Answer
The area of $G\left( {\cal R} \right)$ is approximately $0.06$.
Work Step by Step
We have the domain ${\cal R} = \left[ {1,1.2} \right] \times \left[ {3,3.1} \right]$ and the Jacobian of $G$ at $P = \left( {1,3} \right)$, that is, ${\rm{Jac}}\left( G \right)\left( {1,3} \right) = 3$.
So, the area of ${\cal R}$ is ${\rm{Area}}\left( {\cal R} \right) = \left( {0.2} \right)\left( {0.1} \right) = 0.02$.
For ${\cal R}$ is small we can approximate the area of $G\left( {\cal R} \right)$ using Eq. (7):
${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) \approx \left| {{\rm{Jac}}\left( G \right)\left( P \right)} \right|{\rm{Area}}\left( {\cal R} \right)$
${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) \approx \left( 3 \right)\left( {0.02} \right) = 0.06$
So, the area of $G\left( {\cal R} \right)$ is approximately $0.06$.
