Circle of Scattering for Normal distribution 
A random point $(X,Y)$ has a normal distribution on a plane with circular scattering with
$$ E(X)=E(Y)=0 \quad \text{and} \quad \operatorname{var}(X)=\operatorname{var}(Y)=\sigma^2. $$
The distance of the point $(X,Y)$ from the centre of scattering  is $R$. Find $E(R)$.

I found the joint distribution of $(X,Y)$ and then found out the distribution of $R$ using a simple polar transformation. I am getting that
$$E(R)=\sqrt{\frac{\pi}{2}} \sigma.$$
But, I think this is not correct and I am failing to convince myself regarding the independence assumption of $X,Y$.
Help!
 A: Here is a simulation following my Comment. Using R, we sample a sample a million values $X$ from $\mathsf{Norm}(0,1)$ and, independently, a million values $Y.$
set.seed(822);  m = 10^6
x = rnorm(m);  y = rnorm(m)
r = sqrt(x^2 + y^2)
mean(r);  sqrt(pi/2)
[1] 1.252897          # aprx E(R) = 1.2533
[1] 1.253314          # exact

# 50,000 points for plotting
n = 50000;  X = x[1:n];  Y = y[1:n];  R = r[1:n]
par(mfrow=c(1,3))
 plot(X, Y, pch=".")
 hist(R, prob=T, col="skyblue2", main="Simulated Rayleigh Dist'n")
  curve(x*exp(-.5*x^2), 0, 5, add=T, col="red", lwd=2)
 plot(ecdf(R))
  curve(1-exp(-.5*x^2), 0, 5, add=T, col="red", lwd=3, lty="dashed")
par(mfrow=c(1,1))

The figure below shows 50,000 bivariate normal points, a histogram of
the simulated Rayleigh distribution $(\sigma=1)$ with the exact PDF (red), and an empirical
CDF plot of the simulated Rayleigh distribution along with the exact CDF (dashed red).
Details of the Rayleigh distribution are shown in Wikipedia.
As you say, the derivation of the PDF involves changing to polar coordinates.

