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If we have an integral and want to change coordinates we normally say: $$\iint_\Omega f(x,y)dxdy=\iint_{T(\Omega)}f(u,v)|J|dudv$$ where $|J|$ is the determinant of the Jacobian, I understand why this is here to account for changes of coordinate systems etc. but is there a nice proof of why the Jacobian is of this form

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  • $\begingroup$ A proof is given (for example) in Maxwell Rosenlicht's Introduction to Analysis. However if I recall correctly the proof is rather detailed. $\endgroup$
    – csch2
    Commented Jun 14, 2020 at 22:11
  • $\begingroup$ Thank you! I will have a look and see if I understand haha $\endgroup$
    – Henry Lee
    Commented Jun 14, 2020 at 22:14
  • $\begingroup$ If you already know about volume or measure, the main idea is following: For any rectangle $R$ and linear map $\Phi$ the volume of $\Phi(A)$ is $|\det \Phi|$ times the volume of $R$. Going from there to the change of variables formula is rather technical. $\endgroup$
    – user251257
    Commented Jun 14, 2020 at 22:17
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    $\begingroup$ Just at a first glance this is way above my head, but I will give it a good try tomorrow @csch2 $\endgroup$
    – Henry Lee
    Commented Jun 14, 2020 at 22:17
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    $\begingroup$ no, but if $\Phi$ is degenerate, the volume is simply $0$. $\endgroup$
    – user251257
    Commented Jun 14, 2020 at 22:20

2 Answers 2

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The basic idea is that when you perform the substitution, the diffferentials $dx,\,dy$ are expressed as a linear combination of $du,\,dv$ with the coefficients being the partial derivatives. For example, if $x=f(u,v),$ then we have that $$dx=f_u\,du+f_v\,dv.$$

Hence you have a $2×2$ linear system in $du,\,dv.$ This has a unique solution if and only if the determinant does not vanish.

Does this clarify things?

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This is an attempt to provide intuition in the 2-D case. The determinant can be thought of as the area of a parallelogram, in the standard cartesian co-ordinates, said parallelogram is a square.

When we do linear transformations, essentially we are shifting the 2 unit vectors $i$ & $j$, with restrictions being that the origin remains unchanged. Applying linear transformation to a point $(x,y)$ will keep said point on the same co-ordinate$(x,y)$ after the transformation, under the new co-ordinate system. Try to imagine stacking this new co-ordinate system on top of the standard cartesian co-ordinate system. The new point would have a new co-ordinate on the standard cartesian co-ordinate.

Because of this, we can think of ratios being 'equal' before and after linear transformation, if segment-$\bar{x}$ is twice as long as segment-$\bar{y}$ in the standard Cartesian co-ordinate, it would remain so in the new co-ordinate system.

Next, you can think of length in standard Cartesian as the ratio of said object against the side of a 1x1 box, and area as the ratio of said object against the area of a 1x1 box. Remember that ratios are maintained, so this will also apply to the new co-ordinate system. Think of length and area as ratios of sides and areas of the object against parallelograms.

Thus, if we know the ratio of length or area in the new co-ordinate system, say the value is $a$, the object would also have a length or area of ratio a under standard cartesian co-ordinates. The only thing left is comparing the *ratio of the 1x1 parallelogram against a 1x1 box. The determinant of the transformation matrix gives this ratio (since a 1x1 box is just 1, and anything divided by 1 is itself). That's why we multiply by the determinant of the Jacobian.

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    $\begingroup$ I'm sure in terms of formality this is very horrible, please do comment on areas I can improve to make it more understandable and more accurate. $\endgroup$ Commented Jun 14, 2020 at 22:41
  • $\begingroup$ Thank you, I think that is a good answer $\endgroup$
    – Henry Lee
    Commented Jun 14, 2020 at 23:03

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