PDF of transformed random variable and Surface Element in High Dimension. Consider $k$ functions $T_i(x_1,\dots,x_n)$ on $\mathbb{R}^n$, Let $p(x_1,\dots,x_n)$ be a pdf of random variable $X$, then what is the pdf of
$$T(x)=(T_1,\dots,T_k)?$$
Some books state that
$$p_T(t_1,\dots,t_k)=\int_S p(x_1,\dots,x_n)\left[\displaystyle\sum_{i_1<\cdots<i_k}\left|\frac{\partial(T_1,\dots,T_k)}{\partial(x_{i_1},\dots,x_{x_k})}\right|^2\right]^{-\frac{1}{2}}\,\mathrm{d}S,$$
where $\frac{\partial(T_1,\dots,T_k)}{\partial(x_{i_1},\dots,x_{x_k})}$ is Jacobian determinant, and the $(n-k)$-dimensional surface $S$ is definited by equations
$$T_i(x_1,\dots,x_n)=t_i,\ 1\leq i\leq k.$$
Why? What is the relation between surface element $\mathrm{d}S$ and differential form $\sum f_{i_1,\dots,i_k}\mathrm dx_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}$?
In addition, I know that if $S$ has parameterization (with some abuse of notations)
$$x_i=x_i(u_1,\dots,u_k),\ 1\leq i\leq n.$$
then
$$\begin{align*}
&\quad\;\sum a_{i_1,\dots,i_k}\mathrm{d}x_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}\\
&=\left(\sum a_{i_1,\dots,i_k}\frac{\partial(x_{i_1},\dots,x_{i_k})}{\partial(u_1,\dots,u_k)}\right)\mathrm{d}u_1\wedge\cdots\wedge\mathrm{d}u_k.
\end{align*}$$
But is there exists an inversion of the formula, i.e., can we express
$$a\,\mathrm{d}u_1\wedge\cdots\wedge\mathrm{d}u_k,$$
in form of $\mathrm{d}x_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}$?
 A: This is the coarea formula from geometric measure theory. If we assume that $T$ has constant rank $k$ (which is implicit in the assumption that the level sets $T_i=t_i$, $i=1,\dots,k$, are smooth), then this is just integration over the fiber.
For example, if $k=1$, then we want to write $dV = dx_1\wedge\dots\wedge dx_n= F(x)\,dS\wedge dt$, where $dS$ is the surface area element on the hypersurface $T=t$. We have the classic formula $dS = \iota_{X}dV = dV(X,\cdot)$, where $X$ is the unit normal vector $\nabla T/\|\nabla T\|$ of the level hypersurface. Since $dt(\nabla T)=dT(\nabla T) = \|\nabla T\|^2$, it follows that (up to sign) $dV = \frac1{\|\nabla T\|}dS\wedge dt$. Integration over the fiber gives
$$p_T(t) = \int_{\{T=t\}} p(x_1,\dots,x_n) \frac1{\|\nabla T\|}dS.$$
More generally, if $k>1$, then the hypersurfaces $T_i=t_i$ will not necessarily be orthogonal, and we will have (up to sign)
$$dS = \frac{dV\left(\tfrac{\nabla T_1}{\|\nabla T_1\|},\dots,\tfrac{\nabla T_k}{\|\nabla T_k\|},\cdot\right)}{\|dT_1\wedge\dots\wedge dT_k\|},$$
since  $(dT_1\wedge\dots\wedge dT_k)\left(\tfrac{\nabla T_1}{\|\nabla T_1\|},\dots,\tfrac{\nabla T_k}{\|\nabla T_k\|}\right)$ is the conorm of $dT_1\wedge \dots\wedge dT_k$. This, in turn, is the expression in your formula.
A: One simple example if permitted. I did this mainly to better understand the matter for myself. For
\begin{align}
T(x_1,x_2)=\sqrt{x_1^2+x_2^2}\,,\quad X_1,X_2\sim N(0,1)\,,\text{ independent }
\end{align}
it is easy to see that
\begin{align}
\mathbb P\big(T(X_1,X_2)\le t\big)&=\int_{T(x_1,x_2)\le t}\frac{1}{2\pi}e^{-\frac{x_1^2+x_2^2}{2}}\,dx_1\,dx_2
&
=\int_0^t\underbrace{\int_0^{2\pi}\frac{1}{2\pi}e^{-\frac{r^2}{2}}\,r\,d\varphi}_{(*)}\,dr\\
&=1-e^{-t^2/2}\,,\\
\end{align}
which has density
\begin{align}
p_T(t)&=te^{-t^2/2}\,.
\end{align}
Since,
\begin{align}
\nabla T&=\frac{1}{\sqrt{x_1^2+x_2^2}}\left(\begin{matrix}x_1\\x_2\end{matrix}\right)\,,\quad\frac{1}{||\nabla T||}=1
\end{align}
and
\begin{align}
p(x_1,x_2)=\frac{1}{2\pi}e^{-\frac{x_1^2+x_2^2}{2}}
\end{align}
we can write this in the form given by Ted Shifrin:
\begin{align}
p_T(t)&=\int_{\{T=t\}}p(x_1,x_2)\frac{1}{||\nabla T||}\,dS
=\underbrace{\int_0^{2\pi}\frac{1}{2\pi}e^{-\frac{t^2}{2}}\,t\,d\varphi}_{(*)}\,.
\end{align}
