Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 3y} \right){\rm{d}}x{\rm{d}}y = 80$
Work Step by Step
We have the linear mapping $G\left( {u,v} \right) = \left( {u - 2v,v} \right)$. Notice that the image under $G$ is based at the origin.
Referring to Figure 16, we use the equations $y=1$ and $x+2y=6$ to obtain the vertex $P = \left( {4,1} \right)$. We see from Figure 16, the region ${\cal D}$ is based at $\left( {4,1} \right)$. Thus, we write $a=4$ and $b=1$.
Let $T$ denote the translate of the mapping $G$ based at $\left( {4,1} \right)$. So, we have
$T\left( {u,v} \right) = \left( {4 + u - 2v,1 + v} \right)$
Using the equations $y=1$ and $x+2y=10$, we obtain the vertex $Q = \left( {8,1} \right)$.
Another vertex from Figure 16 is $R = \left( {0,3} \right)$.
Thus, we see that ${\cal D}$ is spanned by the vectors formed by the points $P = \left( {4,1} \right)$, $Q = \left( {8,1} \right)$, $R = \left( {0,3} \right)$.
Using $P = \left( {4,1} \right)$, we find the corresponding point in the $uv$-plane:
$T\left( {u,v} \right) = \left( {4 + u - 2v,1 + v} \right) = \left( {4,1} \right)$
$4+u-2v=4$, ${\ \ \ \ \ }$ $1+v=1$
The second equation above gives $v=0$. Substituting it in the first equation, we get $u=0$. So, $P = \left( {4,1} \right)$ corresponds to $\left( {u,v} \right) = \left( {0,0} \right)$.
Using $Q = \left( {8,1} \right)$, we find the corresponding point in the $uv$-plane:
$T\left( {u,v} \right) = \left( {4 + u - 2v,1 + v} \right) = \left( {8,1} \right)$
$4+u-2v=8$, ${\ \ \ \ \ }$ $1+v=1$
The second equation above gives $v=0$. Substituting it in the first equation, we get $u=4$. So, $Q = \left( {8,1} \right)$ corresponds to $\left( {u,v} \right) = \left( {4,0} \right)$.
Using $R = \left( {0,3} \right)$, we find the corresponding point in the $uv$-plane:
$T\left( {u,v} \right) = \left( {4 + u - 2v,1 + v} \right) = \left( {0,3} \right)$
$4+u-2v=0$, ${\ \ \ \ \ }$ $1+v=3$
The second equation above gives $v=2$. Substituting it in the first equation, we get $u=0$. So, $R = \left( {0,3} \right)$ corresponds to $\left( {u,v} \right) = \left( {0,2} \right)$.
Since $T$ is a linear map, the points $\left( {0,0} \right)$, $\left( {4,0} \right)$ and $\left( {0,2} \right)$ form the vectors that span the domain ${\cal R}$. Thus, ${\cal R}$ is a rectangle given by ${\cal R}:\left[ {0,4} \right] \times \left[ {0,2} \right]$.
Next, we write
$f\left( {x,y} \right) = x + 3y$
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = 4 + u - 2v + 3\left( {1 + v} \right)$
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = 7 + u + v$
Since $T\left( {u,v} \right) = \left( {4 + u - 2v,1 + v} \right)$, so
$x\left( {u,v} \right) = 4 + u - 2v$, ${\ \ \ \ \ }$ $y\left( {u,v} \right) = 1 + v$
Evaluate the Jacobian of $T$:
${\rm{Jac}}\left( T \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&{ - 2}\\
0&1
\end{array}} \right| = 1$
Now, we compute $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 3y} \right){\rm{d}}x{\rm{d}}y$ using the Change of Variables Formula:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( T \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 3y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{v = 0}^2 \mathop \smallint \limits_{u = 0}^4 \left( {7 + u + v} \right){\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{v = 0}^2 \left( {\left( {7u + \frac{1}{2}{u^2} + uv} \right)|_0^4} \right){\rm{d}}v$
$ = \mathop \smallint \limits_{v = 0}^2 \left( {36 + 4v} \right){\rm{d}}v$
$ = \left( {36v + 2{v^2}} \right)|_0^2 = 80$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {x + 3y} \right){\rm{d}}x{\rm{d}}y = 80$.
