Proper parametrization of a closed curve

Question

Let $\gamma:I\to\mathbb{R}^2$ be a closed plane curve, for simplicity, a unit circle. Therefore, we have $$\gamma(\varphi) = (\cos \varphi, \sin \varphi).$$ What is the proper domain of $\varphi$? Wikipedia says it's $\varphi \in [0,2\pi]$ with $\gamma(0) = \gamma(2\pi)$. What is the advantage of this domain rather than a domain $\varphi\in[0,2\pi\rangle$ with the additional condition that $$\lim_{\varphi \to 2\pi}\gamma(\varphi)=\gamma(0)?$$ It seems to me that the second definition is much more natural since no point on the curve is repeating, and the usual angle variable in the polar cordinates uses this domain.

Answer 1

There is no advantage for the study of continuous curves: you can extend the continuous $\gamma \colon [0,2\pi) \to \mathbb{R}^2$ to a unique continuous map on $\Gamma \colon [0,2\pi] \to \mathbb{R}^2$. The problem for many applications (any in particular, for your intended application to Stokes's theorem) is that smoothness of $\gamma$ need not imply smoothness of $\Gamma$. For example, if $\gamma(t)=(x(t),y(t))$ has $y(t) \sim (t-2\pi) \sin(1/(t-2 \pi))$ as $t \to 2 \pi$, then $dy$ is not integrable over $\gamma$.