Double Integral Limits for Triangular Region

[arl146]

Homework Statement

function inside is (x+y)^2 * sin(x^2 - y^2)
R is the triangular region w/ vertices (0,0) , (0,2) , (1,1)

x = (u+v)/2
y = (v-u)/2

What are the correct limits ??

The Attempt at a Solution

Also, when plugging in x and y in the function, i ended up getting (v^2)*(sin(uv)). is that right? for the limits, i have no idea, all my attempts failed =/

 

[HallsofIvy]

There are three boundary lines, x= 0, from (0, 0) to (0, 2), y= x, from (0, 0) to (1, 1), and y= 1- x, from (0, 2) to (1, 1). x= (u+ v)/2= 0 gives u+ v= 0 so v= -u. y= x gives (v- u)/2= (u+ v)/2 so ...