# 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!

• I suspect a typo, as as stated your integrand is $e^0$ – GFauxPas Nov 12 '14 at 22:32
• You're right! I fixed it. – sosoo Nov 12 '14 at 22:57

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
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 $0\le v\le 1$, $-v\le u\le v$.