Step 1: Since $X_1, X_2$ are independent $N(0,1)$, their joint density is\[ f(x_1,x_2) = \frac{1}{2\pi} e^{-(x_1^2+x_2^2)/2} \]
Step 2: We need $P(X_1^2+X_2^2 \le 2)$, the integral of this density over the disk of radius $\sqrt{2}$ centered at the origin.
Step 3: Switch to polar coordinates $x_1 = r\cos\theta$, $x_2 = r\sin\theta$, with $dx_1\,dx_2 = r\,dr\,d\theta$:\[ P = \int_0^{2\pi}\int_0^{\sqrt{2}} \frac{1}{2\pi} e^{-r^2/2}\, r\, dr\, d\theta \]
Step 4: The $\theta$ integral gives $2\pi$, canceling the $\frac{1}{2\pi}$ factor:\[ P = \int_0^{\sqrt{2}} r\, e^{-r^2/2}\, dr \]
Step 5: Substitute $u = r^2/2$, $du = r\,dr$. When $r=0$, $u=0$; when $r=\sqrt{2}$, $u=1$:\[ P = \int_0^1 e^{-u}\, du = \left[-e^{-u}\right]_0^1 = 1 - e^{-1} \]\[\boxed{1-e^{-1}}\]