Suppose I have a 3d solid in ${\bf R}^4$ which can be parametrized by the function $F:W\subset{\bf R}^3\rightarrow{\bf R}^4$. Now suppose I want to calculate the volume of this solid. Then naively I would compute the Jacobian of this map and then compute the following integral

$$\int_VdV=\int_W|\det{\bf J}_F(x,y,z)|dxdydz$$

But of course I can't do this since the Jacobian is not square. My understanding is that the way to do this is to actually compute $\sqrt{\det{\bf J}_F^{\mathrm{T}}{\bf J}_F}$. This of course reduces to $|\det{\bf J}_F|$ when the Jacobian is square, and also agrees with the definition of arc length found here, and the definition of surface integral found here.

And when I tested it for a parametrization ${\bf R}^2\rightarrow{\bf R}^3$ it was equivalent to taking the 2-norm of the vector containing the three square sub-determinants of its associated $3\times 2$ Jacobian. Furthermore it agrees with the way the Riemannian metric changes when you perform the pull-back, since the length of a tangent vector under the pull-back is given by $$\sqrt{\langle{\bf J}v, {\bf J}v\rangle}=\sqrt{v^{\mathrm{T}}{\bf J}^{\mathrm{T}}{\bf J}v}$$ Is all of this correct? Am I missing something? It was maddening to figure this out on my own as it's literally proposed no where in the introductory literature on vector calculus and differential geometry.

I don't know much about differential topology, but I believe this is somehow all connected with the way differential forms operate, and I would conjecture that for a Jacobian, say $m\times n$ with $m\geq n$, we have that $\sqrt{\det{\bf J}^{\mathrm{T}}{\bf J}}$ is equal to the Euclidean norm of the $\binom{m}{n}$ square sub-determinants of $\bf J$.

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    $\begingroup$ I commented already on a related question of yours, but I might as well put it here as well. It seems like this is related to the smooth area/coarea formulas. I'm not sure what the best reference is, but they're discussed for example in section 2.5.3 of this book: springer.com/us/book/9780817645090 $\endgroup$
    – stats_qs
    Commented Jun 5, 2017 at 16:23
  • $\begingroup$ I recently came across this exact same problem. I too have encountered little about this in any of my vector calculus or differential geometry books. Any luck finding references for this? The link in the answer is now dead. +1 for the well written post $\endgroup$ Commented Jul 23, 2022 at 20:42

1 Answer 1


Some care has to be taken to define the measure $dV$ with which you compute the volume of $V = f(W).$ In general, one could/would use the Haussdorff measure and then indeed, what you have obtained is correct and known as the area formula. See e.g. these notes by Hajlasz.

  • $\begingroup$ The link you provided has died. $\endgroup$
    – Galen
    Commented Apr 2, 2022 at 16:30

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