In what follows $ \mathbb{R}^{n+m} = \{(x,y): x \in \mathbb{R}^n, \ y \in \mathbb{R}^m \} \ .$

Suppose $G: U \to V $ is a $C^1$-diffeomorphism from an open subset of a manifold to an open subset of $\mathbb{R}^{n+m}$. We write $$ (\xi,\eta)= G^{-1}(x,y) \ . $$

This is a local parameterization of $U$. We will think of $$ U_\eta: = G^{-1}(\mathbb{R}^n \times \{y\} \cap V) $$ as the "horizontal" fibers, which are $n$-dimensional $C^1$-manifolds. Similarly, we refer to $$ U_\xi = G^{-1}( \{x\} \times \mathbb{R}^m \cap V) $$ as the ``vertical" fibers, which are $m$-dimensional $C^1$-manifolds.

We also use the following notation: $ DG_{|U_\xi} (\xi,\eta)$ is understood as the derivative, at point $(\xi,\eta)$, of the map $G_{|U_\xi}$, i.e. the restriction of $G$ to $U_\xi$. This restriction is from an $m$-dimensional $C^1$-manifold to a subset of $\mathbb{R}^m$.} So its derivative and its Jacobian are defined.

Suppose for some fixed $\eta_0$, the union of vertical fibers along $U_{\eta_0}$ covers the set $U$.

Lemma Under the assumptions above, for any integrable function $f: U \to \mathbb{R} $, $$ \int_U f = \int_{U_{\eta_0}} \left(\int_{U_\xi} f(\xi,\eta) \frac{|\det DG_{U_\xi} (\xi,\eta)| \cdot |\det DG_{U_{\eta_0}} (\xi,\eta_0)|}{|\det DG(\xi,\eta)|} \ d\mathcal{H}^m(\eta)\right) \ d\mathcal{H}^n(\xi) \ . $$

That is saying that to integrate over $U$ simply interate along the fibers and then sum over the "base", which is what Fubini's theorem says in the standard orthogonal coordinates of the Euclidean space. The correcting factor is can be memorized as $$ \frac{\text{ Jacobian along fibers} \times \text {Jacobian along base}}{\text{full Jacobian}} \ .$$

Corollary: Integration in pherical coordinates. There, the fibers are orthogonal to the base, and so, the full Jacobian also factors, cancelling the numerator. Therefore, the correcting factor disapperas (=1), and we get the familiar $$ \int_U f \ d \mathcal{L}^n = \int_0^\infty \left( \int_{\mathbb{S}^{n-1}(r)\cap U} f \ d \sigma \right) dr \ . $$

Example: Let $P$ be a parallelopiped in the plane, resulting from tilting a rectangle so that the acute angle between its edges $A$ and $B$ is $\theta$. Then, foliating $P$ by parallel copies of the edges $B$, indexed $B_x$, leads to $$ \int_P f \ d \mathcal{L}^2 = \int_A \left( \int_{B_x} f \sin(\theta) \ d \mathcal{H}^1 \right) dx \ = \sin(\theta) \int_A \left( \int_{B_x} f \ d \mathcal{H}^1 \right) dx \ . $$