Answer
$-3$
Work Step by Step
The region $R$ in the uv plane is defined as:
$R=${$(u,v) | 0 \leq v \leq 1-u, 0\leq u \leq 1$}
Now, $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 2&1\\1&2\end{vmatrix}=4-1=3$
$I=\iint_R (x-3y) dA=\int_0^1 [\int_0^{1-u} (2u+v) -3(u+2v)] (3) dv du$
and
$=3 \int_0^1 \int_0^{1-u} -u-(5v) dv du=3 \int_0^1 [-uv-(\dfrac{5}{2})v^2]_0^{1-u} du$
Hence, we have $\iint_R (x-3y) dA=(3) \int_0^1 -u(1-u)-(\dfrac{5}{2}) (u^2-2u+1) du=3 [-(\dfrac{5}{2})u-(\dfrac{1}{2})u^3+2u^2]_0^1=-3$
