Jacobian determinant in multiple integration

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The discussion centers on the search for a proof of the theorem related to the Jacobian determinant used in multiple integration. The original poster notes that their calculus textbook presents the theorem but lacks a proof, particularly in the context of transformations for cylindrical, spherical, and polar coordinates. Participants suggest that "Calculus on Manifolds" by Spivak and works by Mary Boas may contain such proofs, although there is some disagreement about the availability of proofs in different editions of Boas. One contributor mentions having derived their own proof in a vector analysis course. The conversation highlights the importance of understanding the Jacobian in simplifying integration problems through transformations.
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In what kind of math course would one learn the proof of the theorem that introduces the Jacobian to computing multiple integrals under various transformations?

My calculus textbook has this theorem, and uses it to derive the triple integral formulas for cylindrical and spherical coordinates, and the double integral formulas for polar coordinates. It also uses it to show how many integration problems can be solved much more easily by applying transformations and using the Jacobian.

But it does not give a proof. In what kind of textbook would I find a proof of this theorem?
Thanks.

BiP
 
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"Calculus on manifolds" by Spivak contains a proof of the theorem.
 
Mary Boas gives one too.
 
rollingstein said:
Mary Boas gives one too.

Not in my version of Boas.
 
micromass said:
Not in my version of Boas.

I might be wrong.
 
Here's my own proof I did in a vector analysis course I took. Cheers.
 

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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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