Is this interpretation of $\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$ correct? Apologies if this is stating the obvious, but I'm a non-physicist trying to understand Griffiths' discussion of the hydrogen atom in chapter 4 of Introduction to Quantum Mechanics. The wave equation for the ground state (I believe) is:$$\psi=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}$$ 
where $a$ is the Bohr radius $0.529\times10^{-10}$m. If I integrate the square of this equation between $r=0$ and $r=x$, am I right in assuming I am calculating the probability of finding the electron in a sphere radius $x$? I've done this for $x=1\textrm{ m}$ and $x=\infty$ and got the answer $1$ (I guess a two metre diameter sphere is pretty big compared to a hydrogen atom). For $x=9\times10^{-11}\textrm{ m}$, the answer is $0.66077$. Is this interpretation correct?
 A: No, you are wrong. Particular for the following statement:

If I integrate the square of this equation between $r=0$ and $r=x$, am I right in assuming I am calculating the probability of finding the electron in a sphere radius x?

The probability density at any points is given by $|\psi(r,\theta,\phi)|^2$. Certainly, the probability is for any region $V$ is given by
$$P=\int_V|\psi|^2 dxdydz$$
However, if you are considering a sphere of radius $R$, you should change the integral into the spherical coordinate using the relation $dxdydz=r^2 \sin(\theta)drd\theta d\phi$ as follow
$$P(R)=\int_V|\psi|^2 dxdydz=\int_0^R \int_0^\pi \int_0^{2\pi}  |\psi|^2 r^2 \sin(\theta) dr d\theta d\phi = 4\pi \int_0^R r^2|\psi|^2 dr$$
Note that there is an extra $r^2$ in the equation and an extra constant $4 \pi$. Simply integrate the $|\psi|^2$ from $0$ to $R$ is wrong. It is very easy to check that this integral lead to the normalization condition of the wavefunction, i.e. $P(R\to\infty)=1$.
A: Yes, you are practically right. The only subtlety is that your wave function argument $\mathbf{r}$ is a relative distance between the nucleus and the electron. Nucleus itself "turns around" the atomic center of mass, but with a shorter radius $(m_e/M_N)\cdot a$.
A: In my opinion, if you want to know the probability of a particle being in a sphere, there should be an integration over a sphere, therefore I would add an additional factor of $4\pi$.
The thing I don't understand is the normalization of the wave function, it  doesn't yield 1 if I integrate it over whole space.
My last issue is just a small one, I wouldn't call this a wave equation, it is already a wave function, the corresponding wave equation is the Schrödinger equation.
