If a curve is defined as $\begin{cases}x=f_x(t)\\y=f_y(t)\\\end{cases},t\in[a,b]$
Smooth are defined as $f_x'(t)+f_y'(t)\ne0$ and $\lim_{t\to t_0}f_x'(t)=f_x'(t_0),\lim_{t\to t_0}f_y'(t)=f_y'(t_0)$ on $[a,b]$
My text book says this smooth curve has zero area.
However, it does not provide a prove. May I ask the proof of this result?
Thanks!
A $C^1$ function ${\bf f}:\>[0,1]\to{\mathbb R}^2$ is automatically Lipschitz continuous: There is an $L>0$ such that $$|{\bf f}(s)-{\bf f}(t)|\leq L|s-t|\qquad\forall\>s,\>t\in [0,1]\ .\tag{1}$$ Choose an $N\gg1$ and partition $I:=[0,1]$ into $N$ equal parts $I_k:=[t_{k-1},t_k]$. Let $\tau_k$ be the midpoint of $I_k$, and put ${\bf z}_k:={\bf f}(\tau_k)$. Then it follows from $(1)$ that ${\bf f}(I_k$) is contained in the disc of diameter ${L\over N}$ centered at ${\bf z}_k$, and this disc is contained in a square of area ${L^2\over N^2}$. Therefore the curve $\gamma={\bf f}(I)$ can be covered with $N$ such squares, of total area ${L^2\over N}$. As $N$ can be chosen arbitrary large the area of $\gamma$ has to be $0$. – Similarly for the volume of a $C^1$ surface in ${\mathbb R}^3$.