I want to solve this Gaussian-like integral
$$ I =\int_0^1 e^{-x^2} \mathrm dx \tag{1} $$
This integral may be solved by considering x as a dummy variable and applying the Fubini's theorem to transform it into a double integral as follows:
$$ I =\int_0^1 e^{-y^2} \mathrm dy \tag{2} $$
$$ I^2 =\left(\int_0^1 e^{-x^2} \mathrm dx\right)\left(\int_0^1 e^{-y^2} \mathrm dy\right) = \int_0^1\int_0^1 e^{-\left(x^2+y^2\right)} \mathrm dx \mathrm dy \tag{3} $$
Here is when things get a bit interesting. The double integral may be solved by a variable change to polar coordinates:
$$ I^2 = \int_0^{\frac{\pi}{2}}\int_0^R e^{-r^2} \mathrm r \mathrm dr \mathrm d\theta \tag{4} $$
The above integral should yield the following result:
$$ I^2 = \frac{\pi}{4} \bigg[\text{erf}\left(1\right)\bigg]^2 \tag{5} $$
Or:
$$ I = \frac{\sqrt\pi}{2} \text{erf}\left(1\right) \tag{6} $$
Now comes my inquiry: I have not managed to figure out how to transform the integration region in (3) so I may find an appropriate value of R for which the integral result is (6). By doing some calculations, i.e., by making (4) = (5), finding the integral and solving for R, I found that:
$$ R = \sqrt{-\text{ln}\bigg[1-\left(\text{erf}\left(1\right)\right)^2\bigg]} \tag{7} $$
Is there an analytical (inductive/non-trial-and-error) procedure to find R as in (7) from a transformation of the integration region in (3) to polar coordinates? I will be most grateful if anyone can help me figure this puzzle.
This is perhaps more of a comment than an answer, because I'm not exactly sure what you're asking. Just by drawing a picture, we have $$R(\theta)=\cases{ \frac1{\cos\theta},&$0\leq\theta\leq\frac\pi4$\\ \frac1{\sin\theta},&$\frac\pi4\leq\theta\leq\frac\pi2$\\ }$$ but you seem to be asking for an $R$ that doesn't depend on $\theta$.
EDIT