I'm trying to evaluate this double integral $\iint_R(x-3y)\,dA$ where $R$ is a triangular region with the vertices $(0,0)$, $(2,1)$, and $(1,2)$ and $x = 2u+v$ and $y=u+2v$. Knowing that $y = 2x$, $u+2v=4u+2v$, so $u=0$. And $y=\frac<1>x$ so $u+2v=u+\frac<1>v$ so $v=0$. But the last relationship, $y=-x+3$, giving us $v=3-3u$. I don't know where to go from here regarding finding the bounds for the integrals.The jacobian here is 3.
$\begingroup$ This is an affine (linear, even) mapping from the $x$-$y$ plane to the $u$-$v$ plane. So a triangle will map to a triangle. If you figure out where the three vertices map to, that'll get you the vertices of the triangular region in the $u$-$v$ plane. $\endgroup$
– user137731 Commented Dec 10, 2015 at 20:04 $\begingroup$ so how do you do that? $\endgroup$ Commented Dec 10, 2015 at 20:11$\begingroup$ Plug your three points into $x$ and $y$ in your two equations and solve for $u$ and $v$. For instance, plugging $(0,0)$ in gets you $$\begin