Calculus proof for the area of a circle I was looking for proofs using Calculus for the area of a circle and come across this one
$$\int 2 \pi r \, dr = 2\pi \frac {r^2}{2} = \pi r^2$$
 and it struck me as being particularly easy. The only other proof I've seen was by a teacher and it involved integrating $x = \sqrt{r^2 - y^2}$ from $-1$ to $1$, using trig substitutions and then doubling the area to get $\pi r^2$ but the above proof seemed much more straight forward.

Is it a valid proof, or is it based on circular logic or some other kind of fallacy?
The above integral seems geometrically as below figure. 



\begin{align}
\int_0^r 2\pi r\, dr&amp; = 2\pi\int_0^r\ r\, dr \\
 &amp; = 2\pi\bigg\lvert_0^r\ \frac{r^2}{2} \\
 &amp; = \bigg(2\pi \frac{r^2}{2}\bigg)-\bigg(2\pi \frac{0^2}{2}\bigg) \\
 &amp; = \frac{2\pi r^2}{2} \\
 &amp; = \require{cancel} \frac{\cancel{2}\pi r^2}{\cancel{2}} \\
 &amp; = \color{red}{\pi r^2}
\end{align}
Consider the standard equation of circle $$x^2 + y^2 = a^2$$ where $a$ is the radius of the circle.