Integrating over an embedded manifold: Jacobian factor? Let's say I want to integrate a function 
$$
f(x,y),\quad x\in\Gamma_1,y\in\Gamma_2
$$ where $\Gamma_1,\Gamma_2$ are both embedded manifolds in $\Bbb{R}^3$.  The dimension of $\Gamma_1$ is 1 (a smooth curve, say), while the dimension of $\Gamma_2$ is 2 (a plane or surface).  Formally, what I want to compute is: 
$$
\int_{\Gamma_1}\int_{\Gamma_2}f(x,y)dxdy
$$
If I choose (global) parametrizations of $\Gamma_1$ and $\Gamma_2$, what does the change of variables Jacobian factor look like?  The matrix of the transformation will not be square (it will be 6x3!)
I think I know the answer, based on this very useful set of lecture notes: if $\Phi:\Bbb{R}^3\rightarrow\Bbb{R}^3\times\Bbb{R}^3$ is my parametrization, the answer should be 
$$
\int_{\Gamma_1}\int_{\Gamma_2}f(x,y)dxdy = \int_{\phi_1}^{\phi_2}\int_{\beta_1}^{\beta_2}\int_{\alpha_1}^{\alpha_2}f(\Phi(\alpha,\beta,\phi))\sqrt{\det (J^T_\Phi J_\Phi)}d\alpha d\beta d\phi
$$
Is this Jacobian factor correct?  Does anyone have a book reference for doing this?  The part I haven't seen before is the non-square Jacobian.  I must have somehow missed this in my calculus courses oh-so long ago.
 A: After a bit more digging, I discovered that this is simply the expression for the volume element on a Riemannian manifold.  The general expression in this setting is 
$$
\int_Mf dV_M = \int_Uf(\Phi(x))\sqrt{\det(g_{ij}(x))}dx 
$$ where $\Phi:U\subset\Bbb{R}^n\rightarrow M$ is the parametrization of $M$, and $g_{ij}$ is the Riemannian metric.  The Riemannian metric in this case takes the form 
$$
g_{ij}(x) = J_\Phi(x)^TJ_\Phi(x)
$$
Reference: Calculus on Manifolds, Spivak.
A: This is called the area formula. Its most general form (except for easy upgrade to manifolds) is as follows.
Theorem (Area Formula) Let $n \leq m$ and $f:\mathbb{R}^n \to \mathbb{R}^m $ be Lipschitz. Then for any Lebesgue measurable subset $A \subset \mathbb{R}^n$, we have
$$
\int_A J_f(x) \, d\mathcal{L}^n(x) = \int_{\mathbb{R}^m} N(f,A,y) \, d\mathcal{H}^n(y) \ ,
$$
where the multiplicity function $N(f,A,y)$ is the number (possibly infinite) of points from $A$ that are mapped to $y$, and
$$
J_f(X) = \sqrt{\det  \left( Df(x)^T \cdot Df(x) \right)} \, .
$$
What this Jacobian does is to see the derivative $Df(x)$ as a linear map from $\mathbb{R}^n$ to the tangent space to the manifold $f(A)$ at $f(x)$, which is a copy of $\mathbb{R}^n$, and ignore the complement to the tangent space. Now you have a linear map between equal dimensional spaces and its determinant is well-defined. It is then left to prove that this equals the one given above.
For proofs of the theorem in this form see Evans and Gariepy: Measure theory and Fine properties of functions
Many versions of the theorem appear in different contexts. Usually $f$ is assumed to be $C^1$ if one does not wish to go through Rademacher's differentiability theorem for Lipschitz maps, and for simplicity $f$ is assumed to be one-to-one, so $N(f,A,y)=1$ (i.e. it disappears).
