How to show $ \int _\Omega |\nabla \phi|=\int _{-\infty} ^{+\infty} {\rm Vol}_{n-1}(\phi^{-1}(t)\cap \Omega ) \ dt $ For open $\Omega\subset R^n$ and smooth function $\phi:\Omega\rightarrow R$, how to show 
$$
\int _\Omega |\nabla \phi|=\int _{-\infty} ^{+\infty}  {\rm Vol}_{n-1}(\phi^{-1}(t)\cap \Omega ) \ dt
$$
 A: On $\Omega \subset \mathbb{E}^n$, consider a function $f$
Consider a vector field $X:=\frac{\nabla f}{|\nabla f|^2}$ Then we
have a hypersurface $S:=f^{-1}(0)\subset \Omega$  
In further we have $c: \Omega':=\mathbb{R}^{n-1}
 \times [0,\infty)\rightarrow \Omega$ s.t.
i) $c|\mathbb{R}^{n-1}\times \{0\}$ is
 diffeomorphism onto $S$ and
ii) $$
 \frac{\partial }{\partial t} c(x,t)=X(c(x,t))$$
Then
$$\int_{t_1}^{t_2} (f\circ c)'(s)\ ds =t_2-t_1\ \ast$$
and
\begin{align*}
  {\rm vol}_n\ \Omega
   &=\int_{\Omega'} |Dc|\ dxdt\end{align*}
To calculate ${\rm vol}_n\ \Omega$
  we want to decompose form $|Dc|\ dxdt$ into
 $\nabla f$-direction element and volume form on $f^{-1} (t)$
By $\ast$ $c(x,t)\in f^{-1} (t)$ for all $x\in \mathbb{R}^{n-1}$ In further
 $$ \frac{\partial }{\partial  x_i}c(x,t)\perp
\frac{\partial }{
 \partial t} c(x,t) $$
So $$ {\rm vol}_n\ \Omega =\int_0^\infty \int_{f^{-1}(t)}
\bigg|\frac{\partial }{
 \partial t} c(x,t)  \bigg|\ dA_t dt$$ where $dA_t$ is volume form on $f^{-1} (t)$
Hence $$ \int_\Omega |\nabla f| =\int_{0}^\infty \int_{f^{-1}(t) }
|\nabla f|\bigg|\frac{\partial }{
 \partial t} c(x,t)  \bigg|\ dA_t dt = \int_{0}^\infty
  \int_{f^{-1} (t)}   1\ dA_t\ dt =   \int_{0}^\infty
 {\rm vol}_{n-1}\ f^{-1}(t)\ dt $$
