Answer
The area of the region enclosed by the ellipse is $ 12\pi $.
Work Step by Step
Let ${\cal D}$ denote the region enclosed by the ellipse ${x^2} + 2xy + 2{y^2} - 4y = 8$.
We have the relations: $u=x+y$, $v=y-2$.
Subtracting $v$ from $u$ gives
$u - v = x + 2$
So, $x=-2+u-v$.
Substituting $x=-2+u-v$ in $u$ gives
$u = - 2 + u - v + y$
$y=2+v$
Thus, we obtain the linear mapping: $G\left( {u,v} \right) = \left( { - 2 + u - v,2 + v} \right)$.
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&{ - 1}\\
0&1
\end{array}} \right| = 1$
Write the equation of the ellipse:
${x^2} + 2xy + 2{y^2} - 4y - 8 = 0$
(1) ${\ \ \ \ \ }$ ${\left( {x + y} \right)^2} - {y^2} + 2{\left( {y - 1} \right)^2} - 10 = 0$
Substituting $x=-2+u-v$ and $y=2+v$ in equation (1) gives
${u^2} - {\left( {2 + v} \right)^2} + 2{\left( {1 + v} \right)^2} - 10 = 0$
${u^2} - 4 - 4v - {v^2} + 2 + 4v + 2{v^2} - 10 = 0$
${u^2} + {v^2} - 12 = 0$
Notice that ${u^2} + {v^2} = 12$ is the equation of the circle of radius $\sqrt {12} $.
Thus, we conclude that the mapping $G\left( {u,v} \right) = \left( { - 2 + u - v,2 + v} \right)$ maps ${{\cal D}_0}$, the disk of radius $\sqrt {12} $ defined by ${u^2} + {v^2} = 12$ to the ellipse ${x^2} + 2xy + 2{y^2} - 4y = 8$ in the $xy$-plane. This is illustrated in the figure attached.
Using the Change of Variables Formula, we evaluate the area of ${\cal D}$:
${\rm{Area}}\left( {\cal D} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} \left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$ = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} {\rm{d}}u{\rm{d}}v$
Since $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} {\rm{d}}u{\rm{d}}v$ gives the area of the disk of radius $\sqrt {12} $, so
${\rm{Area}}\left( {\cal D} \right) = 12\pi $
