Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{9{x^2} + 4{y^2}}}{\rm{d}}x{\rm{d}}y \simeq 2.26 \times {10^{15}}$
Work Step by Step
We learn from Exercise 34, the mapping $G\left( {u,v} \right) = \left( {au,bv} \right)$ maps the disk ${u^2} + {v^2} \le 1$ onto the interior of the ellipse ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} \le 1$. Thus, $G\left( {u,v} \right) = \left( {2u,3v} \right)$ maps the disk ${u^2} + {v^2} \le 1$ onto the interior of the ellipse ${\left( {\frac{x}{2}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} \le 1$.
Evaluate the Jacobian of $G\left( {u,v} \right) = \left( {2u,3v} \right)$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
2&0\\
0&3
\end{array}} \right| = 6$
Write
$f\left( {x,y} \right) = {{\rm{e}}^{9{x^2} + 4{y^2}}}$
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = {{\rm{e}}^{36{u^2} + 36{v^2}}} = {{\rm{e}}^{36\left( {{u^2} + {v^2}} \right)}}$
Let ${{\cal D}_0}$ denote the domain of the disk ${u^2} + {v^2} \le 1$.
Using the Change of Variables Formula, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
(1) ${\ \ \ \ \ }$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{9{x^2} + 4{y^2}}}{\rm{d}}x{\rm{d}}y = 6\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} {{\rm{e}}^{36\left( {{u^2} + {v^2}} \right)}}{\rm{d}}u{\rm{d}}v$
To evaluate this integral, we use polar coordinates. So, the description of ${{\cal D}_0}$ in polar coordinates is
${{\cal D}_0} = \left\{ {\left( {r,\theta } \right)|0 \le r \le 1,0 \le \theta \le 2\pi } \right\}$
Thus, integral (1) becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{9{x^2} + 4{y^2}}}{\rm{d}}x{\rm{d}}y = 6\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {{\rm{e}}^{36{r^2}}}r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{1}{{12}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {{{\rm{e}}^{36{r^2}}}|_0^1} \right){\rm{d}}\theta $
$ = \frac{1}{{12}}\left( {{{\rm{e}}^{36}} - 1} \right)\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta $
$ = \frac{\pi }{6}\left( {{{\rm{e}}^{36}} - 1} \right) \simeq 2.26 \times {10^{15}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {{\rm{e}}^{9{x^2} + 4{y^2}}}{\rm{d}}x{\rm{d}}y \simeq 2.26 \times {10^{15}}$.
