[tex]|\det\mathbf J|=\left|\det\begin{bmatrix}x_s&x_t\\y_s&y_t\end{bmatrix}\right|=\left|\det\begin{bmatrix}2&1\\1&-1\end{bmatrix}\right|=|-3|=3[/tex]
[tex]\mathrm dA=\mathrm dx\,\mathrm dy=|\det\mathbf J|\,\mathrm ds\,\mathrm dt=3\,\mathrm ds\,\mathrm dt[/tex]
The transformation changes the region of integration from a parallelogram to a square with vertices at (0, 0), (2, 0), (2, 2), and (0, 2). Then the integral becomes
[tex]\displaystyle\iint_R(x+y)\,\mathrm dA=3\int_{s=0}^{s=2}\int_{t=0}^{t=2}(2s+t+s-t)\,\mathrm dt\,\mathrm ds[/tex]
[tex]=\displaystyle9\int_{s=0}^{s=2}\int_{t=0}^{t=2}s\,\mathrm dt\,\mathrm ds[/tex]
[tex]=\displaystyle18\int_{s=0}^{s=2}s\,\mathrm ds=36[/tex]
