By using polar coordinates, one can solve Gaussian integral from $-\infty$ to $\infty$. Why can't I use this method for $-\sqrt{t}$ to $\sqrt{t}$ where $t\ge0$ ? I'm sure there are some mistakes in the deduction. Would you please point them out?
Let $$I(t)=\frac 1 {\sqrt{2\pi}}\int_{-\sqrt{t}}^{\sqrt{t}}e^{-x^2/2}dx \;.$$ Therefore, $$ \begin{aligned} I(t)^2&=\frac 1 {2\pi} \int_{\theta=0}^{2\pi}d\theta\int_{r=0}^{\sqrt{t}}e^{-r^2/2}rdr \\ &=1-e^{-t/2} \;, \end{aligned}$$ and then $$I(t)=\sqrt{1-e^{-t/2}} \; .$$ However, the correct answer is: $$I(t)=\text{erf}\left(\sqrt{\frac t 2}\right) \;.$$
Note $$ I(t)^2 = \frac 1 {{2\pi}}\int_{-\sqrt{t}}^{\sqrt{t}}\int_{-\sqrt{t}}^{\sqrt{t}}e^{(-x^2-y^2)/2}\;dx\;dy \;. $$ where we inegrate on a square in the plane with side $2\sqrt{t}$.
But your attempted version $$ \frac 1 {2\pi} \int_{\theta=0}^{2\pi}d\theta\int_{r=0}^{\sqrt{t}}e^{-r^2/2}r\;dr $$ integrates on a disk in polar coordinates: center $0$, radius $\sqrt{t}$. These are not equal.