Independent, Normally Distributed R.V. Working on this:

A shot is fired at a circular target. The vertical and the horizontal
  coordinates of the point of impact (with the origin sitting at the
  target’s center) are independent and normally distributed with $\nu(0, 1)$. Show that the distance of the point of impact from the center is
  distributed with PDF $$p(r) = re^{-r^2/2}, r \geq 0.$$ Find the median of this
  distribution.

So I'm guessing this would be graphed on an X and Y axis. I can intuit that I need to take the integral of the PDF from the lower bound to $m$ (or from $m$ to the upper bound), but I don't know what the normal distribution with $\nu$(0, 1) mean.
Also, how would I show that the point of impact has the desired PDF?
Thank you.

Let $r\ge 0$; put $R = \sqrt{X^2 + Y^2}$, where $X$ and $Y$ are the coordinates of the shot.  Then 
$$P(R\le r) = {1\over 2\pi} \mathop{\int\!\!\int}_{B_r(0)} \exp\{(x^2 + y^2)/2\}\,dx\,dy.$$
Change to polars to get
$$P(R\le r) = {1\over 2\pi}\int_0^r \int_0^{2\pi} \exp(r^2/2)r\,d\theta\, dr
=\int_0^r r\exp\{r^2/2\}\,dr$$
Differentiate and you are there.
