I guess this is something pretty standard in calculus, but I was unable to google the answer.


Assume I have unit hypercube $C\_n = [0,1]^n$. I also have a function $f : \mathbb{R}^n \to \mathbb{R}^{n+1}$. The components $f\_i : \mathbb{R}^n \to \mathbb{R}$ are polynomials $\forall i = 1,...,n+1$, so it's all quite well behaved. I want to find the n-volume (measure) of $f(C\_n)$. How do I do that in general?



How to compute volume of parametric regions?
For $t=(t\_1,\dots,t\_n)$ in the cube $C\_n$, let $A(t)=(a\_{ij}(t))\_{i,j=1}^{n+1}$ be the matrix whose entries $a\_{ij}(t)$ are defined as follows: 


$a\_{ij}(t):=\partial\_i f\_j(t)$ for $i=1,\dots,n$ and $j=1,\dots,n+1$, where $\partial\_k f\_j(t)$ denotes the partial derivative of the function $f\_j$ with respect to its $k$th argument $t\_k$;


$a\_{n+1,j}(t):=e\_j$ for $j=1,\dots,n+1$, where $(e\_1,\dots,e\_{n+1})$ is the standard basis of $\mathbb R^{n+1}$. 


In other words, for $i=1,\dots,n$, the $i$th row of the matrix $A(t)$ is the vector $\partial\_i f(t):=(\partial\_i f\_1(t),\dots,\partial\_i f\_{n+1}(t))$, and the last row of $A(t)$ is $(e\_1,\dots,e\_{n+1})$. 


So, 
\begin{equation\*}
 \det A(t)=\bigwedge(\partial\_1 f(t),\dots,\partial\_n f(t)), 
\end{equation\*}
the generalized $n$-ary cross product of the vectors $\partial\_1 f(t),\dots,\partial\_n f(t)$ in $\mathbb R^{n+1}$; see e.g. Section [Multilinear algebra](https://en.wikipedia.org/wiki/Cross_product). 


It follows that the volume in question is  

$$\int\_{C\_n}\|\det A(t)\|\,dt=\int\_{C\_n}\Big\|\bigwedge(\partial\_1 f(t),\dots,\partial\_n f(t))\Big\|\,dt,$$ 
where $\|\cdot\|$ is the Euclidean norm in $\mathbb R^{n+1}$. 




---


The latter formula is an extension of the formula for the area of a 2D surface in $\mathbb R^3$ in terms of the cross product (in which case $n=2$), and its derivation in general is quite similar to that for $n=2$; cf. e.g. [formulas (16.6.8)--(16.6.12)](https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/16%3A_Vector_Calculus/16.6%3A_Surface_Integrals). The key fact here is as follows. For any vectors $v\_1,\dots,v\_n$ in $\mathbb R^{n+1}$, let $B$ the matrix in $\mathbb R^{(n+1)\times(n+1)}$ with rows $v\_1,\dots,v\_{n+1}$, where $v\_{n+1}:=\bigwedge(v\_1,\dots,v\_n)$. Note that the vector $v\_{n+1}$ is orthogonal to the vectors $v\_1,\dots,v\_n$. 
So, the $(n+1)$-volume $V\_{n+1}(v\_1,\dots,v\_{n+1})$ of the parallelotope defined by the vectors $v\_1,\dots,v\_{n+1}$ is 
\begin{equation\*}
 |\det B|=V\_n(v\_1,\dots,v\_n)\|v\_{n+1}\|; \tag{1}
\end{equation\*}
here, $V\_n(v\_1,\dots,v\_n)$ is of course the $n$-volume of the parallelotope defined by the vectors $v\_1,\dots,v\_n$. 
On the other hand, expanding $\det B$ along its last row, we see that 
\begin{equation\*}
 \det B=v\_{n+1}\cdot\bigwedge(v\_1,\dots,v\_n)=\|v\_{n+1}\|^2, \tag{2}
\end{equation\*}
where $\cdot$ denotes the dot product. 
Comparing (1) and (2), we conclude that 
\begin{equation\*}
 V\_n(v\_1,\dots,v\_n)=\|v\_{n+1}\|=\Big\|\bigwedge(v\_1,\dots,v\_n)\Big\|. 
\end{equation\*}


