the normal just amounts to scaling the radius.

Incidentally, all of the above carries over beautifully to surfaces $M$ enclosing volumes in $\mathbb{R}^3$:

$$V(M + \epsilon N) = V(M) + \epsilon A(M) + \epsilon^2 \int_M H\,dA + \frac{\epsilon^3}{3} \int_M K\,dA$$ Where $V$ is enclosed volume, $A(M)$ is the surface area of $M$, and $H$ and $K$ are mean and Gaussian curvature. By the Gauss-Bonnet theorem, the integral in the last term is always a multiple of $4\pi$ and depends only on the topology of $M$.

Start off with a first principle proof that $\lim_{x\to 0}\frac{\sin x}{x}=1$ ^[1] is true
(we only need to know that the derivative of $sin(x)$ at $0$ is equal to 1).

Using the notation from there, you divide the circle into a $2^n$-gon and approximate the area with $2^n$ times the area of the small isosceles triangle wedges, each with a vertex angle of $\frac{2 \pi}{2^n}$ radians:

$\tag 1 2^n .5 \; a \sqrt{r^2 - (.25) a^2}$

Since $a(\theta)=r\, 2\sin\frac{\theta}{2}$, you can rewrite (1) as follows
$2^n (.5)\, r\; 2sin(\frac{\pi}{2^n}) \sqrt{r^2 - .25\, {(r\; 2sin(\frac{\pi}{2^n}))}^2}$
$2^n \, r^2\; sin(\frac{\pi}{2^n})\; cos(\frac{\pi}{2^n}) $
$\pi \, r^2\; \frac{sin(\frac{\pi}{2^n})}{\frac{\pi}{2^n}}\; cos(\frac{ \pi}{2^n}) $

and as $n$ goes to infinity you get the result,

AREA OF CIRCLE = $\pi r^2$

Note: Using properties of the sagitta you can show that this increasing sequence of area approximations gives you the only 'area definition' that makes sense.

the integral of r(d(theta)) from 0 to 2π is 2πr (the circumference of a circle with radius r), now integrate 2πr(dr) from 0 to r and the answer is πr^2 (the area of a circle of radius r). This is my derivation.

If you have a regular polygon with n sides and r being the distance from the center to a vertex, and take the area of that, you get: A = nr^2sin(180(n - 2)/2n)cos(180(n - 2)/2n). As n goes to infinity, the polygon turns into a circle. The sine term goes to 1 so you can take that out and r^2 is a constant so we can put that back in at the end. We're left with ncos(180(n - 2)/2n)as n goes to infinity. The cosine term goes to 0 and n goes to infinity but the product turns out to be pi so we have pi*r^2.

Simplest solution: I, Myself wrote a proof for the area of the circle. The idea is that a polygon is created by triangles. For example, a square is created by 4 triangles and its area can be calculated by summing up the area of the triangles. $ r = \frac12 a $ then $S = 4(\frac12ra) = 2ra = a^2$ For hexagonal it is $ S = 6 (\frac12 r a) $ and for n sided polygon it is $ S = n (\frac12 r a) = \frac{n r a}{2} $. By adding the triangles, the polygon is going to be a circle: $$S = \lim_{n\to ∞} \frac{n r a}{2} = = \lim_{n\to \infty} nr^2 \tan\tfrac{\pi}{n} = \pi r^2$$

For full description and visualization, you can see this post ^[1].

Consider the standard equation of circle $$x^2 + y^2 = a^2$$ where $a$ is the radius of the circle.