To find the volume dilation, integrate the determinant of the Jacobian On the road toward proving the change of variables theorem in several variables, is there a painless way to show that 
$$\text{Vol}(\phi(U))=\int_{U}|\text{det}(d\phi)|,$$
where $\phi$ is $C^1$, $d\phi$ is non-singular, and $U$ is some reasonable set (say open and connected)? Having done the single-variable version of change of variables when $f$ is discontinuous (as presented in baby Rudin, for example), I wanted to adapt the same method to this, but so far I haven't been successful. Many proofs I've seen have passed through several lemmas, but I think there should be a straightforward way.
I am comfortable taking for granted that $|\text{det}(M)|$ is how much $M$ dilates the area of a parallelipiped.
 A: I can think of a rather low-tech proof in the case when $U = [0,1]^d$ is the unit cube in $\mathbb{R}^d$ and $\phi : U \to \mathbb{R}^d$ is $C^1$ and one-to-one. Basically, you take advantage of the fact that $U$ can be evenly divided into cubes of side length $1/N$ for $N$ large. 
Denote $P_N(x) = \{y \in U : \|x-y\|_{\infty} \leq 1/2N\}$, where $\|v\|_{\infty} = \max_{1 \leq i \leq d} |v_i|$ is the max norm on vectors in $\mathbb{R}^d$. It is not hard to show the following: for any $\epsilon > 0$, there exists $N$ sufficiently large such that for any $x \in U$ (let's assume the distance from $x$ to the boundary of $U$ is $\geq 1/2N$ for simplicity), we have
\begin{align}
\frac{1}{1 + \epsilon} d \phi_x P_N(x) \subset \phi(P_N(x)) \subset (1 + \epsilon) d \phi_x P_N(x).
\end{align}
For each $N$, let $S_N \subset [0,1]^d$ be the collection of vectors of the form $(k_1/2N, k_2/2N, \cdots, k_d/2N)$, where $1 \leq k_i \leq 2N-1$ are odd. Then $U = \cup_{x \in S_N} P_N(x)$, and (modulo an issue I'll address soon)
\begin{align}
\frac{1}{(1 + \epsilon)^d} \frac{1}{N^d} \sum_{x \in S_N} |\det d \phi_x| \leq Vol(\phi(U)) \leq (1 + \epsilon)^d \frac{1}{N^d} \sum_{x \in S_N} |\det d \phi_x|.
\end{align}
Finally, you note that $\frac{1}{N^d} \sum_{x \in S_N} |\det d \phi_x|$ is a Riemann sum, hence the limit as $N \to \infty$ equals $\int_U |\det d \phi_x| dx$.
The issue: one must know that the images under $\phi$ of the sides of the cubes $P_N$ have zero $d$-dimensional volume. I think that this doesn't present a problem if you modify the definition of the $P_N$'s so as to prevent the double counting of sides.
One can now use a completely analogous technique to prove that for continuous $f : \phi(U) \to \mathbb{R}$,
\begin{align}
(*) ~~ \int_{\phi(U)} f (x) dx = \int_{U} (f\circ \phi) (x)\cdot |\det d \phi_x| dx.
\end{align}
This can be shown by using the fact that for any $\epsilon > 0$, there exists $N$ sufficiently large so that
\begin{align}
(1 + \epsilon)^{-d} |\det d \phi_x| \cdot \inf_{y \in P_N(x)} (f \circ \phi)(y) \leq \int_{\phi(P_N(x))} f(y) dy \leq (1 + \epsilon)^d |\det d \phi_x| \cdot \sup_{y \in P_N(x)} (f \circ \phi)(y),
\end{align}
and using Riemann sums as before.
To complete the proof for a more general class of $U$, one can use $(*)$ to conclude by assuming that the domain $U$ has a $C^1$ diffeomorphism to $[0,1]^d$.
A: Fix an $x_0$ where $df(x_0)$ is nonsingular. Then for sufficiently small $r$, on the ball $B(x_0,r)$, the map $f(y)$ is little $o(|y-x|)$ approximated by the affine map $L = x_0+df(x_0)(y-x)$.
The image $L(B(x_0,r)$ is an ellipsoid of volume $\det (df(x_0)) \times vol(B(x_0,r))$. By the argument above, this also little $o(r)$ approximates the volume of $f(B(x_0,r))$. If $f$ is $C^1$ over open $U$ and $E$ is a compact subset of it, then by covering $E$ by such balls, we can get control over how well  $ \sum _i \det (df(x_i)) \times vol(B(x_i,r) \cap E)$ estimates $vol(f(E))$. And in limit one obtains the equality.
Lots of details to be added, of course!
