Answer
Please see the figure attached.
(a) ${\rm{Jac}}\left( G \right) = {\rm{Jac}}{\left( F \right)^{ - 1}} = \frac{1}{3}$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{x + y}}{\rm{d}}x{\rm{d}}y \simeq 86.47$
Work Step by Step
(a) We sketch the domain:
${\cal D} = \left\{ {\left( {x,y} \right)|1 \le x + y \le 4, - 4 \le y - 2x \le 1} \right\}$
We have the map $F\left( {x,y} \right) = \left( {x + y,y - 2x} \right)$.
So, $u=x+y$ and $v=y-2x$.
Evaluate the Jacobian of $F$:
${\rm{Jac}}\left( F \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}\\
{\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&1\\
{ - 2}&1
\end{array}} \right| = 3$
By Eq. (14),
${\rm{Jac}}\left( G \right) = {\rm{Jac}}{\left( F \right)^{ - 1}} = \frac{1}{3}$
(b) From the description of ${\cal D}$, we obtain:
$1 \le x + y \le 4$, ${\ \ \ \ \ }$ $ - 4 \le y - 2x \le 1$
Let ${\cal R}$ denote the domain in the $uv$-plane such that $F:{\cal D} \to {\cal R}$.
Since $u=x+y$ and $v=y-2x$, so the domain description of ${\cal R}$:
$1 \le u \le 4$, ${\ \ \ \ \ }$ $ - 4 \le v \le 1$
Using the Change of Variables Formula, we compute:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{x + y}}{\rm{d}}x{\rm{d}}y = \frac{1}{3}\mathop \smallint \limits_{v = - 4}^1 \mathop \smallint \limits_{u = 1}^4 {{\rm{e}}^u}{\rm{d}}u{\rm{d}}v$
$ = \frac{1}{3}\left( {\mathop \smallint \limits_{v = - 4}^1 {\rm{d}}v} \right)\left( {\mathop \smallint \limits_{u = 1}^4 {{\rm{e}}^u}{\rm{d}}u} \right)$
$ = \frac{1}{3}\left( 5 \right)\left( {{{\rm{e}}^u}|_1^4} \right) = \frac{5}{3}\left( {{{\rm{e}}^4} - {\rm{e}}} \right) \simeq 86.47$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{x + y}}{\rm{d}}x{\rm{d}}y \simeq 86.47$.
