I struggle with following problem. A two-dimensional Gaussian function $f(x,y) = \frac{1}{2 \pi s^2} e^{-(x^2+y^2)/(2s^2)}$ is given and shall be integrated within the limits of $a=-\sqrt{2\mathrm{ln}(2)}s\,$ to $\,b=\sqrt{2\mathrm{ln}(2)}s$ for both coordinates. These limits define the FWHM of the function.
Now, I know that for a one-dimensional Gaussian $g(x) = \frac{1}{\sqrt{2 \pi} s} e^{-x^2/(2s^2)}$ one obtains $I = \int_a^b g(x) \mathrm{d}x = 0.760968$.
For the problem at hand, I thought that
$$ \int_a^b \int_a^b f(x,y) \mathrm{d}x \mathrm{d}y = \int_a^b g(x) \mathrm{d}x \int_a^b g(y) \mathrm{d}y = 0.760968 \cdot 0.760968 = 0.57907, $$ with rewriting $f(x,y)=g(x)\cdot g(y)$.
However, if the problem is recalculated in Polar coordidnates, I obtain a different result, namely
$$ \int_a^b \int_a^b f(x,y) \mathrm{d}x \mathrm{d}y = \int_0^{2\pi} \int_0^b \frac{1}{2 \pi s^2} e^{-r^2/(2s^2)} r \,\mathrm{d}r \mathrm{d} \phi = \int_0^b \frac{1}{s^2} e^{-r^2/(2s^2)} r \,\mathrm{d}r = 0.5. $$
Where is the error and which one is correct?
You have an error in the logic of the solution in polar coordinates. The original expression: $$\int_{a}^{b} \int_{a}^{b} f(x,y)\,dx\,dy $$ is integrated over a square region from $a$ to $b$ in both dimensions.
You convert this to: $$\int_{0}^{2\pi} \int_{0}^{b} \frac{1}{2\pi s^2}e^{-r^2/(2s^2)}r\,dr\,d\phi$$
This is incorrect as now the integration is a circular region, centered at the origin, with radius $b$.
Converting to polar form seems inappropriate with a square region unless some other transformation is involved.