Another Method to Prove the Circumference of a Circle with Arc Length Integration

Question

I'm trying to prove that the circumference of a circle is $C=2\pi r$. There are other simple ways to prove it, as shown here, such as just giving a definition to $\pi$, but I thought I would overcomplicate it and employ the arc length integral, as shown below.

$$s=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$

Note that my strategy is to find the length of half the circle, then double the result. The derivative of the graph of a semi-circle, $y=\sqrt{{1+x}^2}$ is, according to WolframAlpha,

$$\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}}$$

So $1+\left(\frac{dy}{dx}\right)^2 = \frac{1}{1-x^2}$. Plugging this into the integral, I expected to get $\pi r$. Instead I got

$$\int_{-r}^r \frac{dx}{\sqrt{1-x^2}} = 2\sin^{-1}(r)$$

Could someone please explain where I went wrong? Thanks.

Answer 1

Since $$\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}}$$

$$\int_{-r}^r \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx=\int_{-r}^r \frac{r\,dx}{\sqrt{r^2-x^2}}=\int_{-r}^r \frac{r\,dx}{\sqrt{1-(x/r)^2}}$$ We will use the change of variables

$t=\frac xr \implies \frac{dt}{dx}=\frac 1r$. After $rdt=dx$ and if $x = −r → u = −1$, $x = r → u = 1$. Hence $$\int_{-r}^r \frac{r\,dx}{\sqrt{1-(x/r)^2}}=\int_{-1}^1 \frac{r\,du}{\sqrt{1-u^2}}=r\arcsin u\Bigl|_{-1}^{1}=r\left( \frac {\pi}2+ \frac {\pi}2\right)=r\pi$$

Addendum: In your case $r=1$ hence $$\int _{-1}^1 \frac{dx}{\sqrt{1-x^2}}=\left[\arcsin \left(x\right)\right]_{-1}^1=\pi.$$

Answer 2

This problem borders on the trivial when viewed in the complex plane. The arc length is given by

$$ s=\int |\dot z|\ du $$

Now, for you case we have

$$ z=re^{i\theta}\\ \dot z=ire^{i\theta}\\ |\dot z|=r $$

So that $$ s=\int_0^{2\pi} |\dot z|\ d\theta=\int_0^{2\pi} r\ d\theta=2\pi r $$