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.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}