Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{3x - 2y}}{\rm{d}}x{\rm{d}}y = \frac{{14}}{{39}}\left( {{{\rm{e}}^{13}} - 1} \right)\left( {1 - {{\rm{e}}^{ - 3}}} \right)$
Work Step by Step
Step 1. Find a linear mapping:
$G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$,
where $A$, $B$, $C$, $D$ are constants, such that $G$ maps a rectangle ${\cal R}$ in the $uv$-plane to the parallelogram ${\cal D}$ in the $xy$-plane as is shown in Figure 7.
In the $uv$-plane, the rectangle is spanned by $\left( {1,0} \right)$ and $\left( {0,1} \right)$. So,
$G\left( {1,0} \right) = \left( {A,B} \right)$ ${\ \ \ }$ and ${\ \ \ }$ $G\left( {0,1} \right) = \left( {C,D} \right)$
Referring to Figure 7, we have the vertices of the parallelogram in the $xy$-plane:
$\left( {0,0} \right)$, ${\ \ }$ $\left( {5,1} \right)$, ${\ \ }$ $\left( {1,3} \right)$, ${\ \ }$ $\left( {6,4} \right)$
From Figure 7, we see that the parallelogram is spanned by the vectors $\left( {5,1} \right)$ and $\left( {1,3} \right)$.
Since $G$ is linear, it maps the rectangle ${\cal R}$ spanned by $\left( {1,0} \right)$ and $\left( {0,1} \right)$ in the $uv$-plane to the parallelogram spanned by the images $G\left( {1,0} \right)$ and $G\left( {0,1} \right)$. Therefore,
$G\left( {1,0} \right) = \left( {A,B} \right) = \left( {5,1} \right)$
$G\left( {0,1} \right) = \left( {C,D} \right) = \left( {1,3} \right)$
Thus, we obtain the linear mapping: $G\left( {u,v} \right) = \left( {5u + v,u + 3v} \right)$.
So, the domain description of ${\cal R}$:
${\cal R} = \left\{ {\left( {u,v} \right)|0 \le u \le 1,0 \le v \le 1} \right\}$
Step 2. Evaluate the Jacobian of $G$:
Using $x = 5u + v$ and $y = u + 3v$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
5&1\\
1&3
\end{array}} \right| = 14$
We have $f\left( {x,y} \right) = {{\rm{e}}^{3x - 2y}}$. So,
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = {{\rm{e}}^{3\left( {5u + v} \right) - 2\left( {u + 3v} \right)}} = {{\rm{e}}^{13u - 3v}}$
Evaluate $f\left( {x,y} \right) = {{\rm{e}}^{3x - 2y}}$ over ${\cal D}$ using the Change of Variables Formula:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{3x - 2y}}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {{\rm{e}}^{13u - 3v}}\left| {Jac\left( G \right)} \right|{\rm{d}}v{\rm{d}}u$
$ = 14\mathop \smallint \limits_{u = 0}^1 \mathop \smallint \limits_{v = 0}^1 {{\rm{e}}^{13u - 3v}}{\rm{d}}v{\rm{d}}u$
$ = 14\left( {\mathop \smallint \limits_{u = 0}^1 {{\rm{e}}^{13u}}{\rm{d}}u} \right)\left( {\mathop \smallint \limits_{v = 0}^1 {{\rm{e}}^{ - 3v}}{\rm{d}}v} \right)$
$ = - \frac{{14}}{{39}}\left( {{{\rm{e}}^{13u}}|_0^1} \right)\left( {{{\rm{e}}^{ - 3v}}|_0^1} \right)$
$ = - \frac{{14}}{{39}}\left( {{{\rm{e}}^{13}} - 1} \right)\left( {{{\rm{e}}^{ - 3}} - 1} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{3x - 2y}}{\rm{d}}x{\rm{d}}y = \frac{{14}}{{39}}\left( {{{\rm{e}}^{13}} - 1} \right)\left( {1 - {{\rm{e}}^{ - 3}}} \right)$.
