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In Stewart's Calculus book, an argument is given to intuitively explain why the theorem for change of variables, using Jacobian matrix, is valid without giving the full proof.

However, there are parts that I have little intuition over and would like to understand more coherently. I would appreciate it if someone explains them to me and preferably, if it is not too hard, by giving the exact proof.

  1. What is it about $C^1$ Transformation that is of interest to us?
  2. Why should $T$ (the $C^1$ transformation) map one region to another region (and not to multiple one)? Is this because of $C^1$ transformation?
  3. In Example 1: "The transformation maps the boundary of S into the boundary of the image" - I know this is not generally true. But where is it true? And how do we know that it is in this particular example?
  4. By moving around the boundary in the counterclockwise direction, we also move around the boundary of the image in the counterclockwise direction. Why is this true?

from text_1

from text_2

Thank you in advance.

Aria
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  • In the formula you have a derivative, and the product of that derivative and another function inside an integral. Therefore, you also need that that product can be integrated. These two conditions don't require that the change of variable has a derivative everywhere, and it doesn't require that that derivative is continuous. The condition can be relaxed somewhat, and even a bit more than in that link. The existence of continuous derivatives ...
    • –  Apr 05 '18 at 13:24
  • ... gives you control on the derivative through the values of the function $f(a)-f(b)=f'(c)(a-b)$. And when the derivative is continuous you get even more, since that $f'(c)$ is bounded, uniformly continuous, etc. –  Apr 05 '18 at 13:27
  • Thanks. Now I understand why using $C^1$ transformation is desirable. But what about 3 & 4? – Aria Apr 05 '18 at 13:31
  • This is a property of all continuous functions. You don't need the $C^1$ property. If you have a connected set, and you apply a continuous function, then the image is connected. This has been proven in this site several times. Here is one
    • –  Apr 05 '18 at 13:33
  • In Example that property of mapping the boundary to the boundary can be checked directly. You can compute the images of the boundary, as they did there, and you could compute the image of the interior points directly and verify that they fall inside the previously computed image of the boundary. That part, being Stewart's, they didn't do. Since you are using that awful book I know your calculus course is doomed to be a bad one. There reason behind that property is the open mapping theorem. The $T$ in that example ...
    • –  Apr 05 '18 at 13:45