Formula for finding the N-dimensional hypervolume bounded by a parametric (N-1)-dimensional hypersurface? Say we have an $N$-dimensional hypervolume, which is bounded by a parametric $N-1$-dimensional hypersurface. We can state the dimensions as $x_1\dots x_n$ and the parameters as $t_1\dots t_{n-1}$.
Would the formula:
$\frac{1}{N}\idotsint\begin{vmatrix}x_1 & \dots & x_N \\\frac{\partial x_1}{\partial t_1} & \dots & \frac{\partial x_N}{\partial t_1} \\\vdots & \ddots & \vdots \\\frac{\partial x_1}{\partial t_{N-1}} & \dots & \frac{\partial x_N}{\partial t_{N-1}} \\\end{vmatrix}dt_1 \dots dt_{N-1}$
allow me to find its hypervolume?
 A: No, that is not quite right. The actual formula is hard to express in closed form.
Let's see where the original formula comes from $\int y(t)x'(t)dt$. This comes from taking $\int y(t)dx$ and substituting in $x'(t)dt$ for $dx$. 
First of all, the variable that is used in the integral is not arbitrary. Doing this in n dimensions requires choosing one variable. This is the variable that the area "under" is measured. I will call this variable $y$, and all the other variables $x_1$ through $x_n$. In the case with two dimensions, the formula becomes $\int y(t_1)x_1'(t_1)dt_1$. 
We can create an equivalent formula for more dimensions, an example with 3:
$$\iint y(t_1, t_2)dx_1dx_2$$
Now we apply a change of variables with multiple integrals. An explanation of how to do this can be found here: https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables. 
Basically, substitute $dx_1...dx_n$ with 
$$\left|d\begin{bmatrix}
\frac{dx_1}{dt_1} & \dots & \frac{dx_1}{dt_n}\\
\vdots & \ddots & \vdots\\
\frac{dx_n}{dt_1} & \dots & \frac{dx_n}{dt_n}
\end{bmatrix}\right|$$
Where $d(m)$ is the discriminant of matrix $m$. This gives the formula:
$$ \int\dots\int y \left|d\begin{bmatrix}
\frac{dx_1}{dt_1} & \dots & \frac{dx_1}{dt_n}\\
\vdots & \ddots & \vdots\\
\frac{dx_n}{dt_1} & \dots & \frac{dx_n}{dt_n}
\end{bmatrix}\right|dt_1\dots dt_n$$
Note that this function doesn't always work as nicely as the integral works in 2 dimensions. It may count the volume of certain areas multiple times. Strictly, it calculates the total volume between the function and the $y=0$ plane, counting volume below the plane as negative. Removing the absolute value bars from the Jacobian discriminant may make the integral work like the 2-dimensional version, but I can't confirm that.
