Use the change of variables $x=u^2-v^2$, $y=2uv$ to evaluate $$\iint_{R}y dA$$ where $R$ is the region bounded by the x-axis, the parabolas $y^2=4-4x$ and $y^2=4+4x, y\geq0$
I'm following along with this solution:
I don't understand why the preimage of the domain is $[0,1]\times [0,1]$
I understand how we get $u=\pm 1$ and $v=\pm 1$ but I don't know how the bounds are from $0$ to $1$?
Why isn't it $$\int_{-1}^{1}\int_{-1}^{1} 2uv(4u^2+4v^2)dudv$$
EDIT: verify that there is some typo for the Jacobian, Because it should be
$\begin{vmatrix}
2u & -2v \\
2v & 2u
\end{vmatrix}=4(u^2+v^2)$
Best Answer
It may be helpful to break the boundary of $R$ into four pieces and look at their preimages:
Since $y=2uv$ and $y\ge 0$ imply that $u$ and $v$ must have the same sign, you could either use $[0,1]\times[0,1]$ or $[-1,0]\times [-1,0]$, but not both because you need the transformation to be one-to-one in order to calculate the double integral.
[Added.] Note that the mapping $G(u,v)=(u^2-v^2,2uv)$ is not injective: $G(u,v)=G(-u,-v)$. In order to use it in the double integral, one should restrict its domain to a subset of $\mathbf{R}^2$ in order to get an injection. For instance, $\{(u,v):v\ge 0\}$.
Now, consider $$ R_2=\{(x,y): 0\le y\le 2,\quad x=1-\frac{y^2}{4}\} $$ The second condition in terms of $(u,v)$ is $$ x=1-\frac{y^2}{4}\Longleftrightarrow u^2-v^2=1-u^2v^2 \Longleftrightarrow (u^2-1)(v^2+1)=0 \Longleftrightarrow u^2=1\tag{1} $$ The first condition $$ 0\le y\le 2 \Longleftrightarrow 0\le 2uv \le 2 \Longleftrightarrow 0\le uv\le 1\tag{2} $$ Combining (1) and (2) together you get that the preimage of the piece $R_2$ is $$ S_2=\{(u,v): u=1, 0\le v\le 1\}. $$
You can similarly figure out how to get the preimage of the other pieces.