Answer
(a) we show that the mapping $u = xy$, $v = x - y$ maps ${\cal D}$ to the rectangle ${\cal R} = \left[ {2,4} \right] \times \left[ {0,3} \right]$.
(b) $\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = - \frac{1}{{x + y}}$
(c) we show that $I$ is equal to the integral of $f\left( {u,v} \right) = v$ over ${\cal R}$.
$I = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} v{\rm{d}}u{\rm{d}}v$
$I=9$
Work Step by Step
(a) From the domain description of ${\cal D}$, we obtain:
$2 \le xy \le 4$, ${\ \ \ \ }$ $0 \le x - y \le 3$, ${\ \ \ \ }$ $x \ge 0$, ${\ \ \ \ }$ $y \ge 0$
Since $u = xy$, $v = x - y$, so
$2 \le u \le 4$, ${\ \ \ \ \ }$ $0 \le v \le 3$
Hence, the mapping $u = xy$, $v = x - y$ maps ${\cal D}$ to the rectangle ${\cal R} = \left[ {2,4} \right] \times \left[ {0,3} \right]$.
(b) Denote the mapping $u = xy$, $v = x - y$:
$F\left( {x,y} \right) = \left( {xy,x - y} \right)$
Evaluate the Jacobian of $F$:
${\rm{Jac}}\left( F \right) = \frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}} = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}\\
{\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
y&x\\
1&{ - 1}
\end{array}} \right| = - \left( {x + y} \right)$
By Eq. (14),
${\rm{Jac}}\left( G \right) = {\rm{Jac}}{\left( F \right)^{ - 1}} = - \frac{1}{{x + y}}$
So, ${\rm{Jac}}\left( G \right) = \frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = - \frac{1}{{x + y}}$.
(c) Using the Change of Variables Formula, write
$I = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} - {y^2}} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {{x^2} - {y^2}} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {{x^2} - {y^2}} \right)\left( {\frac{1}{{x + y}}} \right){\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {x - y} \right){\rm{d}}u{\rm{d}}v$
Since $v = x - y$, so $I = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} v{\rm{d}}u{\rm{d}}v$.
This implies that $I$ is equal to the integral of $f\left( {u,v} \right) = v$ over ${\cal R}$.
Evaluate $I$:
$I = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{x^2} - {y^2}} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} v{\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{v = 0}^3 \mathop \smallint \limits_{u = 2}^4 v{\rm{d}}u{\rm{d}}v$
$ = \left( {\mathop \smallint \limits_{v = 0}^3 v{\rm{d}}v} \right)\left( {\mathop \smallint \limits_{u = 2}^4 {\rm{d}}u} \right) = \frac{1}{2}\left( {{v^2}|_0^3} \right)\left( 2 \right) = 9$
So, $I=9$.
