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 $e^{x^2+y^2}$.
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}