## Thomas' Calculus 13th Edition

$-3$
$\int \int_R 2(x-y)dx dy$ =$\int \int_G (-2v)|\frac{\sigma(x,y)}{\sigma(u,v)}|$ du dv =$\int \int_G(-2v) du dv$ =$\int \int_G(-2v)$ du dv =$\int^1_0 \int^{3-3v}_{-3v}(-2v)$ du dv =$\int^1_0 (-2v)(3-3v+3v)dv$ =$\int^1_0 (-6v)dv$ =$[-3v^2]^1_0$ =-3