Double Integral $\iint\limits_Ry\,dx\,dy$ The question is as follows:

Use the transformation in Example 3 to evaluate the integral $$\iint_R y\  dx\  dy$$
  where $R$ is the region bounded by the $x$-axis and the parabolas $y^2 = 4-4x$ and $y^2=4+4x$.

The transformation from the example 3 is $x = u^2 - v^2$ and $y = 2uv$.
So here is what I have tried. Let $x = u^2 - v^2$ and $y = 2uv$.
The equation for the bounds become:
$$(2uv)^2 = 4-4(u^2-v^2)$$
After some simplification we get:
$$ -(u^2 -1)(v^2 +1)=0 $$
So $u=\pm 1$.
I do the same with the second bounds equation and get:
$$(u^2+1)(1-v^2)=0$$
Therefore $v=\pm1$
So the $u$ bounds is $-1\rightarrow1$ and same for the $v$ bounds.
We know $f(x,y)=y$ so $f(u,v) = 2uv$.
And we calculate the Jacobian and get $|J| = 4u^2+4v^2$
So the integral becomes:
$$\int^1_{-1} \int^1_{-1} 2uv(4u^2+4v^2)du\ dv 
 = 8\int^1_{-1} \int^1_{-1} u^3v + uv^3du\ dv
 \\= 8\int^1_{-1} \left.\bigg(\frac{u^4}{4}v + \frac{u^2}{2}v^3\bigg)\right\rvert^1_{u=-1} dv
= 8\int^1_{-1}  0\,dv.$$
I am pretty sure this is wrong... where did I go wrong?
 A: No, you are correct if you ignore the bound on the $x$-axis.
Without using transformations or such, we immediately see that you have the correct result:
$$\begin{align}\int_{-2}^2 y \int^{1-y^2/4}_{-1+y^2} \mathrm d\,x\,\mathrm d\, y ~=&~ \tfrac 1 2\int_{-2}^2 y(4-y^2)\,\mathrm d\, y \\[1ex] =&~ 0 \end{align}$$
[ Ever quicker: If you like, sketch out the graph of the parabolas and compare bounded areas where $y>0$ and $y<0$ and ask: what happens if I integrate $y$ over these areas? ]
Also your working appears to check out mostly okay.
Now try again with the bound on the x-axis.   Your target is:
$$\begin{align}\int_{0}^2 y \int^{1-y^2/4}_{-1+y^2} \mathrm d\,x\,\mathrm d\, y ~=&~ 2 \end{align}$$

Addendum:
To preserve content you need a bijection.   Every point in the original interval, $R$, must map onto only one point in the image and every point in the image must be mapped by only one point in the interval.   Else if we map $R$ into two points $(U,V)$ for every point $(X,Y)$ we double the corresponding intergral.   So we use only the absolute square root in the transformation : $$(x, y)=(u^2+v^2, 2uv)~\mathop{\iff}\limits_{(x,y)\in R}~(u, v)=\left(\left\lvert\sqrt{\frac{x-\sqrt{x^2-y^2}}2}\right\rvert, \frac{y}{\left\lvert\sqrt{2(x-\sqrt{x^2-y^2})}\right\rvert}\right)$$
Hence $R\hookrightarrow [0;1]^2$.
