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

[bobby2k]

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?

If we have the original area in the x-y-plane, described by m inequalities, we call this area A:
<br /> F_{1}(x,y)\geq 0\\<br /> F_{2}(x,y)&gt;0\\<br /> .\\<br /> .\\<br /> F_{m}(x,y)\geq 0<br />

We also have a continous, invertible transformation:
u=\hat{u}(x,y), v=\hat{v}(x,y)

To find the area in the u-v plane I invert these functions.
x=\hat{x}(u,v), y = \hat{y}(u,v)

And put this into the inequalities we started with:

<br /> F_{1}(\hat{x}(u,v),\hat{y}(u,v))\geq 0\\<br /> F_{2}(\hat{x}(u,v),\hat{y}(u,v))&gt;0\\<br /> .\\<br /> .\\<br /> F_{m}(\hat{x}(u,v),\hat{y}(u,v))\geq 0<br />

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.

 

[HallsofIvy]

I am puzzled by your question "is this trivial or can it be proven". If something is "trivial" then it can be easily proven! Of course, what is "easy" or "trivial" is often in the eye of the beholder!

(I want to put in an old story: A mathematics professor is delivering a proof in class. At one point he says "it is trivial that ..." when he suddenly stops, thinks for a minute, then sits down at his desk and starts writing. After scribbling for about 15 minutes he jumps up and says "Yes, it is trivial!")

But what you are asking follows directly from the fact that transformation is "invertible". That means that the transformation is "one to one" and "onto" (in technical language, "surjective" and "injective"). Every point in A is mapped to a single point in B, every point in B is mapped to a single point in A.

 

[bobby2k]

I am puzzled by your question "is this trivial or can it be proven". If something is "trivial" then it can be easily proven! Of course, what is "easy" or "trivial" is often in the eye of the beholder!

(I want to put in an old story: A mathematics professor is delivering a proof in class. At one point he says "it is trivial that ..." when he suddenly stops, thinks for a minute, then sits down at his desk and starts writing. After scribbling for about 15 minutes he jumps up and says "Yes, it is trivial!")

But what you are asking follows directly from the fact that transformation is "invertible". That means that the transformation is "one to one" and "onto" (in technical language, "surjective" and "injective"). Every point in A is mapped to a single point in B, every point in B is mapped to a single point in A.

Hehe, I laughed from that story! :)

Thank you for your answer, but I am not sure if it was the answer I was looking for, I was maybe not precice enough.
What I mean is that the area B is the area we got when we solve the inequalities, and let's call the mapping of A to the u-v plane C. Another way to ask the question is to ask if C is completely in B, and if B is completely in C?
I mean, how do we know that working with the inequalities is enough?, why do we not have to go over every point in A to get C, we can instead describe A with inequalities, and then do what I described, and then we get the area B, which we know is containted in C, and does not have any points C does not have.

 

[bobby2k]

Did you understand my question, or was it too weird of a question? I'd be happy to try to explain it better if what I meant is still unclear.

 

[Office_Shredder]

Are you looking for something like this? If I have a point in B, say (u,v), then I can get (x,y) = (x(u,v),y(u,v)). By definition since (u,v) is in B, F(x(u,v),y(u,v)) > 0 which is the same as writing F(x,y) > 0. So (x,y) is in A

Given a point (x,y) in A, IF it is mapped to by a point in the u,v plane, then that point has to be in B, because if (x,y) = (x(u,v),y(u,v)) then F(x,y) > 0 because (x,y) is in A, and therefore F(x(u,v),y(u,v)) > 0 which is exactly what is required for (u,v) to be in B. So the only remaining question is given a point (x,y), can I find a point (u,v) which maps to it? The answer is yes - if I apply my transformation from the x-y plane to the u-v plane, any specific point (x,y) gets mapped to some point (u,v), and that point will get mapped back to (x,y) when I take the inverse

 

[bobby2k]

Thank you very much Office_Shredder, this was exactly what I was looking for. It was not clear at all for me that it was safe to transform areas with the help of inequalities, but now you have proved that it is completely safe.

I just have four very easy follow-up questions to make sure that I got the point.

Your proof:

If I have a point in B, say (u,v), then I can get (x,y) = (x(u,v),y(u,v)). By definition since (u,v) is in B, F(x(u,v),y(u,v)) > 0 which is the same as writing F(x,y) > 0. So (x,y) is in A

proves my question?:

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.

And your proof:

Given a point (x,y) in A, IF it is mapped to by a point in the u,v plane, then that point has to be in B, because if (x,y) = (x(u,v),y(u,v)) then F(x,y) > 0 because (x,y) is in A, and therefore F(x(u,v),y(u,v)) > 0 which is exactly what is required for (u,v) to be in B.

proves my question?:

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

The last point you made was also interesting:

So the only remaining question is given a point (x,y), can I find a point (u,v) which maps to it? The answer is yes - if I apply my transformation from the x-y plane to the u-v plane, any specific point (x,y) gets mapped to some point (u,v), and that point will get mapped back to (x,y) when I take the inverse

It was not enough with the two proofs above? You need this statement as well to justify that we can use this method when we double-integrate?, so that we know that every point in the original area is covered by the new area?

Lastly I see you made the proof with only one inequality, but I guess it is no problem to say that they hold for m inequalities aswell?

EDIT: Also I wonder, does the proof hold if the function is invertible, but not continious?