CDF of standardized normal R.V. I'm attemtping to solve this problem:

Suppose a shot is fired at a circular target.  The vertical and the
  horizontal coordinates  of the point of impact (taking the center of
  the target as origin) are independent random variables, each
  distributed according to the standardized normal distribution. 
a. Write down the PDFs of the two coordinates.
  b. Write down the joint
  PDF of the coordinates of the point of impact.
  c. What is the PDF of
  the radius of the point of impact $r = \sqrt{x^2 + y^2}$ ?  Show all  steps you used to
  derive this expression.
Hint: you can use the CDF method for finding the PDF of a transformed RV.
  d. What is the probability that the
  point of impact will land in the ring of radii 2-3?

MY attempt at a solution:
a)
P.D.F. of x is: $$\frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{2}}$$
P.D.F. of y is: $$\frac{1}{2\sqrt{\pi}}e^{-\frac{y^2}{2}}$$
b)
joint PDF is:
PDF(x) * PDF(y) = $$\frac{1}{2\sqrt{\pi}}e^{-\frac{y^2}{2}} * \frac{1}{2\sqrt{\pi}}e^{-\frac{x^2}{2}} = 2\pi^{-1}e^{\frac{-(x^2 + y^2)}{2}} $$
c) this is where I'm getting hung up on...
So the C.D.F. of a std. normal R.V. $x$ is:
$$\phi (x) = \frac{1}{\sqrt{2 \pi}}\int e^{\frac{-t^{2}}{2}}dt$$ 
correct?
since $r = \sqrt{x^2 + y^2}$  is given, then is the CDF of $r$ just
$$\phi (r) = \frac{1}{\sqrt{2 \pi}}\int e^{\frac{-r^{2}}{2}}dr$$
And then would the PDF of the radius just be the first derivative of this?
 A: Note that you have the pdf of the standard normal not quite right, or at least not consistently right. The constant in front should be $\frac{1}{\sqrt{2\pi}}$. So the constant in front of the joint density is $\frac{1}{2\pi}$.
Let random variable $R$ denote the distance to the origin. Then $R=\sqrt{X^2+Y^2}$. We want the cumulative distribution function (cdf) of $R$, so we want $P(R\le r)$. Clearly this is $0$ for $r \lt 0$. So from now on assume $r\ge 0$.
We have
$$P(R \le r)=\iint_{D_r} \frac{1}{2\pi}e^{-(x^2+y^2)/2}dx\,dy,$$
where $D_r$ is the disk of radius $r$, centre the origin. We need to evaluate this integral.  
To do this, change to polar coordinates. (If that has been forgotten, look back to your several variable calculus course.)  Because the letter $r$ is already taken, I will use $t$ for the polar coordinate distance to the origin. Let $x=t\cos\theta$, $y=t\sin\theta$.  The integral becomes 
$$\int_0^{2\pi} \int_0^r \frac{1}{2\pi}e^{-t^2/2}t\,dt\, d\theta.$$
The integration is straightforward. The inner integral is $\frac{1}{2\pi}\left(1-e^{-r^2/2}\right)$, and for the outer integral you just multiply by $2\pi$, getting $1-e^{-r^2/2}$.
Now we have the cumulative distribution function of $R$. The rest follows without much trouble.  For (c), differentiate. For (d), if the cdf is $F_R(r)$, we want $F_R(3)-F_R(2)$.
