Answer
We use the change of variables: $u=x+y$ and $v=x-y$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {x + y} \right)^2}{{\rm{e}}^{{x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y = \frac{2}{{\rm{e}}}$
Work Step by Step
To simplify the integrand, we choose the mapping $F$: $u=x+y$ and $v=x-y$.
Using $u=x+y$ and $v=x-y$ we find the corresponding vertices of $\left( {1,0} \right)$, $\left( {0,1} \right)$, $\left( { - 1,0} \right)$, $\left( {0, - 1} \right)$ in the $uv$-plane. The results are
$\left( {1,1} \right)$, $\left( {1, - 1} \right)$, $\left( { - 1, - 1} \right)$, $\left( { - 1,1} \right)$
Since $F$ is linear, the domain in the $uv$-plane is a square with vertices $\left( {1,1} \right)$, $\left( {1, - 1} \right)$, $\left( { - 1, - 1} \right)$, $\left( { - 1,1} \right)$. We denote this square ${{\cal R}_0}$.
Subtracting $v$ from $u$ gives
$u - v = 2y$, ${\ \ \ \ \ }$ $y = \frac{{u - v}}{2}$
Adding $v$ to $u$ gives
$u + v = 2x$, ${\ \ \ \ \ }$ $x = \frac{{u + v}}{2}$
So, the mapping $G\left( {u,v} \right) = \left( {\frac{{u + v}}{2},\frac{{u - v}}{2}} \right)$ maps the square ${{\cal R}_0}$ to the square ${\cal R}$ in the $xy$-plane.
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{\frac{1}{2}}\\
{\frac{1}{2}}&{ - \frac{1}{2}}
\end{array}} \right| = - \frac{1}{2}$
Since $x = \frac{{u + v}}{2}$ and $y = \frac{{u - v}}{2}$, we have
$f\left( {x,y} \right) = {\left( {x + y} \right)^2}{{\rm{e}}^{{x^2} - {y^2}}}$
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = {u^2}{{\rm{e}}^{uv}}$
Using the Change of Variables Formula, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal R}_0}}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}v{\rm{d}}u$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {x + y} \right)^2}{{\rm{e}}^{{x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y = \frac{1}{2}\mathop \smallint \limits_{u = - 1}^1 \mathop \smallint \limits_{v = - 1}^1 {u^2}{{\rm{e}}^{uv}}{\rm{d}}v{\rm{d}}u$
$ = \frac{1}{2}\mathop \smallint \limits_{u = - 1}^1 {u^2}\left( {\frac{1}{u}{{\rm{e}}^{uv}}|_{ - 1}^1} \right){\rm{d}}u$
$ = \frac{1}{2}\mathop \smallint \limits_{u = - 1}^1 u\left( {{{\rm{e}}^u} - {{\rm{e}}^{ - u}}} \right){\rm{d}}u$
$ = \frac{1}{2}\mathop \smallint \limits_{u = - 1}^1 u{{\rm{e}}^u}{\rm{d}}u - \frac{1}{2}\mathop \smallint \limits_{u = - 1}^1 u{{\rm{e}}^{ - u}}{\rm{d}}u$
Using Integration by Parts Formula (Section 8.1),
$\smallint s{\rm{d}}t = st - \smallint t{\rm{d}}s$
we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {x + y} \right)^2}{{\rm{e}}^{{x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y = \frac{1}{2}\left( {u{{\rm{e}}^u}|_{ - 1}^1 - \mathop \smallint \limits_{u = - 1}^1 {{\rm{e}}^u}{\rm{d}}u} \right)$
$ - \frac{1}{2}\left( { - \left( {u{{\rm{e}}^{ - u}}|_{ - 1}^1} \right) + \mathop \smallint \limits_{u = - 1}^1 {{\rm{e}}^{ - u}}{\rm{d}}u} \right)$
$ = \frac{1}{2}\left( {{\rm{e}} + {{\rm{e}}^{ - 1}} - \left( {{{\rm{e}}^u}|_{ - 1}^1} \right)} \right) - \frac{1}{2}\left( { - \left( {{{\rm{e}}^{ - 1}} + {\rm{e}}} \right) - \left( {{{\rm{e}}^{ - u}}|_{ - 1}^1} \right)} \right)$
$ = \frac{1}{2}\left( {{\rm{e}} + {{\rm{e}}^{ - 1}} - {\rm{e}} + {{\rm{e}}^{ - 1}}} \right) - \frac{1}{2}\left( { - {{\rm{e}}^{ - 1}} - {\rm{e}} - {{\rm{e}}^{ - 1}} + {\rm{e}}} \right)$
$ = {{\rm{e}}^{ - 1}} + {{\rm{e}}^{ - 1}} = 2{{\rm{e}}^{ - 1}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {x + y} \right)^2}{{\rm{e}}^{{x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y = \frac{2}{{\rm{e}}}$.
