TITLE: Verify the circumference of the circle formula for a circle of radius r using the arc length formula.
QUESTION [2 upvotes]: Verify the circumference of the circle formula for a circle of radius r using the arc length formula.
How do I complete this problem? Do I use the ellipse equation or the proper circle equation with (x, h) and (y,k)?

REPLY [5 votes]: Do it with half a circle that we can assume is centered at the origin:
$$x^2+y^2=r^2\implies y=\sqrt{r^2-x^2}\;,\;\;y'=-\frac x{\sqrt{r^2-x^2}}$$
Thus (the function's even)
$$I:=\int_{-r}^r\sqrt{1+\frac{x^2}{r^2-x^2}}\;dx=2\int_0^r\frac{dx}{\sqrt{1-\left(\frac xr\right)^2}}$$
Substitution
$$x=r\sin u\implies dx=r\cos u\implies$$
$$I=2\int_0^{\pi/2}\frac{r\cos u\;du}{\cos u}=\pi r$$
Using parametrization in two variables: we can write the same circle above as
$$(x(t)\,,\,y(t))\;,\;\;\text{with}\;\;x(t)=r\cos t\;,\;\;y=r\sin t\;,\;\;0\le t\le2\pi$$
and thus the arclength is given by
$$\int_0^{2\pi}\sqrt{x'(t)^2+y'(t)^2}dt=\int_0^{2\pi}\sqrt{r^2\sin^2t+r^2\cos^2t}dt=r\int_0^{2\pi}dt=2\pi r$$
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