# How to compute volume of parametric regions?

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?

• Just to clarify, do you want the volume of the parameterized surface or of its image? E.g., if $f(x)=(4 (x-1/2)^2,0)$ do you want $f(C_1)$ to have volume $2$ (the length parameterized curve) or $1$ (the length of the image)? If it is the first one, then you can use the Area formula (e.g., see Theorem 1.8 of these notes: pitt.edu/~hajlasz/Teaching/Math2304Spring2014/Part%202.pdf ). – RBega2 Mar 8 at 13:03
• I think in my application we can also assume that $f$ is injective (or injective up to zero measure set), but if not, I would be interested in analysis for the later which gives 1. – user1747134 Mar 8 at 13:10

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.

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). 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 $$\begin{equation*} V_n(v_1,\dots,v_n)=\|v_{n+1}\|=\Big\|\bigwedge(v_1,\dots,v_n)\Big\|. \end{equation*}$$

• Gives the expected results on examples. Thanks! – user1747134 Mar 8 at 14:04
• Where I could learn more about how this is derived? The thing with mixing vectors and scalars as matrix elements looks kind of weird to me. Is this only abuse of notation? If yes, what would be a more correct way of writing it? – user1747134 Mar 10 at 11:29
• Is it true that here $\mathrm{det} A(t) = \bigwedge(\nabla f_1, ..., \nabla f_n)$ where $\bigwedge$ is the generalized cross product? – user1747134 Mar 10 at 12:16
• @user1747134 : I have provided details addressing your comments. – Iosif Pinelis Mar 10 at 19:28