How to derive formula for circumference of circle using integration??

Question

Can using the formula of arc of a curve be of use here...as in:

$$L = \int_{{\,a}}^{{\,b}}{{\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx}}$$

Maybe we could consider the semicircle, find the length of its curve and multiply it by 2. But are then any other approaches?

Answer 1

Another way would be to write the equation of the circle in parametric form, i.e. $$x=r\cos\theta, y=r\sin\theta$$ and then use the arc length formula in the form$$L=\int\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$$ So the integral is $$\int_0^{2\pi}rd\theta=2\pi r$$

Answer 2

The circle $x^2+y^2=r_0^2$ in polar coordinates is $r=r_0$. So, $dr=0$ and $ds=\sqrt{dr^2+r^2d\theta^2}=r_0d\theta$. Thus,

$$L=\int_0^{2\pi}r_0d\theta=2\pi r_0.$$

Answer 3

Since $\pi$ is defined to be $C/D = C/(2r)$, we get $C = 2\pi r$ without calculus. But that skips over an argument explaining why $C/D$ is the same for all circles.

In the file here, if we declare $2\pi$ to be the circumference of the unit circle, then integration by parts is used to show the area of a unit circle is $\pi$.

Answer 4

Consider a circumference with center $C = (0, 0)$ that is at the origin, and of radius $R$:

$$x^2 + y^2 = R^2$$

The formula for finding the lenght of a given curve is indeed the one you wrote. Thence we find $$y = + \sqrt{R^2 - x^2}$$ where I chose the plus sign to denote the semi-circumference in the upper half plane. If you choose the minus sign it won't change anything (just the lower half plane).

Now because of symmetry, here we have $a = -R$ and $b = R$ (try to mentally picture a semi circumference in the upper half plane.)

We then have:

$$\ell = \int_a^b \sqrt{1 + \left(\dfrac{\text{d}y}{\text{d}x}\right)^2}\ \text{d}x = \int_{-R}^{R} \sqrt{1 + \dfrac{x^2}{R^2-x^2}}\ \text{d}x = \int_{-R}^{R} \sqrt{\dfrac{R^2}{R^2-x^2}}\ \text{d}x = R\int_{-R}^R \dfrac{1}{\sqrt{R^2-x^2}}\ \text{d}x$$

Now the integration is rather easy, and it fires out the arctangent; remember that $\lim_{z\to \pm\infty} \arctan(z) = \pm \frac{\pi}{2}$ so:

$$\ell = R \tan ^{-1}\left(\frac{x}{\sqrt{R^2-x^2}}\right)\bigg|_{-R}^R = \pi R$$

Now since we integrated in the upper half plane, we got only HALF of the total lenght of the circumference. Thence

$$L = 2\ell = 2\pi R$$

Answer 5

We calculate the length $|\gamma|$ of the circumference of radius $R ≥ 0$. The parametric coordinates are $$\gamma(t)=\begin{cases} x (t)=R\cos t, & t\in[0,2\pi[ \\ y(t)=R\sin t \end{cases}$$ and the tangent vector has components $$\gamma'(t)=\begin{cases} x'(t)=-R\sin t, & t\in[0,2\pi[ \\ y'(t)=R\cos t \end{cases}$$

Then using

$$|\gamma| = \int_{{\,a}}^{{\,b}}{{\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx}}$$$$\iff |\gamma|=\int_\gamma ds=\int_a^b ||\gamma'(t)||dt=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}dt$$ we will have: $$|\gamma|=\int_0^{2\pi}\sqrt{(-R\sin t)^2+(R\cos t)^2}\,dt=\int_0^{2\pi}\sqrt{R^2(-\sin^2 t+\cos^2 t)}\,dt=\int_0^{2\pi}Rdt$$$$=\color{red}{2\pi R}$$