Integral transformation

I'm familiar with the transformation theorem in $\mathbb{R}^n$: given $\varphi : \Omega \rightarrow \mathbb{R}^n$ which is a diffeomorphism, $\Omega$ open, then $$\int_{\varphi(\Omega)} f(y) dy = \int_{\Omega} f(\varphi(x)) |\det \varphi'(x)| dx$$

but this fails when $\Omega$ is a subset of $\mathbb{R}^m$, $m < n$.

In this case the formula is rather $$\int_{\varphi(\Omega)} f(y) dy = \int_{\Omega} f(\varphi(x)) \sqrt{\det (\varphi'(x)^T \varphi'(x))} dx$$

For example the arclength formula: let $f:[0,1] \rightarrow \mathbb{R}^2$ be differentiable and consider $\phi : [0,1] \rightarrow \mathbb{R}^2, t \rightarrow \begin{pmatrix} t \\ f(t) \end{pmatrix}$, then $$\int_{\varphi([0,1])} dy = \int_0^1 \sqrt{\det(\begin{pmatrix} 1 & f'(x) \end{pmatrix} \begin{pmatrix} 1 \\ f'(x) \end{pmatrix})} dx = \int_0^1 \sqrt{1 + f'(x)^2} dx$$

and I'd like to ask what the geometric motivation for this is and if it follows easily from the transformation theorem I gave at the top.

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For me these two formula look equivalent. –  Fabian Aug 4 '12 at 18:55
@Fabian If $m=n$ they are equivalent but otherwise $\varphi'(x)$ is not a square matrix and you can't take the determinant. I guess the question is why this is the correct generalization to that case –  doyouevenlift Aug 4 '12 at 18:57
I think you need to consider integration on manifolds here. Since $\phi(\Omega)$ will have 'dimension' $m$ in general, and $m<n$, then right hand side integral above will be zero, whatever meaning you ascribe to $\det$ in this case. –  copper.hat Aug 4 '12 at 19:10