Let's make it simpler. You have particle on a 2d plane with an Hamiltonian that depends on the distance to some center $r$ and also to the angle with respect to this center $\theta$. You can imagine your system as a cylinder with 0 height.
The probability of finding a particle at a point $(r_0, \theta_0)$ of the plane is:
$P(r=r_0, \theta=\theta_0)=e^{-\beta H(r=r_0,\theta=\theta_0)}/Z$
Where $Z$ is a normalisation function. This is the probability of finding a particle at a given point. Now you may ask, what is the probability of finding a particle at a distance $r_0$ of my center irrespective on its angle. That is: what is the probability of finding my particle at any point that is at distance $r_0$ to my center (a particle sitting on a circle of radius $r_0$). See the difference between this question and the one above?
In this case, the probability is:
$$P(r=r_0)=\int_0^{2\pi}e^{-\beta H(r=r_0, \theta=\theta_0)}r_0 d\theta_0/Z$$
Which has to be understood as the sum over all angles of the probabilities of being at ONE point at distance $r_0$: the probability of finding a particle at a distance $r_0$ from the center is equal to the probability of finding a particle at a distance $r_0$ from the center and angle 0 + the probability of finding a particle at a distance $r_0$ from the center and angle 0.000000..001 + the probability of finding a particle at a distance $r_0$ from the center and angle 0.0000...0002 , etc
For simple Hamiltonian, you will obtain the right radius dependence if you don't perform the integration. For example if $H$ can be written as $H=V_1(r) +V_2(\theta)$ then, the probabilities you will obtain by looking at $P(r,\theta)$ or $P(r)$ will simply differ by a constant and (in the case of $r$) a linear $r$. However, this will not be the case if the radius and the angle are coupled:
$H=\cos(\theta)r$.
In general, the process of "integrating" a degree of freedom in a probability distribution is called marginalisation.
