Calculus proof for the area of a circle

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".

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 $$\frac{1}{2}\int_0^{2\pi}r^2(\cos^2t+\sin^2t)\,dt=\frac{r^2}{2}\int_0^{2\pi}dt=\pi r^2$$