Let x=u+v, y=u-v.
If D is the region 0 is less than or equal to y which is less than or equal to x which is less than or equal to 1, what is the region D* that maps onto D?
I know that if you are taking a 1-1 map of D* onto D, then you have the double integral of (u+v) times (u-v), but I'm not quite sure the role that the Jacobian plays in this? Can someone guide me through the steps to this problem. Thank you!
$D=\{(x,y): 0<y<x<1\}$ is the region in red below.
The vertices are $(0,0), (1,1)$ and $(1,0)$
$D^*:$ In u,v space these vertices map to $(0,0), (1,0), (\frac 12, \frac 12)$ respectively.
Plotting this region on the same set of axes, the red region in $x,y$ space maps to the green region in $u,v$ space.
The jacobian $\left|\begin{array}{} \frac{\partial x}{\partial u} & \frac {\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac {\partial y}{\partial v}\end{array}\right| = 2$
$\int_0^1\int_0^x xy \ dy\ dx = \int_0^{\frac 12} \int_v^{1-v} 2 (u^2 - v^2) \ du\ dv$