Answer
$ \dfrac{e^6}{4}-\dfrac{7}{4}$
Work Step by Step
Plug $u=x-y$ and $v=x+y$
Here, we have: $Jacobian (x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=1 \cdot (1) -(-1)(1)=2$
Then, we have $Jacobian(u,v)=\dfrac{1}{2}$
Now, $\iint_R (x+y) e^{x^2-y^2} dA=\iint_{R} (x+y) e^{(x+y)(x-y)} dA$
This implies that
$=(\dfrac{1}{2}) \int_0^{3} \int_0^2 v e^{uv} du dv$
$=(\dfrac{1}{2}) \int_0^3 (e^{2v}-1) dv$
and $\iint_R (x+y) e^{x^2-y^2} dA=(\dfrac{1}{2}) [(\dfrac{1}{2}) (e^{2v}-v)]_0^3 =(\dfrac{1}{2}) [\dfrac{e^6}{2}-3-\dfrac{1}{2}]$
Hence,
$\iint_R (x+y) e^{x^2-y^2} dA= \dfrac{e^6}{4}-\dfrac{7}{4}$
