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Given a double integral, I want to find out what should I prove for the equality: $$ \int \int _\Omega f(x,y) dx dy = \int \int _{\Omega_{new}} f(x(u,v),y(u,v))\cdot J dudv $$

My dilemma is as follows: I know that the condition $J\neq 0$ in the domain $\Omega$ implies the validity of the inverse function theorem , and in particular that my mapping $u=u(x,y), v=v(x,y)$ is injective. But, if so, why does all the statements that I find of the theorem of changing variables has the two conditions: $ J\neq 0 $ and our mapping in injective ?

In addition, why does for linear mappings $u,v$ it is enough to check the Jacobian does not vanish ?

Clara
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The condition $J\ne 0$ only ensures local invertibility of the map (the inverse function theorem is all about "neighborhoods", isn't it?). For example, the map given in polar coordinates by $(r,\theta)\to(r,2\theta)$ has nonzero Jacobian in the region $1<r<2$, but it's not invertible, being 2-to-1.

For linear maps, there's a linear algebra theorem saying that a matrix with nonzero determinant has an inverse. That works globally.

  • Thanks a lot, I completely understand this, but , as you said - since $J\neq 0 $ ensures local invertibility , why can't we just partition our domain of integration into separate neighberhoods , such that in each neighberhood the inverse function theorem applies ? Hope I made my self clear now. Thanks a lot ! – Clara Jun 23 '15 at 06:33
  • You can have, say, neighborhoods $A$ and $B $ such that $f$ is injective in $A$ and is injective in $B$. But if $f(A)$ overlaps $f(B)$, you are double-counting things. –  Jun 23 '15 at 12:48
  • so can't I then ommit $f(A) \cap f(B)$ ? – Clara Jun 23 '15 at 13:25
  • Think of what you just wrote and what it means in the context of integrals. (Hint: nothing) –  Jun 23 '15 at 13:36
  • Got it ! Thanks !!!! – Clara Jun 23 '15 at 14:08