Answer
(a) ${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) = 105$
(b) ${\rm{Area}}\left( {G\left( {\cal R} \right)} \right)= 126$
Work Step by Step
We have the mapping $G\left( {u,v} \right) = \left( {3u + v,u - 2v} \right)$.
So, $x\left( {u,v} \right) = 3u + v$ and $y\left( {u,v} \right) = u - 2v$.
Compute the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
3&1\\
1&{ - 2}
\end{array}} \right| = - 7$
Since $G$ is a linear map, we use Eq. (6) of Theorem 1 to evaluate the area of $G\left( {\cal R} \right)$:
${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {\cal R} \right)$
(a) We have ${\cal R} = \left[ {0,3} \right] \times \left[ {0,5} \right]$. So, ${\rm{Area}}\left( {\cal R} \right) = \left( 3 \right)\left( 5 \right) = 15$.
${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) = \left( 7 \right)\left( {15} \right) = 105$
(b) We have ${\cal R} = \left[ {2,5} \right] \times \left[ {1,7} \right]$. So, ${\rm{Area}}\left( {\cal R} \right) = \left( 3 \right)\left( 6 \right) = 18$.
${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) = \left( 7 \right)\left( {18} \right) = 126$
