square root of (x^2+y^2)==> expand it using binomial theorem then take expectation value of every term, use the following equations

E(XY)=E(X)E(Y)

E(X^2)=Var(X)+[E(X)] ^2

i hope, it will be helpful 

If they are Gaussian and independent, then the sum of the square will be chi-square distribution. You can follow solution on:

https://math.stackexchange.com/questions/1059938/whats-the-expectation-of-square-root-of-chi-square-variable

Yes, when each varibale is N(0,1) then X^2 is chi(1), then X^2+y^2 is aslo chi(2)

or

E(X^2+Y^2)=E(X^2)+E(y^2)

and E(x^2)=var(x)+[E(X)]^2  this =1+0=1 and the same result of E(y^2)

then

E(X^2+Y^2)=E(X^2)+E(y^2)=1+1=2   as above result of chi.

  Best regards

If X is N(0, sigma12) and Y is N(0,sigma22) then E(X2+Y2) = sigma12+sigma22. This is irrespective of whether X and Y are independent or not. Also sigma1 and sigma2 do not have to be equal to 1 each.

How to find the expectation of square root of (X^2+X^2) where X and Y can be independent continuous or discrete random variables, not just standard normals?