Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {\left( {x - y} \right)\sin \left( {x + y} \right)} \right)^2}{\rm{d}}x{\rm{d}}y = \frac{{{\pi ^4}}}{3}$
Work Step by Step
From the linear mapping $G\left( {u,v} \right) = \left( {\frac{{u + v}}{2},\frac{{u - v}}{2}} \right)$, we obtain
$x = \frac{{u + v}}{2}$, ${\ \ \ \ \ }$ $y = \frac{{u - v}}{2}$
From these equations, we solve for $u$ and $v$ in terms of $x$ and $y$. So,
$u = x + y$, ${\ \ \ \ \ }$ $v = x - y$
We have the domain ${\cal R}$ in the $xy$-plane defined by the square with vertices $\left( {\pi ,0} \right)$, $\left( {2\pi ,\pi } \right)$, $\left( {\pi ,2\pi } \right)$, and $\left( {0,\pi } \right)$.
Using $u = x + y$, $v = x - y$, we find the corresponding vertices in the $uv$-plane. Let ${{\cal R}_0}$ denote the domain in the $uv$-plane. We list the results in the following table:
$\begin{array}{*{20}{c}}
{{\cal R}{\ }{\rm{vertices}}}&{{{\cal R}_0}{\ }{\rm{vertices}}}\\
{\left( {x,y} \right)}&{\left( {u,v} \right)}\\
{\left( {\pi ,0} \right)}&{\left( {\pi ,\pi } \right)}\\
{\left( {2\pi ,\pi } \right)}&{\left( {3\pi ,\pi } \right)}\\
{\left( {\pi ,2\pi } \right)}&{\left( {3\pi , - \pi } \right)}\\
{\left( {0,\pi } \right)}&{\left( {\pi , - \pi } \right)}
\end{array}$
We sketch the domain and see that ${{\cal R}_0}$ is defined by the rectangle:
${{\cal R}_0} = \left\{ {\left( {u,v} \right)|\pi \le u \le 3\pi , - \pi \le v \le \pi } \right\}$
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{\frac{1}{2}}\\
{\frac{1}{2}}&{ - \frac{1}{2}}
\end{array}} \right| = - \frac{1}{4} - \frac{1}{4} = - \frac{1}{2}$
Now, using $u = x + y$, $v = x - y$, and the Change of Variables Formula, we compute:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {\left( {x - y} \right)\sin \left( {x + y} \right)} \right)^2}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal R}_0}}^{} {\left( {v\sin u} \right)^2}\left| {Jac\left( G \right)} \right|{\rm{d}}v{\rm{d}}u$
$ = \frac{1}{2}\mathop \smallint \limits_{u = \pi }^{3\pi } \mathop \smallint \limits_{v = - \pi }^\pi {v^2}{\sin ^2}u{\rm{d}}v{\rm{d}}u$
(1) ${\ \ \ \ }$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {\left( {x - y} \right)\sin \left( {x + y} \right)} \right)^2}{\rm{d}}x{\rm{d}}y = \frac{1}{2}\left( {\mathop \smallint \limits_{u = \pi }^{3\pi } {{\sin }^2}u{\rm{d}}u} \right)\left( {\mathop \smallint \limits_{v = - \pi }^\pi {v^2}{\rm{d}}v} \right)$
Consider the integral $\mathop \smallint \limits_{u = \pi }^{3\pi } {\sin ^2}u{\rm{d}}u$.
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
$\mathop \smallint \limits_{u = \pi }^{3\pi } {\sin ^2}u{\rm{d}}u = \frac{1}{2}\mathop \smallint \limits_{u = \pi }^{3\pi } \left( {1 - \cos 2u} \right){\rm{d}}u$
$ = \frac{1}{2}\left( {u - \frac{1}{2}\sin 2u} \right)|_\pi ^{3\pi }$
$ = \frac{1}{2}\left( {3\pi - \pi } \right) = \pi $
Substituting the result back in equation (1), we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {\left( {x - y} \right)\sin \left( {x + y} \right)} \right)^2}{\rm{d}}x{\rm{d}}y = \frac{1}{2}\left( \pi \right)\left( {\frac{1}{3}{v^3}|_{ - \pi }^\pi } \right)$
$ = \frac{1}{2}\left( \pi \right)\left( {\frac{2}{3}{\pi ^3}} \right) = \frac{{{\pi ^4}}}{3}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {\left( {\left( {x - y} \right)\sin \left( {x + y} \right)} \right)^2}{\rm{d}}x{\rm{d}}y = \frac{{{\pi ^4}}}{3}$.
