A calculus of Jacobian on a surface This question might be quite easy for someone used to this kind of object.
Let us consider an open bounded domain $\Omega$ with $C^1$ boundary, $\Omega \subset \mathbb{R}^n$, and a vector $v \in \mathbb{R}^n$. For any point $x \in \Omega$, let us define 
$$\sigma(x) = \inf \{s > 0, x + sv \in \partial \Omega\}$$
and
$$y(x) = x + \sigma(x) v $$
I want to prove an equality of this kind, for a function having enough regularity so that everything is well defined
$$ \int_{\Omega} f(\sigma(x), y(x)) dx = \int_{z \in \partial \Omega, n_z \cdot v > 0} \int_{A_z} f(t,z) |v \cdot n_z| dt dz $$
(i.e., I want to parametrize every point $x \in \Omega$ as the associated point at the boundary obtained by following $v$ and the distance to that point). Any clue on how to compute the Jacobian ? 
I believe I should have something as $A_z = (0, \sigma(z,-v))$ but this is not the main thing I am interested in here. 
Edit: following the answer from Christian Blatter (thanks a lot for your help), the integral is only over the half space $n_z \cdot v > 0$, where $n_z$ is the normal vector at the surface at the point $z$. I am still looking for a rigourous derivation of the Jacobian in the general case (the case of a half ball is quite clear).
 A: I suggest assuming $v$ to be a unit vector. If ${\rm d}z$ denotes the scalar surface element on $\partial\Omega$ then the factor $|v\cdot n_z|$ already is the Jacobian. 
First try to understand whats happening in the case when $\Omega$ is a ball in ${\mathbb R}^3$ and $v=e_3$. A surface element ${\rm d}z$ then "covers" a slab of cross section area  $|v\cdot n_z|{\rm d}z$ and length the distance from the point $z\in\partial\Omega$ to the other point $z- \lambda(z) v\in\partial\Omega$, if such a point exists. The $t$-integral then is over the interval $[0,\lambda(z)]$.
The main defect of the proposed formula is that the integral is not over the full boundary $\partial\Omega$ but only over the part of it where $v\cdot n_z>0$. Determining this part in a general situation is not easy.  
In order to simplify matters assume that $\Omega$ is convex, and that $v=e_n$. Write the points $x\in{\mathbb R}^n$ as $x=(x',z)$ with $x'=(x_1,\ldots, x_{n-1})$. Let $\Omega'$ be the projection of $\Omega$ to the $x'$-base. Then there are two functions $a$, $b:\ \Omega'\to{\mathbb R}$ such that
$$\Omega=\bigl\{(x',z)\bigm|x'\in\Omega', \ a(x')\leq z\leq b(x')\bigr\}\ .$$The left hand side of the intended formula is then given by
$$\int_\Omega f(\sigma(x), y(x))\>{\rm d}(x)=\int_{\Omega'}\int_{a(x')}^{b(x')} f\bigl(\sigma(x',z), y(x',z)\bigr)\>dz\>{\rm d}(x')\ .$$
Now work your way to the RHS whereby ${\rm d}(x')$ is replaced by the scalar surface element ${\rm d}\omega$ on the upper boundary of $\Omega$. Note that integrating with respect to ${\rm d}\omega$ is just "symbolic". In reality we have an integral over the $(n-1)$-dimensional domain $\Omega'$ and the $x_n$-variable, denoted by $z$ here.  
