# How can one derive the circumference of a circle using integrals?

Many proofs for the area of a circle start with something like

$$A(r) = \int_0^r 2 \pi t dt$$

such as at https://en.wikipedia.org/wiki/Area_of_a_disk#Onion_proof , but I don't understand how to get to this starting point.

Using the same notation at the wikipedia article, we can imagine the radial line made up of many infintesimally small fragments $h$ each at a distance $t$ from the center of the circle. Considering just one of these fragments, the area swept out by this fragment around the circle (i.e., the circumferential line length) is

$$L = \int_0^{2\pi} ([t+h]-t) d \theta = \int_0^{2\pi} h d \theta = 2 \pi h$$

But obviously the circumference of a line should be $2 \pi t$.

I see that there are various ways to derive this, but I'm specifically trying to understand where my logic is wrong in applying this method of reasoning. I realize this is an elementary question, but I appreciate any kind help.

Thank you.

• You shouldn't be considering a small change of length $h$ unless you plan to sum up infinitely many of them, and I do not see any sums here – ASKASK Dec 17 '15 at 1:33
• @ASKASK If h is infinitesimally small, isn't the "height" of the area effectively small enough that I have a line? Or is there a better approach? – Struggling snowman Dec 17 '15 at 1:33
• What do you think you are measuring the "height" of? – ASKASK Dec 17 '15 at 1:34
• @ASKASK The height h is the height of the infinitesimally small piece of the radial line. It seems to me that “infinitesimally small” can be the height of a circumferential line (i.e., essentially zero). But are you saying this cannot be the case? Is it possible to mathematically explain why not? – Struggling snowman Dec 17 '15 at 1:44
• @Strugglingsnowman I think it would be helpful if you could add a sketch to your question, showing what you mean by "height of the infinitesimally small piece of the radial line", what $t$ is, etc. In particular it's not clear what the expression $[(t+h)-t]$ refers to geometrically. – mweiss Dec 17 '15 at 2:43

The area of the black ring is $2 \pi t\ dt$. You add those up, from $t = 0$ to $t = r$.