Understanding change of variables in double integrals In this libre text chapter, in example 15.7.1B, the author illustrates the change of variables using the following example:

The statement near it is written:
For the vertical length  $A:u=0,0≤v≤1$  transforms to  $x=−v^2,y=0$  so this is the horizontal length $A′ $ that joins  $(−1,0)$  and $ (0,0)$
I find it confusing that regardless of whatever function of $x$ could have been on $v$, it seems that all of them would only have a straight line of $A'$ like suppose it was $x=-v^3$, that'd also denote the same line
If I understood correctly, we derive $A'$ from $A$ by feeding constraints of $A$ into the dependencies of the new variables on the old. Under that idea, does the kind of function $x$ is on the old coordinates not matter in sketching the region(only the bounds does)?
 A: *

*When you transform a range to another range via a transformation, the relation between old and new variables are fixed by that transformation.

*When you transform a range to another range via a transformation, the only thing you need is to find the boundary of new range.

*When you transform a range to another range via a transformation, it has not yet involve any functions.

Let us see a function $f(v,u)$ defined on a domain $D$, your only task is to transform the domain of the function $f(v,u)$ to a new function $\tilde f(x,y)$ on a new domain $\tilde D$. You do not need to know the exact form of the function, only thing you need is to find the boundary of $\tilde D$, based on knowing the boundary of $D$ and the transformation.
A: Note the mapping
\begin{align*}
x&=u^2-v^2\tag{1}\\
y&=uv\tag{2}
\end{align*}
from the $u,v$-plane in the $x,y$-plane is also given in (2) by $y=uv$. This means whenever we set $\color{blue}{u=0}$, we have $y=0$ regardless of the setting of $v$.

*

*Nevertheless the mapping to parabolic coordinates given by (1) and (2) is often used since it is a one-to-one mapping with very nice geometrical properties:

*

*The curves $x = \mathrm{constant}$ give rise in the $u,v$-plane to the rectangular hyperbolas $u^2-v^2=\mathrm{constant}$. They have asymptotes which are $u=v$ and $u=-v$. The lines $y = \mathrm{constant}$ also correspond to a family of rectangular hyperbolas. Here the asymptotes are the coordinate axes. The hyperbolas of each family cut those of the other family at right angles.


*The lines parallel to the axes in the $u,v$-plane correspond to two families of parabolas in the $x,y$-plane. The parabolas $y^2=c^2(c^2-x)$ correspond to the lines $u=c$ and the parabolas $y^2=k^2(k^2+x)$ correspond to the lines $v=k$. All these parabolas have the origin as focus and the $x$-axis as axis. They form a famliy of confocal  and coaxial parabolas.
This explanation can be found for instance in the classic Introduction to Calculues and Analysis II by R. Courant and F. John.
I also like the following application of this kind of bijective mapping given by the authors. In the following we denote in OP's example the region bounded by the line segments $A,B$ and $C$ with $R$ and the region bounded by the curve segments $A^{\prime}, B^{\prime}$ and $C^{\prime}$ with $R^{\prime}$.

*

*One-one transformations have an important interpretation and application in the representation of deformations or motions of continuously
distributed substances, such as fluids. If we think of such a substance as spread out at a given time over a region $R$ and then deformed by a motion, the substance originally spread over $R$ will in general cover a region $R^{\prime}$ different from $R$. Each particle of the substance can be distinguished at the beginning of the motion by its coordinates $(u,v)$ in $R$ and at the end of the motion by its coordinates $(x,y)$ in $R^{\prime}$. The $1$-$1$ character of the transformation obtained by bringing $(u,v)$ into correspondence with $(x,y)$ is simply the mathematical expression of the physically obvious fact that separate particles remain separate.

