Given a scalar field $f: \mathbb R^n \supseteq V \to S \subseteq \mathbb R, \vec x\mapsto f(\vec x)$, what is the $n$ dimensional hypersurface (or volume, however you want to call this submanifold of $S\times V$) of $\{(y,\vec x) \in S\times V \big|\ y=f(\vec x)\}$?
The case $n=1$ leading to the arc-length of $f:[a,b]\to\mathbb R$, $\int_a^b\sqrt{1+|f'(x)|^2}\,dx$ is well-known, but what about $n>1$?
If it is just about the $n$-dimensional area $\omega_n(G)$ of graphs $G$ lying in ${\mathbb R}^n\times{\mathbb R}$ you can argue as follows: Such a graph is given by an equation of the form $$x_{n+1}-f(x_1,x_2,\ldots, x_n)=0\ .$$ The surface normal at a point $p=(p',p_{n+1})\in G$ is given by $$n(p)=\bigl(-f_{.1}(p'),\ldots, -f_{.n}(p'),1\bigr)=\bigl(-\nabla f(p'),1\bigr)\ .$$ Denote the angle enclosed between $n(p)$ and the positive $x_{n+1}$-axis by $\theta$. Then $$\cos\theta={n(p)\cdot e_{n+1}\over|n(p)|}={1\over|n(p)|}\ .$$ An "$n$-dimensional surface element" ${\rm d}\omega$ at $p$ projects $1:1$ to an $n$-volume element ${\rm d}(x)$ at $p'$ and is thereby multiplied by the factor $\cos\theta$. Conversely: One has $${\rm d}\omega={1\over\cos\theta}{\rm d}(x)=|n(p)|\>{\rm d}(x)=\sqrt{1+(\nabla f(p')\bigr)^2}{\rm d}(x)\ .$$ This geometric argument then leads to $$\omega_n(G)=\int_V \sqrt{1+(\nabla f(p')\bigr)^2}{\rm d}(x)\ .\tag{1}$$ The formula for the arc length of a graph in the $(x,y)$-plane is the case $n=1$ of $(1)$.
The above geometric argument is of course a "quickie" derivation. Depending on the available or desired level of sophistication one may use a partition of $V$ into tiny cubical boxes and use some approximation of $G$ as a union of $n$-dimensional parallelotopes, or one might plunge into more involved machinery, using Gram determinants, or even Hausdorff measures.
I forgot the theorem's name, but "it" states that $V$ can be expressed as a product (well, not soundly formulated, but the reader might know what is meant)
$$V = [a_1(x_2, ..., x_n), b_1(x_2, ..., x_n)] \times [a_2(x_3, ..., x_n), b_2(x_3, ..., x_n)] \times \cdots \times [a_n, b_n]$$
Let $\vec x_k := (x_k, ..., x_n)$, $f_1(\vec x_1) := f(\vec x)$ and
$$f_{k+1}(\vec x_{k+1}) := \int\limits_{a_k(\vec x_{k+1})}^{b_k(\vec x_{k+1})} \sqrt{1 + (\partial_kf_k(\vec x_k))^2}\, dx_k\quad \text{for}\quad k \ge 1.$$
Then $f_n$ is the sought hyperarcsurfacevolumething quantity: $f_{k+1}$ integrates over the $x_k$-direction in an arc-length sense, so after this arc-length-integration over all directions, the sought hypersurface is obtained.
From now on let's drop the variable and shorten $\partial_l f_k(\vec x_k) =: f_{k,l}$ and $f_k' := f_{k,k}$. Then by definition,
$$f_{k+1, l} = \sqrt{1+f_k'^2}\Big|_{a_{k,l}}^{b_{k,l}} + \int_{a_k}^{b_k} \frac{f_k' f_{k,l}'}{\sqrt{1+f_k'^2}} \,dx_k$$
and so on - this is getting messy, isn't it?
At first I thought this simplifies to
$$f_n = \int_V \sqrt{1 + \Big|\vec\nabla f(\vec x)\Big|^2}\, dx^n$$
but at the moment I am not so sure anymore. This sounds actually more logical than the upper result, and results from using $df_{k+1}^2 := dx_k^2 + df_k^2$ instead - to be continued...
$\int\sqrt(Tr(J^TJ))d\mathbf{x}$
