Answer
We prove the area of the ellipse is $\pi ab$.
Work Step by Step
Write $x = au$ and $y = bv$. So, $u = \frac{x}{a}$ and $v = \frac{y}{b}$.
Then the interior of the ellipse ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} \le 1$ correspond to the disk ${u^2} + {v^2} \le 1$ in the $uv$-plane. Thus, we can use $G\left( {u,v} \right) = \left( {au,bv} \right)$ to map the disk ${u^2} + {v^2} \le 1$ onto the interior of the ellipse ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} \le 1$.
Evaluate the Jacobian of $G\left( {u,v} \right) = \left( {au,bv} \right)$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
a&0\\
0&b
\end{array}} \right| = ab$
Let ${{\cal D}_0}$ denote the domain of the disk and ${\cal D}$ denote the region of the ellipse. Using the Change of Variables Formula, we evaluate the area of ${\cal D}$:
${\rm{Area}}\left( {\cal D} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} \left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$ = ab\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 ${{\cal D}_0}$, a unit disk; we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} {\rm{d}}u{\rm{d}}v = \pi $
Thus, ${\rm{Area}}\left( {\cal D} \right) = \pi ab$.
Hence, the area of the ellipse is $\pi ab$.
