Double integral change of variables Using an appropriate change of variables, evaluate
$$
\iint_{B}\exp\left(\,{y - x \over y+x}\,\right)\,{\rm d}x\,{\rm d}y
$$ 
Where $B$ is the interior of the triangle with vertices at
$\left(\, 0,0\,\right), \left(\, 0,1\,\right)\ \mbox{and}\ \left(\, 1,0\,\right)$.
Attempt:
I used $u = y - x\,,\ v = y + x$ as my new variables. I also found that the old bounds were $0\ \leq\ x\ \leq\ 1$ and $0\ \leq\ y\ \leq\ 1 - x$.
However, I don't understand how to put my bounds in terms of $u$ and $v$.
I tried setting $y = u + x = v - x$, then solving for $x$ to get
$x = \left(\, v - u\,\right)/2$. I substituted that into the bound for $x$ and got $0\ \leq\ v\ \leq\ 2 + u$.
Is that correct ?. How do I get the constant bounds for $u$ ?.
Thank you!
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\iint_{B}\exp\pars{y - x \over y + x}\,\dd x\,\dd y}
=\int_{0}^{1}\ \overbrace{%
\int_{0}^{1 - x}\exp\pars{y - x \over y + x}\,\dd y}
^{\ds{\color{#c00000}{{y - x \over y + x}\equiv t\ \imp\
y={1 + t \over 1 - t}\,x}}}\ \,\dd x
\\[5mm]&=\int_{0}^{1}\int_{-1}^{1 - 2x}\expo{t}\,{2x \over \pars{1 - t}^{2}}
\,\dd t\,\dd x
=2\expo{}\int_{0}^{1}x\int_{-2}^{-2x}{\expo{t} \over t^{2}}\,\dd t\,\dd x
\\[5mm]&=2\expo{}\int_{0}^{1}x\int_{1}^{x}{\expo{-2t} \over 4t^{2}}\,\pars{-2}
\,\dd t\,\dd x
=\expo{}\int_{0}^{1}x\int_{x}^{1}{\expo{-2t} \over t^{2}}\,\dd t\,\dd x
=\expo{}\int_{0}^{1}{\expo{-2t} \over t^{2}}\int_{0}^{t}x\,\dd x\,\dd t
\\[5mm]&=\half\,\expo{}\int_{0}^{1}\expo{-2t}\,\dd t
=\half\,\expo{}\,{\expo{-2} - 1 \over - 2}
=\color{#66f}{\large{\expo{2} - 1 \over 4\expo{}}}\approx{\tt 0.5876}
\end{align}
A: The points $(0,0), (1,0), (0,1)$ in the xy-plane map to the points $(0,0), (-1,1), (1,1)$ in the uv-plane; so 
the region should be described by the inequalities $0\le v\le 1$, $-v\le u\le v$.
