Answer
(a) we show that the image of the horizontal line $v=c$ is summarized in the following table:
$\begin{array}{*{20}{c}}
{uv{\rm{ - plane}}}&G&{xy{\rm{ - plane}}}\\
{v = c \ne 1}& \to &{y = \frac{c}{{1 - c}}x}\\
{v = c = 1}& \to &{y{\rm{ - axis}}}
\end{array}$
(b) the image of the vertical line $u=c$ is summarized in the following table:
$\begin{array}{*{20}{c}}
{uv{\rm{ - plane}}}&G&{xy{\rm{ - plane}}}\\
{u = c \ne 0}& \to &{y = c - x}\\
{u = c = 0}& \to &{\left( {0,0} \right)}
\end{array}$
(c) ${\rm{Jac}}\left( G \right) = u$
(d) Using geometry, we obtain ${\rm{Area}}\left( {\cal D} \right) = \frac{1}{2}\left( {{b^2} - {a^2}} \right)$
Using the Change of Variable Formula we obtain:
${\rm{Area}}\left( {\cal D} \right) = \frac{1}{2}\left( {{b^2} - {a^2}} \right)$
The two results agree.
(e) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}x{\rm{d}}y = \frac{1}{{24}}\left( {{b^4} - {a^4}} \right)$
Work Step by Step
We have the mapping $G\left( {u,v} \right) = \left( {u - uv,uv} \right)$.
(a) The image of the horizontal line $v=c$ is given by
(1) ${\ \ \ \ \ }$ $G\left( {u,c} \right) = \left( {u - cu,cu} \right)$
So, $x = u\left( {1 - c} \right)$ and $y = cu$.
1. Case $c \ne 1$
In this case, we can write $u = \frac{x}{{1 - c}}$. Substituting it in $y$ gives $y = \frac{c}{{1 - c}}x$.
Thus, the image of the horizontal line $v=c$ for $c \ne 1$ is the line $y = \frac{c}{{1 - c}}x$.
2. Case $c=1$
In this case, equation (1) becomes $G\left( {u,1} \right) = \left( {0,u} \right)$.
So, $x=0$ and $y=u$. Thus, the image of the horizontal line $v=c$ for $c=1$ is the $y$-axis.
In summary, the image of the horizontal line $v=c$:
$\begin{array}{*{20}{c}}
{uv{\rm{ - plane}}}&G&{xy{\rm{ - plane}}}\\
{v = c \ne 1}& \to &{y = \frac{c}{{1 - c}}x}\\
{v = c = 1}& \to &{y{\rm{ - axis}}}
\end{array}$
Notice that for $v=c=0$, we obtain $y=0$, which is the $x$-axis.
(b) The image of the vertical line $u=c$ is given by
$G\left( {c,v} \right) = \left( {c - cv,cv} \right)$
1. Case $c \ne 0$
So, $x = c - cv$ and $y = cv$. It follows that $y = c - x$.
Thus, the image of the vertical lines $u=c$ for $c \ne 0$ is the lines $y = c - x$.
2. Case $c=0$
The image of the vertical lines $u=c$ for $c=0$ is the origin in the $xy$-plane.
In summary, the image of the vertical line $u=c$:
$\begin{array}{*{20}{c}}
{uv{\rm{ - plane}}}&G&{xy{\rm{ - plane}}}\\
{u = c \ne 0}& \to &{y = c - x}\\
{u = c = 0}& \to &{\left( {0,0} \right)}
\end{array}$
(c) From the map $G\left( {u,v} \right) = \left( {u - uv,uv} \right)$, we obtain
$x\left( {u,v} \right) = u - uv$, ${\ \ \ \ \ }$ $y\left( {u,v} \right) = uv$
Compute the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{1 - v}&{ - u}\\
v&u
\end{array}} \right| = u\left( {1 - v} \right) + uv = u$
(d) We see from Figure 14, the area of the shaded region ${\cal D}$ is
${\rm{Area}}\left( {\cal D} \right) = \frac{1}{2}\left( b \right)\left( b \right) - \frac{1}{2}\left( a \right)\left( a \right) = \frac{1}{2}\left( {{b^2} - {a^2}} \right)$
Notice that ${\cal D}$ is bounded by the $x$-axis ($y=0$), the $y$-axis ($x=0$), $y=a-x$, and $y=b-x$.
Using our previous results in part (a), we see that the $x$-axis ($y=0$) and the $y$-axis ($x=0$) correspond to the horizontal lines $v=0$ and $v=1$ in the $uv$-plane, respectively.
Whereas, the results in part (b) tells us that $y=a-x$ and $y=b-x$ correspond to the vertical lines $u=a$ and $u=b$ in the $uv$-plane, respectively.
Therefore, the domain ${{\cal D}_0}$ in the $uv$-plane corresponds to ${\cal D}$ is given by
${{\cal D}_0} = \left\{ {\left( {u,v} \right)|a \le u \le b,0 \le v \le 1} \right\}$
Using ${\rm{Jac}}\left( G \right) = u$ from part (c) and the General Change of Variables Formula, we evaluate the area of ${\cal D}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} \left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = a}^b u{\rm{d}}u{\rm{d}}v$
$ = \frac{1}{2}\mathop \smallint \limits_{v = 0}^1 \left( {{u^2}|_a^b} \right){\rm{d}}v$
$ = \frac{1}{2}\left( {{b^2} - {a^2}} \right)\mathop \smallint \limits_{v = 0}^1 {\rm{d}}v = \frac{1}{2}\left( {{b^2} - {a^2}} \right)$
Thus, ${\rm{Area}}\left( {\cal D} \right) = \frac{1}{2}\left( {{b^2} - {a^2}} \right)$. The two results agree.
(e) We have $f\left( {x,y} \right) = xy$. So,
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = \left( {u - uv} \right)\left( {uv} \right) = {u^2}v\left( {1 - v} \right)$
Using the General Change of Variables Formula:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} {u^2}v\left( {1 - v} \right)Jac\left( G \right){\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = a}^b {u^3}v\left( {1 - v} \right){\rm{d}}u{\rm{d}}v$
$ = \left( {\mathop \smallint \limits_{v = 0}^1 v\left( {1 - v} \right){\rm{d}}v} \right)\left( {\mathop \smallint \limits_{u = a}^b {u^3}{\rm{d}}u} \right)$
$ = \left( {\left( {\frac{1}{2}{v^2} - \frac{1}{3}{v^3}} \right)|_0^1} \right)\left( {\frac{1}{4}{u^4}|_a^b} \right)$
$ = \frac{1}{{24}}\left( {{b^4} - {a^4}} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} xy{\rm{d}}x{\rm{d}}y = \frac{1}{{24}}\left( {{b^4} - {a^4}} \right)$.
