I have a random vector $(X_1,X_2)$ which has a uniform distribution on the triangle of vertices $(-1,0)$, $(0,1)$ and $(1,0)$. In other words, $(X_1,X_2)\sim \text{Uniform}(T)$ where $$T=\{(x_1,x_2)\in \mathbb{R}^2:-1<x_1<1,\ 1+x_1<x_2<1-x_1\}.$$ I'm asked to find the density function of the transformed vector $$(Y_1,Y_2)=\big(X_1^2,2X_2\big).$$
My attempt:
The transformation $(x_1,x_2)\mapsto\big(x_1^2,2x_2\big)$ has two inverses, namely $$h_1(y_1,y_2)=\left(\sqrt{y_1},\frac{y_2}{2}\right),\ h_2(y_1,y_2)=\left(-\sqrt{y_1},\frac{y_2}{2}\right).$$ On the other hand, since $T$ has area $1$ it follows that the density function of $(X_1,X_2)$ is $$f(x_1,x_2)=1,\ (x_1,x_2)\in T.$$ Now, I have all the ingredients to apply the formula I have for the transformed density when the transformation has more than one inverse: \begin{align*}g(y_1,y_2)&=f(h_1(y_1,y_2))|\text{J}_{h_1}(y_1,y_2)|+f(h_2(y_1,y_2))|\text{J}_{h_2}(y_1,y_2)|\\[1ex] &=\frac{1}{4\sqrt{y_1}}\left(f\!\left(\sqrt{y_1},\frac{y_2}{2}\right)+f\!\left(-\sqrt{y_1},\frac{y_2}{2}\right)\right).\end{align*} (Here, $g$ is the density function of $(Y_1,Y_2)$ and $\text{J}$ is the jacobian).
In order to determine whether $f\!\left(\pm\sqrt{y_1},\frac{y_2}{2}\right)$ is $1$ or $0$, I study the inequalities of the triangle: $$-1<\pm\sqrt{y_1}<1,\ 1\pm\sqrt{y_1}<\frac{y_2}{2}<1\mp\sqrt{y_1}.$$ For $\color{blue}{+}\sqrt{y_1}$ these inequalities have no solution. Thus, I get that $$g(y_1,y_2)=\frac{1}{4\sqrt{y_1}},\ (y_1,y_2)\in\{(y_1,y_2)\in \mathbb{R}^2:0<y_1<1,\ 2\big(1-\sqrt{y_1}\big)<y_2<2\big(1+\sqrt{y_1}\big)\}.$$ This region looks like this
Is my approach correct? Thanks!