How can ${\iint\limits_{D}{{e^{x^2+y^2}}}}dxdy$ be found? How can  ${\iint\limits_{D}{{e^{x^2+y^2}}}}dxdy $  be found, if $D$ is $x$ O $y$ axis?
So far I have done it this far:
$$\iint\limits_D{e^{x^2+y^2}}dxdy=\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}e^{x^2+y^2}dy=\int_{-\infty}^{\infty}dx\lim_{b\to\infty}\int_{-b}^{b}e^{x^2+y^2}dy=\cdots$$
And then I am stuck there as $e^{x^2+y^2}$, appears to have no integral in terms of elementary functions.
How do I approach this problem?
EDIT:
In my country $x$ O $y$ is used equivalently to $\Bbb{R}^{2}$. The rest of integral and functions is as they should be $e^{x^2+y^2}$.
 A: The integral can be easy evaluated by using polar coordinate, where $r^2=x^2+y^2$ and $dx\ dy=r\ dr\ d\theta$. The region of integration is $0<r<\infty$ and $0<\theta<2\pi$. Therefore
\begin{align}
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{\large -(x^2+y^2)}\ dx\ dy&=\int_{\theta=0}^{2\pi}\int_{r=0}^\infty e^{\large -r^2}r\ dr\ d\theta\\
&=\int_{\theta=0}^{2\pi}\ d\theta\int_{r=0}^\infty e^{\large -r^2}r\ dr\\
&=2\pi\int_{u=0}^\infty e^{\large -u}\ \frac{du}{2}\quad\Rightarrow\quad u=r^2\;\rightarrow\,du=2r\ dr\\
&=-\pi\ \left.e^{\large-u}\right|_{u=0}^\infty\\
&=\Large\color{blue}\pi.
\end{align}

If the integrand is $e^{\large  x^2+y^2 }$, then
\begin{align}
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{\large  x^2+y^2}\ dx\ dy&=\int_{\theta=0}^{2\pi}\int_{r=0}^\infty e^{\large r^2}r\ dr\ d\theta\\
&=\int_{\theta=0}^{2\pi}\ d\theta\cdot\lim_{r\to\infty}\int_{0}^r e^{\large r^2}r\ dr\\
&=2\pi\cdot\lim_{r\to\infty}\int_{r=0}^r e^{\large u}\ \frac{du}{2}\quad\Rightarrow\quad u=r^2\;\rightarrow\,du=2r\ dr\\
&=\pi\ \left.\cdot\lim_{r\to\infty}e^{\large r^2}\right|_{=0}^r\\
&\to\Large\color{blue}\infty.
\end{align}
A: $$\iint\limits_D{e^{-(x^2+y^2)}}dxdy=\iint\limits_D{e^{-r^2}}rdrd\theta=\int_{0}^{\infty}e^{-r^2}rdr\int_{-\pi}^{\pi}d\theta$$
A: Firstly, As PhoemueX says, it should probably be $-x^2-y^2$ in the exponent, as otherwise the integral diverges.
Hint: make the substitution $x^2+y^2=r^2$, $dxdy=rdrd\theta$.
