This is about what we do when we find the new area in double integration. I am wondering if the reason the method under work, is supposed to be trivial, or can it be proved that it works?(adsbygoogle = window.adsbygoogle || []).push({});

If we have the original area in the x-y-plane, described by m inequalities, we call this area A:

[itex]

F_{1}(x,y)\geq 0\\

F_{2}(x,y)>0\\

.\\

.\\

F_{m}(x,y)\geq 0

[/itex]

We also have a continous, invertible transformation:

[itex] u=\hat{u}(x,y), v=\hat{v}(x,y)[/itex]

To find the area in the u-v plane I invert these functions.

[itex]x=\hat{x}(u,v), y = \hat{y}(u,v)[/itex]

And put this into the inequalities we started with:

[itex]

F_{1}(\hat{x}(u,v),\hat{y}(u,v))\geq 0\\

F_{2}(\hat{x}(u,v),\hat{y}(u,v))>0\\

.\\

.\\

F_{m}(\hat{x}(u,v),\hat{y}(u,v))\geq 0

[/itex]

Now, if we call the new area that these inequalities describe in the u-v plane B. Is it then trivial, that:

1. All points in A are mapped to B, there is not a point in A that is mapped to a point outside B.

2. The points in B only contains points that are mapped from A, that is, the area B does not contain any "extra" points, that do not come from A.

Are these two statesments trivial? I have seen them do this in many calculus books when they change variables in integration. But it is never discussed why we can do this.

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# Is this supposed to be trivial?, or can it be proved?

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