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I am learning more about the change of variables formula and was confused about the role of the determinant of the jacobian. I came across this post, in which I found this particularly interesting:

"The higher dimensional version of the rectangle and interval is the $n$ dimensional rectangle defined by $[a_1,b_1] \ times ... \times [a_n,b_n]$ which has hypervolume $\prod_{i=1}^{n} (b_i-a_i)$ If we consider an invertible transformation $T: [a_1,b_1] \times ... \times [a_n,b_n] \rightarrow \mathbb{R}^n,$ by which $T(x)=Ax$ and $A$ is some $n \times n$ matrix that has nonzero determinant, we have the general formula

$$\text{Vol}(T([a_1,b_1] \times ... \times [a_n,b_n])) = |\det(A)| \text{Vol}([a_1,b_1] \times ... \times [a_n,b_n])."$$

My question is if we can derive a similar formula for the determinant of the jacobian to establish a similar rule for general $C^1$ maps

  • @Invinciible: Oh, I see. Anyway, your suggested formula doesn't make much sense, since the volume of $B$ can't depend on $x$. You should instead integrate $|\det DF(x)| , dx$ over $A$ to get the volume of $B$. (You also need to assume that $F$ is one-to-one.) – Hans Lundmark Oct 30 '22 at 18:47
  • @Invinciible: This is a good question. It would be a sort of mean value theorem if true. I deleted my previous comment in light of the changed question. –  Oct 30 '22 at 19:06

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Assuming $A$ is connected, yes, since the function is $C^1$, the determinant of the Jacobian is continuous, and this follows from the change of variables theorem and the mean value theorem for (multiple) integrals. If $A$ has two (or more) connected pieces, as usual it may well fail.

Ted Shifrin
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  • I see the MVT application here, but would you mind explaining the role of change of variables here in establishing this? –  Oct 30 '22 at 19:28
  • Just apply the change of variables theorem directly. Integrating the function $1$ over $B=F(A)$. (You assumed $F$ injective. Depending on the statement of the theorem you use, you may want to add the hypothesis that the Jacobian determinant is everywhere nonzero.) – Ted Shifrin Oct 30 '22 at 19:56
  • Then that would prove the statement on its own correct? So thenI suppose that means statement can be proved using either change of variables or MVT, but both in combination aren't necessary? –  Oct 30 '22 at 20:01
  • @Invinciible you need both. The change of variables theorem tells you (under suitable hypotheses) $\text{vol}(B)=\int_B1,d\eta=\int_A1\cdot |\det DF(\xi)|,d\xi$. This is an integral formula. To relate the integral on the right to a pointwise one, i.e to say that integral is equal to $|\det DF(x)|\cdot\text{vol}(A)$ for some point $x\in A$, you need the mean-value theorem (which again has its own set of hypotheses). So you need both. – peek-a-boo Oct 30 '22 at 20:16