In section 1.3 of "A Geometric Approach to Differential Forms", we want to integrate the function $f(x,y) = y^2$ on the top half of the unit circle,
For the 2 parameterizations of the top half of the unit circle:
$\phi_1 (a) = (a, \sqrt{1-a^2}), \space a \in [-1,1] $
$\phi_2 (t) = (\cos t, \sin t), \space t \in [0, \pi].$
We end up needing to do: $\int_{-1}^{1} f( \phi_1 (a)) \space |\frac{d\phi_1}{da}| \space da.$
Which is equivalent to: $\int_{0}^{\pi} f( \phi_2 (t)) \space |\frac{d\phi_2}{dt}| \space dt.$
As opposed to the naive approach of not including the absolute value factor in the integral (and this is how the author introduces a 1-form).
However it turns out that already the second parameterization gives the correct answer without the "correction" factor (because $|\frac{d\phi_2}{dt}|$ = 1), and this seems to be (as far as I can tell) because the second parameterization preserves distances along the arc of the circle, so for 2 points $ t_1 \& t_2 \in [0, \pi]$ with $\Delta t = t_2 - t_1$, then $\Delta \phi_2 (t) = r(t_2-t_1) = t_2-t_1$ (Because $r=1$) after $\phi_2$ transformation, whereas $\phi_1$ doesn't have this property.
So my question is does this property of preserving the integral hold for all isometric embeddings? So ones that preserve only the metric and not just the distance as well? (So that you can find a parameterization for (intrinsically) curved surfaces like the top half of the sphere that would preserve the integral?)