Answer
The area of the image of ${\cal R}$ is $156$.
Work Step by Step
A linear mapping has the form:
$G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$
where $A$, $B$, $C$, $D$ are constants.
Since $G$ maps the unit square to the parallelogram in the $xy$-plane spanned by the vectors $\left( {3, - 1} \right)$ and $\left( {1,4} \right)$, so
$G\left( {1,0} \right) = \left( {A,B} \right) = \left( {3, - 1} \right)$
$G\left( {0,1} \right) = \left( {C,D} \right) = \left( {1,4} \right)$
Thus, we obtain the linear mapping:
$G\left( {u,v} \right) = \left( {3u + v, - u + 4v} \right)$
So,
$x = 3u + v$, ${\ \ \ \ \ }$ $y = - u + 4v$
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
3&1\\
{ - 1}&4
\end{array}} \right| = 13$
Now, we have the rectangle ${\cal R} = \left[ {0,4} \right] \times \left[ {0,3} \right]$ in the $uv$-plane. The area of ${\cal R}$ is
${\rm{Area}}\left( {\cal R} \right) = 4 \times 3 = 12$
Using Eq. (6) in Section 16.6, the area of the image of ${\cal R}$ is
${\rm{Area}}\left( {G\left( {\cal R} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {\cal R} \right) = 13 \times 12 = 156$
So, the area of the image of ${\cal R}$ is $156$.
