Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}x{\rm{d}}y = \frac{{21}}{2}$
Work Step by Step
We have the mapping $G\left( {u,v} \right) = \left( {\frac{u}{{v + 1}},\frac{{uv}}{{v + 1}}} \right)$.
From Figure 17, we see that ${\cal D}$ is bounded by the lines $y=x$, $y=2x$, $y=3-x$, and $y=6-x$.
1. The image of the vertical line $u=c$ is given by
$G\left( {c,v} \right) = \left( {\frac{c}{{v + 1}},\frac{{cv}}{{v + 1}}} \right)$
So, $x = \frac{c}{{v + 1}}$ and $y = \frac{{cv}}{{v + 1}}$.
From the first equation, we obtain
$v = \frac{c}{x} - 1 = \frac{{c - x}}{x}$
Substituting $v = \frac{{c - x}}{x}$ in $y = \frac{{cv}}{{v + 1}}$ gives
$y = \left( {\frac{c}{{\frac{{c - x}}{x} + 1}}} \right)\left( {\frac{{c - x}}{x}} \right)$
$y = \left( {\frac{{cx}}{c}} \right)\left( {\frac{{c - x}}{x}} \right) = c - x$
So, the image of the vertical line $u=c$ is the line $y=c-x$ in the $xy$-plane.
Thus, the two lines $y=3-x$ and $y=6-x$ correspond to $u=3$ and $u=6$, respectively. So, the range of $u$ is $3 \le u \le 6$.
2. The image of the horizontal line $v=c$ is given by
$G\left( {u,c} \right) = \left( {\frac{u}{{c + 1}},\frac{{cu}}{{c + 1}}} \right)$
So, $x = \frac{u}{{c + 1}}$ and $y = \frac{{cu}}{{c + 1}}$. It follows that $y = cx$.
Thus, the two lines $y=x$ and $y=2x$ correspond to $v=1$ and $v=2$, respectively. So, the range of $v$ is $1 \le v \le 2$.
So, the domain in the $uv$-plane is ${\cal R} = \left\{ {\left( {u,v} \right)|3 \le u \le 6,1 \le v \le 2} \right\}$.
Evaluate the Jacobian of $G\left( {u,v} \right) = \left( {\frac{u}{{v + 1}},\frac{{uv}}{{v + 1}}} \right)$:
$Jac\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\frac{1}{{v + 1}}}&{ - \frac{u}{{{{\left( {v + 1} \right)}^2}}}}\\
{\frac{v}{{v + 1}}}&{\frac{u}{{{{\left( {v + 1} \right)}^2}}}}
\end{array}} \right| = \frac{u}{{{{\left( {v + 1} \right)}^3}}} + \frac{{uv}}{{{{\left( {v + 1} \right)}^3}}} = \frac{{u\left( {v + 1} \right)}}{{{{\left( {v + 1} \right)}^3}}}$
$Jac\left( G \right) = \frac{u}{{{{\left( {v + 1} \right)}^2}}}$
Write
$f\left( {x,y} \right) = x + y$
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = \frac{u}{{v + 1}} + \frac{{uv}}{{v + 1}} = \frac{{u\left( {v + 1} \right)}}{{v + 1}} = u$
Using the Change of Variables Formula, we compute
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{v = 1}^2 \mathop \smallint \limits_{u = 3}^6 \frac{{{u^2}}}{{{{\left( {v + 1} \right)}^2}}}{\rm{d}}u{\rm{d}}v$
$ = \left( {\mathop \smallint \limits_{v = 1}^2 \frac{1}{{{{\left( {v + 1} \right)}^2}}}{\rm{d}}v} \right)\left( {\mathop \smallint \limits_{u = 3}^6 {u^2}{\rm{d}}u} \right)$
$ = - \frac{1}{3}\left( {\frac{1}{{v + 1}}|_1^2} \right)\left( {{u^3}|_3^6} \right)$
$ = - \frac{1}{3}\left( {\frac{1}{3} - \frac{1}{2}} \right)\left( {216 - 27} \right) = \frac{{21}}{2}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}x{\rm{d}}y = \frac{{21}}{2}$.
