How do can you derive the equation for a circle's circumference using integration?

Question

Ezekiel Alexander · Accepted Answer

Let assume that the  circle is centered at  the origin hence its
equation is #x^2+y^2=r^2# Using parametrization in two variables: we can write the same circle above as #(x(t),y(t))#,with #x(t)=rcost#,#y=rsint#,#0≤t≤2π# and thus the arclength is given by (integration) #\int_0^{2\pi}\sqrt{x'(t)^2+y'(t)^2}dt=\int_0^{2\pi}\sqrt{r^2\sin^2t+r^2\cos^2t}dt=r\int_0^{2\pi}dt=2\pi r#

Luke Alvarez · Answer

undefined To derive the equation for a circle's circumference using integration, we start by considering a quarter of a circle in the first quadrant. Let $ y = \sqrt{r^2 - x^2} $ be the equation of the upper half of the circle, where $ r $ is the radius.

We integrate this function from $ x = 0 $ to $ x = r $ to find the arc length of this quarter circle. The arc length $ S $ is given by the integral:

$$ S = \int_{0}^{r} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx $$

$$ = \int_{0}^{r} \sqrt{1 + \left(\frac{d}{dx}\sqrt{r^2 - x^2}\right)^2} dx $$

$$ = \int_{0}^{r} \sqrt{1 + \left(\frac{-x}{\sqrt{r^2 - x^2}}\right)^2} dx $$

$$ = \int_{0}^{r} \sqrt{1 + \frac{x^2}{r^2 - x^2}} dx $$

$$ = \int_{0}^{r} \sqrt{\frac{r^2 - x^2 + x^2}{r^2 - x^2}} dx $$

$$ = \int_{0}^{r} \sqrt{\frac{r^2}{r^2 - x^2}} dx $$

Now, let $ x = r\sin(\theta) $, then $ dx = r\cos(\theta)d\theta $, and the integral becomes:

$$ S = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{r^2}{r^2 - r^2\sin^2(\theta)}} r\cos(\theta)d\theta $$

$$ = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{r^2}{r^2(1 - \sin^2(\theta))}} r\cos(\theta)d\theta $$

$$ = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{r^2}{r^2\cos^2(\theta)}} r\cos(\theta)d\theta $$

$$ = \int_{0}^{\frac{\pi}{2}} \frac{r}{|\cos(\theta)|} r\cos(\theta)d\theta $$

$$ = r \int_{0}^{\frac{\pi}{2}} d\theta $$

$$ = r\left[\theta\right]_{0}^{\frac{\pi}{2}} $$

$$ = r\left(\frac{\pi}{2} - 0\right) $$

$$ = \frac{\pi r}{2} $$

Since the quarter circle has symmetry, the circumference $ C $ of the full circle is four times this value:

$$ C = 4 \times \frac{\pi r}{2} = 2\pi r $$

Thus, we have derived the equation for the circumference of a circle using integration.