If you integrate the gaussian integral using polar coordinates you can convert into the following integral:
$$\int_{\theta=0}^{\theta=2\pi} \int _{r=0}^{r=\infty} e^{-r^2}r dr d\theta=I^2$$
I get the conversion from -$\infty$ and $\infty$, to 0 and $\infty$ because radius is always positive but what about the theta bounds? The minimum angle of 0, if you draw out the graph of $e^{-x^2}$ the lowest angle you can go to is 0(As x approaches $\infty$ which is valid because one of the bounds are $\infty$) using this same logic shouldn't the other bound should be $\pi$?
Only the domain of the function that you are integrating matters here. In this case, that domain is $\Bbb R^2$. In particular, there are points of the domain all around $(0,0)$, and that's why we take $0\leqslant\theta\leqslant2\pi$. The same thing would have occurred if the domain was, say, the closed disk centered at $(0,0)$ with radius $1$. But if it was, say, $\{(x,y)\in\Bbb R^2\mid y\geqslant0\}$ (that is, the upper half-plane), then we would take $0\leqslant\theta\leqslant\pi$.