Determine double integral over region, use change of variables A seemingly simple computation but I'm not quite sure how to proceed.
The question says to determine $\iint_S \frac{(x+y)^4}{(x-y)^5} \,dA$ where S = $\{-1 \leq x + y \leq 1, 1 \leq x - y \leq 3\}$.
My incomplete answer proceeds as follows:
This region suggests that we take a change of variables of form $u = x + y$ and $v = x - y$ so that setting $T = \{-1 \leq u \leq 1, 1 \leq v \leq 3\}$ implies $G: S -> T$ given by $(u,v) = G(x,y) = (x+y, x - y)$ is a diffeomorphism.
Now $|det\ DG(x,y)| = \bigl(\begin{smallmatrix}
1&1 \\ 1&-1
\end{smallmatrix} \bigr) = -1 -1 = -2$.
Thus $dudv = -2dxdy$ and our integral becomes: ...
Any clarification on where to go from here would be appreciated or if there's an error in my steps above.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\iint_{S}{\pars{x + y}^{4} \over \pars{x - y}^{5}}\,\dd x\,\dd y & =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{\pars{x + y}^{4} \over \pars{x - y}^{5}}\bracks{-1 \leq x + y \leq 1}
\bracks{1 \leq x - y \leq 3}\dd x\,\dd y
\\[5mm] & =
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{x^{4} \over \pars{x - 2y}^{5}}\bracks{-1 \leq x \leq 1}
\bracks{1 \leq x - 2y \leq 3}\dd x\,\dd y
\\[5mm] & =
-\,{1 \over 2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{x^{4} \over \pars{y - x}^{5}}\bracks{-1 \leq x \leq 1}
\bracks{-3 \leq y - x \leq -1}\dd x\,\dd y
\\[5mm] & =
-\,{1 \over 2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{x^{4} \over y^{5}}\bracks{-1 \leq x \leq 1}
\bracks{-3 \leq y \leq -1}\dd x\,\dd y
\\[5mm] & =
-\,{1 \over 2}\pars{\int_{-1}^{1}x^{4}\dd x}
\pars{\int_{-3}^{-1}{\dd y \over y^{5}}} = \bbx{\ds{4 \over 81}}
\end{align}
