Let $W=X^2+Y^2$. We find the cumulative distribution function $F_W(w)$ of $w$.
This is the probability that $W\le w$. Clearly $F_W(w)=0$ if $w\le 0$. Now we deal with $w\gt 0$.
So we want $\Pr(X^2+Y^2 \le w)$. This is the integral over the disk with centre the origin, and radius $\sqrt{w}$, of the joint density function of $X$ and $Y$. So for positive $w$, $F_W(w)=\iint_D \frac{1}{2\pi}e^{-(x^2+y^2)/2}\,dx\,dy.$ Change to polar coordinates. Then $dx\,dy=r\,dr\,d\theta$, and our integral is $\int_0^{2\pi}\frac{1}{2\pi}\left(\int_0^{\sqrt{w}}re^{-r^2/2}\,dr \right)\,d\theta.$ The integration is easy. For the inner integral, let $u=r^2$. The outer integral is even simpler. Quickly we get $\dfrac{1}{2}e^{-w/2}$, which is the cdf of the exponential with parameter $\dfrac{1}{2}$.