Why is arc length independent of parametrization? For example, say a curve $C$ can be parametrized as $x_1(t)$, $y_1(t)$ over an interval $A$ and $x_2(t), y_2(t)$ over an interval $B$. Why is the arc length computed using the first parameterization the same as the length computed using the second parameterization? Is this always the case, or are there any exceptions?
 A: Assuming $A = [c,d]$ and $B=[e,f]$, there is a change of parameter diffeomorphism $h : [c,d] \to [e,f]$ such that $x_1(t)=x_2(h(t))$ and $y_2(t)=y_2(h(t))$, $t \in [c,d]$. Letting $s = h(t) \in [e,f]$ we get
\begin{align*}
\int_{s=e}^{s=f} \sqrt{(dx_2/ds)^2 + (dy_2/ds)^2} \, ds &= \int_{t=c}^{t=d}  \sqrt{\left(\frac{dx_1/dt}{h'(t)}\right)^2 + \left(\frac{dy_1/dt}{h'(t)}\right)^2} \, h'(t) \, dt \\
  &= \int_{t=c}^{t=d} \sqrt{(dx_1/dt)^2 + (dy_1/dt)^2} \, dt
\end{align*}
A: Let $\phi = (x_1, y_1)$, $\psi = (x_2, y_2)$. Note the definition of the surface integral claims that,
$$\int_{C}1\,dS := \int_{A}\sqrt{\det(D\phi(t)^TD\phi(t))}\,dt$$
is well defined, independent of the parameterization $(A, \phi)$. That is, it claims that
$$\int_{A}\sqrt{\det(D\phi(t)^TD\phi(t))}\,dt = \int_{B}\sqrt{\det(D\psi(s)^TD\psi(s))}\,ds.$$
Set $F = \psi^{-1} \circ \phi \colon A \to B$. Note that $F$ is a diffeomorphism.
By the change of variables formula for integration substituting $s = F(t)$, we have
\begin{align}
\int_{B}\sqrt{\det(D\psi(s)^TD\psi(s))}\,ds = \int_{A}\sqrt{\det(D\psi(F(t))^TD\psi(F(t)))}|\det(DF(t))|\,dt.
\end{align}
We have $\phi(t) = \psi(F(t))$, so $D\phi(t) = D\psi(F(t))DF(t)$. Thus
\begin{align}
\sqrt{\det(D\phi(t)^TD\phi(t))} &= \sqrt{\det(DF(t)^TD\psi(F(t))^TD\psi(F(t))DF(t))} \\
&= \sqrt{\det(D\psi(F(t))^TD\psi(F(t)))}|\det(DF(t))|.
\end{align}
Thus $\int_{A}\sqrt{\det(D\phi(t)^TD\phi(t))}\,dt = \int_{B}\sqrt{\det(D\psi(s)^TD\psi(s))}\,ds$, so the surface integral is well defined, independent of the parameterization. The above proof works for surfaces of arbitrary dimension, and with small modification, works for integrals of any function $f \colon C \to \mathbb{R}$, not just $f = 1$. Technically the sets $A$ and $B$ should be open though.
