Answer
Please see the figure attached.
(a) the image of ${\cal D}$ under $F$ is the rectangle ${\cal R}$ in the $uv$-plane:
${\cal R} = \left\{ {\left( {u,v} \right)|\frac{1}{2} \le u \le 2,\frac{1}{2} \le v \le 2} \right\}$
(b) we show that
$\left| {{\rm{Jac}}\left( G \right)} \right| = \frac{1}{{2\left| v \right|}}$
(c) we prove the formula:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {\frac{y}{x}} \right){\rm{d}}x{\rm{d}}y = \frac{3}{4}\mathop \smallint \limits_{1/2}^2 \frac{{f\left( v \right){\rm{d}}v}}{v}$
(d) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{y}{x}{{\rm{e}}^{y/x}}{\rm{d}}x{\rm{d}}y \simeq 4.31$
Work Step by Step
Please see the figure attached for the region ${\cal D}$ bounded by the curves $y = 2/x$, $y = 1/\left( {2x} \right)$, $y = 2x$, $y = x/2$ in the first quadrant.
From these curves, we obtain the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {x,y} \right)|\frac{1}{2} \le xy \le 2,\frac{1}{2} \le \frac{y}{x} \le 2} \right\}$
(a) We have the map $F$: $u = xy$, $v = y/x$ from the $xy$-plane to the $uv$-plane. So, we can write
$F\left( {x,y} \right) = \left( {xy,\frac{y}{x}} \right)$
Since $u = xy$, $v = y/x$, so the inequalities in ${\cal D}$ become
$\frac{1}{2} \le u \le 2,\frac{1}{2} \le v \le 2$
Let ${\cal R}$ denote the image of ${\cal D}$ under $F$. Thus, the description of ${\cal R}$:
${\cal R} = \left\{ {\left( {u,v} \right)|\frac{1}{2} \le u \le 2,\frac{1}{2} \le v \le 2} \right\}$
So, the image of ${\cal D}$ under $F$ is the rectangle ${\cal R}$ in the $uv$-plane.
(b) In part (a), we obtain
$F\left( {x,y} \right) = \left( {xy,\frac{y}{x}} \right)$
Evaluate the Jacobian of $F$:
${\rm{Jac}}\left( F \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}\\
{\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
y&x\\
{ - \frac{y}{{{x^2}}}}&{\frac{1}{x}}
\end{array}} \right| = \frac{y}{x} + \frac{y}{x} = 2\frac{y}{x}$
Using Eq. (14) in Section 16.6:
${\rm{Jac}}\left( G \right) = {\rm{Jac}}{\left( F \right)^{ - 1}}$, ${\ \ \ \ }$ where $G = {F^{ - 1}}$
we get ${\rm{Jac}}\left( G \right) = \frac{x}{{2y}}$.
Since $v = y/x$, so $\left| {{\rm{Jac}}\left( G \right)} \right| = \frac{1}{{2\left| v \right|}}$.
(c) Using $v = y/x$, and the Change of Variables Formula in Section 16.6, we evaluate $f\left( {\frac{y}{x}} \right)$ over ${\cal D}$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {\frac{y}{x}} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( v \right)\left| {{\rm{Jac}}\left( G \right)} \right|{\rm{d}}v{\rm{d}}u$
Using $\left| {{\rm{Jac}}\left( G \right)} \right| = \frac{1}{{2\left| v \right|}}$ from part (b), we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {\frac{y}{x}} \right){\rm{d}}x{\rm{d}}y = \frac{1}{2}\mathop \smallint \limits_{u = 1/2}^2 \mathop \smallint \limits_{v = 1/2}^2 \frac{{f\left( v \right)}}{{\left| v \right|}}{\rm{d}}v{\rm{d}}u$
$ = \frac{1}{2}\left( {\mathop \smallint \limits_{u = 1/2}^2 {\rm{d}}u} \right)\left( {\mathop \smallint \limits_{v = 1/2}^2 \frac{{f\left( v \right)}}{{\left| v \right|}}{\rm{d}}v} \right)$
$ = \frac{1}{2}\left( {\frac{3}{2}} \right)\left( {\mathop \smallint \limits_{v = 1/2}^2 \frac{{f\left( v \right)}}{{\left| v \right|}}{\rm{d}}v} \right)$
Since $\frac{1}{2} \le v \le 2$ ($v$ is positive), we can drop the absolute value sign, and hence
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {\frac{y}{x}} \right){\rm{d}}x{\rm{d}}y = \frac{3}{4}\mathop \smallint \limits_{1/2}^2 \frac{{f\left( v \right){\rm{d}}v}}{v}$
(d) Write $f\left( {\frac{y}{x}} \right) = \frac{y}{x}{{\rm{e}}^{y/x}}$.
Since $v = y/x$, so $f\left( v \right) = v{{\rm{e}}^v}$.
Using the result in part (c):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {\frac{y}{x}} \right){\rm{d}}x{\rm{d}}y = \frac{3}{4}\mathop \smallint \limits_{1/2}^2 \frac{{f\left( v \right){\rm{d}}v}}{v}$
we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{{y{{\rm{e}}^{y/x}}}}{x}{\rm{d}}x{\rm{d}}y$:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{y}{x}{{\rm{e}}^{y/x}}{\rm{d}}x{\rm{d}}y = \frac{3}{4}\mathop \smallint \limits_{1/2}^2 \frac{{v{{\rm{e}}^v}{\rm{d}}v}}{v}$
$ = \frac{3}{4}\left( {{{\rm{e}}^v}|_{1/2}^2} \right) = \frac{3}{4}\left( {{{\rm{e}}^2} - {{\rm{e}}^{1/2}}} \right) \simeq 4.31$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \frac{y}{x}{{\rm{e}}^{y/x}}{\rm{d}}x{\rm{d}}y \simeq 4.31$.
