Volume of 3-dimensional region Let $G$ be bounded finitely connected domain in $\mathbb{R}^3$ with 2-smooth boundary $\partial G$ each connected component of which has positive Gaussian curvature.
Each sufficiently small open subset of $\partial G$ can be represented as the graph $\Gamma$ of a 2-smooth function $\phi(x)$ for $x \in V$, where V is an open subset of the hyperplane perpendicular to one of the basis vectors e of the space $\mathbb{R}^3$.
Let $t(\sigma)$ be the cosine of the angle between $e$ and $n(\sigma)$, where $n(\sigma)$ is the unit outward normal vector. We suppose that $t(\sigma)>0$ for $\sigma \in \Gamma$.
Let $\Pi \subset V$ be an open 2-dimensional parallelogram with edges parallel to the basis vectors of $\mathbb{R}^3$. Let $\xi$ the Lebesgue measure of $\Pi$ for dimension $2$.
Let $a,b>0$. We introduce the following 3-dimensional region
$$
F=\{\sigma+n(\sigma) u \; | \;\sigma\in \Gamma(\Pi);\; at(\sigma) < u < bt(\sigma)\}
$$
In the two dimensional case the region $F$ (with $a=0$ and $b=1$) looks like 
The research paper I am currently reading claims without proof that the set F has measure $\xi(b-a)$. This seems consistent with the sketch in the image.
Can anyone give me a hint on how to prove this claim.
 A: $\newcommand\si\sigma$The expression $\Gamma(\Pi)$ makes no sense, and hence the definition of $F$ in your post makes no sense. Replacing $\Gamma$ by $\phi$, we get a definition which does makes sense:
$$F:=\{\si+n(\si) u\colon\si\in \phi(\Pi),\, at(\si)<u<bt(\si)\}.$$
However, the formula
$$|F|=\xi(b-a)\tag{1}\label{1}$$
for the Lebesgue measure $|F|$ of $F$ is of course incorrect in general.
E.g., let $\Pi:=[0,1]\times[0,1]$, $e=(0,0,1)$, and $\phi(x,y):=(x,y,1-x^2)$ for $(x,y)\in\Pi$. Then $n(\phi(x,y))=\frac{(2x,0,1)}{\sqrt{4x^2+1}}$, $t(\phi(x,y))=\frac1{\sqrt{4x^2+1}}$, and
$$F=g(\Pi_{a,b}),$$
where
$$g(x,y,u):=\left(x+\frac{2 u x}{\sqrt{4 x^2+1}},y,1-x^2+\frac{u}{\sqrt{4 x^2+1}}\right),$$
$$\Pi_{a,b}:=\{(x,y,u)\colon(x,y)\in\Pi,\,at(\phi(x,y))<u<bt(\phi(x,y))\}.$$
The Jacobian determinant of transformation $g$ is
$$J(x,y,u)=\frac{2 u \sqrt{4 x^2+1}+16 x^4+8 x^2+1}{\left(4 x^2+1\right)^{3/2}}>0$$
for $(x,y,u)\in\Pi_{a,b}$ if $a=0<b$. So, then
$$|F|=\int_{\Pi_{a,b}} J
=b+\frac{1}{20} b^2 \left(2+5 \tan ^{-1}(2)\right)
\approx b + 0.376787 b^2,$$
which is quadratic in $b$, in contrast with the purported expression \eqref{1}, which is affine in $b$.

However, in this example -- and in general, the expression \eqref{1} holds "in the first approximation": If $a=0$ and $b\downarrow0$, then
$$|F|=\xi b+o(b).\tag{2}\label{2}$$
Indeed, in general $\phi(x,y):=(x,y,f(x,y))$ for some smooth enough function $f$, and then
$$J(x,y,0)=\frac1{t(\phi(x,y))}>0$$
and hence
$$
\begin{aligned}
|F|&=\int_{\Pi_{a,b}} |J| \\ 
&=\int_\Pi dx\,dy\, \int_0^{bt(\phi(x,y))} du\,|J(x,y,u)| \\ 
&=\int_\Pi dx\,dy\, \int_0^{bt(\phi(x,y))} du\,(J(x,y,0)+o(1)) \\ 
&=\int_\Pi dx\,dy\, bt(\phi(x,y))\,(J(x,y,0)+o(1)) \\ 
&=\int_\Pi dx\,dy\, b\,(1+o(1)) \\ 
&=\xi\, b\,(1+o(1))=\xi b+o(b),
\end{aligned}$$
as claimed.

Geometrically/heuristically, \eqref{2} is almost obvious. Indeed, for infinitesimally small $b>0$, consider the body
$$G(b):=\{(x,y,z)\colon(x,y)\in\Pi,\, f(x,y)\le z\le f(x,y)+b\}$$
between the graphs of the functions $f$ and $f+b$ (over $\Pi$). Clearly,
$$|G(b)|=\xi b.\tag{3}\label{3}$$
Note that the distance from most points $P$ on the graph of $f$ (except for some $\asymp b$-area fraction of them), the distance from $P$ to the graph of $f+b$ along the outer normal $n(P)$ to the graph of $f$ at $P$ will be $\sim bt(P)$ (you may want to draw a simple picture here for yourself). So, $G(b)\approx F$ if $a=0$ and $b>0$ is infinitesimally small. Now \eqref{2} "approximately follows" from \eqref{3}.
