Answer
(a) ${\rm{Jac}}\left( G \right) = 3$
(b) Please see the figure attached.
(c) ${\rm{Area}}\left( {\cal D} \right) = 27$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}x{\rm{d}}y \simeq 174.67$
Work Step by Step
We have the domain ${\cal R} = \left[ {1,4} \right] \times \left[ {1,4} \right]$ in the $uv$-plane. Let ${\cal D} = G\left( {\cal R} \right)$.
(a) Evaluate the Jacobian of $G\left( {u,v} \right) = \left( {\frac{{{u^2}}}{v},\frac{{{v^2}}}{u}} \right)$.
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\frac{{2u}}{v}}&{ - \frac{{{u^2}}}{{{v^2}}}}\\
{ - \frac{{{v^2}}}{{{u^2}}}}&{\frac{{2v}}{u}}
\end{array}} \right| = 4 - 1 = 3$
(b) We have ${\cal R} = \left\{ {\left( {u,v} \right)|1 \le u \le 4,1 \le v \le 4} \right\}$.
Step 1. Find the image of the vertical lines $u=1$ and $u=4$
For $u=1$:
$G\left( {1,v} \right) = \left( {\frac{1}{v},{v^2}} \right)$
For $1 \le v \le 4$, this is a curve parametrized by
${f_1}\left( t \right) = \left( {\frac{1}{t},{t^2}} \right)$, ${\ \ \ \ }$ for $1 \le t \le 4$
For $u=4$:
$G\left( {4,v} \right) = \left( {\frac{{16}}{v},\frac{{{v^2}}}{4}} \right)$
For $1 \le v \le 4$, this is a curve parametrized by
${f_2}\left( t \right) = \left( {\frac{{16}}{t},\frac{{{t^2}}}{4}} \right)$, ${\ \ \ \ }$ for $1 \le t \le 4$
Thus, the region ${\cal D}$ is bounded by ${f_1}\left( t \right)$ and ${f_2}\left( t \right)$ corresponding to the vertical lines $u=1$ and $u=4$, for $1 \le v \le 4$.
Step 2. Find the image of the horizontal lines $v=1$ and $v=4$
For $v=1$:
$G\left( {u,1} \right) = \left( {{u^2},\frac{1}{u}} \right)$
For $1 \le u \le 4$, this is a curve parametrized by
${g_1}\left( t \right) = \left( {{t^2},\frac{1}{t}} \right)$, ${\ \ \ \ }$ for $1 \le t \le 4$
For $v=4$:
$G\left( {u,4} \right) = \left( {\frac{{{u^2}}}{4},\frac{{16}}{u}} \right)$
For $1 \le u \le 4$, this is a curve parametrized by
${g_2}\left( t \right) = \left( {\frac{{{t^2}}}{4},\frac{{16}}{t}} \right)$, ${\ \ \ \ }$ for $1 \le t \le 4$
Thus, the region ${\cal D}$ is bounded by ${g_1}\left( t \right)$ and ${g_2}\left( t \right)$ corresponding to the horizontal lines $v=1$ and $v=4$, for $1 \le u \le 4$.
Using these parametric equations, we sketch ${\cal D}$. Please see the figure attached.
(c)
1. Compute ${\rm{Area}}\left( {\cal D} \right)$:
${\rm{Area}}\left( {\cal D} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$ = 3\mathop \smallint \limits_{v = 1}^4 \mathop \smallint \limits_{u = 1}^4 {\rm{d}}u{\rm{d}}v$
$ = 3\left( {\mathop \smallint \limits_{v = 1}^4 {\rm{d}}v} \right)\left( {\mathop \smallint \limits_{u = 1}^4 {\rm{d}}u} \right) = 27$
So, ${\rm{Area}}\left( {\cal D} \right) = 27$.
2. Compute $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$, where $f\left( {x,y} \right) = x + y$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}x{\rm{d}}y = 3\mathop \smallint \limits_{v = 1}^4 \mathop \smallint \limits_{u = 1}^4 \left( {\frac{{{u^2}}}{v} + \frac{{{v^2}}}{u}} \right){\rm{d}}u{\rm{d}}v$
$ = 3\mathop \smallint \limits_{v = 1}^4 \left( {\left( {\frac{{{u^3}}}{{3v}} + {v^2}\ln u} \right)|_1^4} \right){\rm{d}}v$
$ = 3\mathop \smallint \limits_{v = 1}^4 \left( {\frac{{64}}{{3v}} + {v^2}\ln 4 - \frac{1}{{3v}}} \right){\rm{d}}v$
$ = 3\mathop \smallint \limits_{v = 1}^4 \left( {\frac{{21}}{v} + {v^2}\ln 4} \right){\rm{d}}v$
$ = 3\left( {\left( {21\ln v + \frac{1}{3}{v^3}\ln 4} \right)|_1^4} \right)$
$ = 3\left( {21\ln 4 + \frac{{64}}{3}\ln 4 - \frac{1}{3}\ln 4} \right)$
$ = 3\left( {21\ln 4 + 21\ln 4} \right) = 126\ln 4 \simeq 174.67$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + y} \right){\rm{d}}x{\rm{d}}y \simeq 174.67$.
