3D harmonic oscillator- expected value of distance

[Rorshach]

Homework Statement

Hey! I got this problem about 3D harmonic oscillator, here it goes:
A particle can move in three dimensions in a harmonic oscillator potential
$V(x,y,z)=\frac{1}{2}m\omega^2(x^2+y^2+z^2)$. Determine the ground state wave function. Check by explicitly counting that it is properly normalized.
Calculate the expectation value $<r>$, where $r=\sqrt{x^2+y^2+z^2}$ is a distance to the origin.

Homework Equations

$\alpha=\sqrt{\frac{m\omega}{\hbar}}$
$\omega=\sqrt{\frac{k}{m}}$
$V(x,y,z)=\frac{1}{2}m\omega^2(x^2+y^2+z^2)$
$\psi(x,y,z)==X(x)Y(y)Z(z)$
$-\frac{\hbar^2}{2m}[\frac{\xi^2\psi}{\xi x^2}+\frac{\xi^2\psi}{\xi y^2}+\frac{\xi^2\psi}{\xi z^2}]+\frac{k}{2}(x^2+y^2+z^2)\psi=E\psi$
$E_x+E_y+E_z=E$

The Attempt at a Solution

I pretty much calculated the wave function for the ground state, and it came out to be equal to $\psi_0(x,y,z)=(\frac{\alpha}{\pi})^\frac{3}{4} exp(-\alpha (x^2+y^2+z^2))$
with energy equal to $\frac{3}{2}\hbar\omega$. I don't understand what do they mean by explicitly calculating proper normalization, and I have trouble with calculating the expectation value for the distance from the point of origin. My first guess was to do this:
$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \psi^{*}(x,y,z) r \psi(x,y,z)\,dx\,dy\,dz.$ But it was quite a while since I've done multpile integrals, so I used wolframalpha for this:
http://www.wolframalpha.com/input/?...)^(3/4))*exp(-a(x^2+y^2+z^2)(1/2)))^2)+dxdydz
I am pretty shure that notation is wrong, but you probably get my point. I don't know if my way of thinking is correct. Help?

 

[tannerbk]

I don't understand what do they mean by explicitly calculating proper normalization.

In general: $< ψ_{i} | ψ_{j} > = δ_{ij}.$
Use the normalization condition $∫^{\infty}_{-\infty}dxdydz|ψ(x,y,z)|^{2}=1$ to check the ground state is properly normalized.

To calculate $< r >$ use spherical coordinates.