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I understand the need for a factor to account for change of units of length/area/volume in multiple integration up to triple integration - & understand why the Jacobian is the appropriate factor, but cannot find any simple intuitive explanation or inductive proof that shows that the Jacobian gives the appropriate factor for change of variable in multiple integration in 4 variables or higher. Why is the volume of an element in the new variables J times the volume of the old? Would be grateful for an outline proof without any measure theory - happy to assume that the change of variables is 1:1 in the region of interest. David Kault, retired maths lecturer, JCU, Townsville, Australia

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The way of thinking of the proof is not really inductive. The following might help. First we show that when a linear transformation(matrix) is applied, the volume of some region changes by a factor of the determinant. I suppose this is proved in Rudin's "Principles of Mathematical Analysis", in the chapter on analysis of several variables; but I am not sure. The proof of this would require the fact that a matrix is a product of elementary matrices, and we prove it for elementary matrices, and scalar multiplications. Now having done that, we continue in the next comment. –  Regenbogen Feb 28 '10 at 4:45
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A differentiable function of one variable can be approximated at every point by its tangent line. Similarly, a smooth function of several variables at each point can be approximated with a linear transformation(plus a constant). Now, the volume of an extremely small area changes by a factor of the determinant, under a linear transformation. So at each point, looking infinitesimally close, the volume change is by a linear transformation. And this relevant linear transformation is the Jacobian at the point. This is the rough outline. It is proved with the missing steps in Rudin's book. –  Regenbogen Feb 28 '10 at 4:47
    
Yes, I checked, it is proved in Rudin's book, Chaps 9 and 10, without measure theory. –  Regenbogen Feb 28 '10 at 4:50
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Other than that it cannot be directly visualized, how is the case of 4 or more variables any different from the cases of 2 and 3 variables? –  Pete L. Clark Feb 28 '10 at 6:03
    
It is not too hard to show that the area of a parallellogram spanned by two vectors $u$, $v$ (with a sign attached to account for orientation effects) satisfies the axiom for a $2\times2$ determinant (multilinear, alternating, $\det I=1$). Once the students get it, it is not a long stretch to extend that to three dimensions. And so it may not be altogether unreasonable to wave your arms about a bit and say it goes the same way in higher dimensions too. (But when it's time to get rigorous, look at the other comments.) –  Harald Hanche-Olsen Feb 28 '10 at 13:45
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