How to calculate Gaussian integral under cartesian coordinate?

Question

Recently, I encounted a probelm. Look at the Gaussian function below. $$ \text{Gaussian function}= e^{-\frac{2r^2}{\omega^2} } $$ where, $r=\sqrt{x^2+y^2}$, actually, I do know how to calculate Gaussian integral under polar coordinate, like below, which is quite simple. $$ \int_{0}^{a} \int_{0}^{2\pi} e^{-\frac{2r^2}{\omega^2} }r\mathrm{d}\theta\mathrm{d}r=\frac{\pi\omega^2}{2} \left ( 1-e^{-\frac{2a^2}{\omega^2} } \right ) $$ However, I really do not know how I can calcualte integral of this Gaussian function under cartesian coordinate. I even do not know how the form of "dx" and "dy" look like. Can someone teach me this? That will be nice if you can also tell me how to deal with this integral by using parametric equation.

Answer 1

$$ \int_0^a rdr \int_0^{2\pi}e^{-2r^2/\omega^2} d\theta = \int_{-a}^a dx \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}dy e^{-2(x^2+y^2)/\omega^2} $$ $$ = \int_{-a}^a dx e^{-2x^2/\omega^2} \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}dy e^{-2y^2/\omega^2} = 2\int_{-a}^a dx e^{-2x^2/\omega^2} \int_0^{\sqrt{a^2-x^2}}dy e^{-2y^2/\omega^2} $$ $$ = 2\int_{-a}^a dx e^{-2x^2/\omega^2} \frac{1}{4}\sqrt{2\pi} \omega \,\mathrm{erf}(\sqrt 2 \sqrt{a^2-x^2}/\omega) = \sqrt{2\pi} \omega \int_0^a dx e^{-2x^2/\omega^2} \,\mathrm{erf}(\sqrt 2 \sqrt{a^2-x^2}/\omega) $$ where erf is the error function. Substitute $\sqrt{a^2-x^2}=t$, $dx/dt= -t dt/\sqrt{a^2-t^2}$, $$ ...= -\sqrt{2\pi} \omega \int_a^0 \frac{t}{\sqrt{a^2-t^2}}dt e^{-2(a^2-t^2)/\omega^2} \,\mathrm{erf}(\sqrt 2 t/\omega) $$ Obviously it's a strategic error not to use polar coordinates for that type of symmetry adapted integrals. I explicitly do not attempt to simplify that integral over the error function...