I have the integral $$\int\int_R\left(2x+y\right)dA$$ Where $R$ is the region bounded by $$x+y=-1, x+y = 3, 2x=y,2x-4=y$$ So my first though was drawing the region, which gave me this odd region, so I thought I'd probably need to apply a transformation. My problem in this process is going about finding the appropriate transformation.
The transformation ended up being $$u = x+y, v = 2x-y$$ Which led to me getting $$u+v=3x\rightarrow x = \frac{1}{3}(u+v)$$ Using that value of $x$ $$u = \frac{1}{3}(u+v)+y\rightarrow y = \frac{1}{3}(2u-v)$$ So that process gave me $x = \frac{1}{3}(u+v)$ and $y = \frac{1}{3}(2u-v)$.
And after applying the constraints of the region to see my bounds of $u$ and $v$, and finding the Jacobian, simplified the integral greatly.
So my question is, is there a process for finding the transformation needed to simplify, or is it all trial and error, which is how I solved this problem.
Thanks for all the help in advance.
Your procedure generalizes to linear transformations aand regions enclosed by parallelograms. You map mapped the lines with slope -1 to vertical lines (in the uv plane) and lines of slope 2 to horizontal lines. In general, the transformations are ad hoc. However, there a coordinate systems and transformation that are adapted to various type of regions. For example, the polar coordinate conversion for $\int\int_D xy \ dx \, dy,$ where $D$ is the annulus $1\leq x^2 + y^2 \leq 4,$ is easily converted to an integral in "box" in $(r,\theta)$ space: $$ \int^{2\pi}_0 \int^2_1 r^3 \cos \theta \sin \theta \ dr \ d\theta $$