Given that a particle is inside a sphere of radius 1 , and that it has equal probabilities of bcing found in any two volume elements of the same size, find the distribution function \(F(r)\) for the spherical coordinate \(r\), and from it find the density function \(f(r)\).

In this problem, we are told that the particle has equal probabilities of being found in any two volume elements of the same size within the sphere. This means that the particle's distribution is **uniform**. A uniform distribution means that every part of the sphere is equally likely to contain the particle. Unlike other distributions where some areas might have higher probabilities, a uniform distribution ensures that the particle is equally likely to be anywhere inside the sphere.
This idea of uniform distribution simplifies calculations, as it implies that probabilities only depend on the size of the region and not its specific location.

To solve problems involving spheres, it's often easiest to use **spherical coordinates**. This coordinate system uses three parameters: the radial distance \(r\), the polar angle \(\theta\), and the azimuthal angle \(\phi\). In spherical coordinates, a point in space is represented as (r, \(\theta\), \(\phi\).
The volume element in spherical coordinates is given by:
\[ dV = r^2 \sin(\theta) \, dr \, d\theta \, d\phi \]
Here, \(dr\) is a tiny change in the radial distance, \(d\theta\) is a tiny change in the polar angle, and \(d\phi\) is a tiny change in the azimuthal angle. This tiny volume element is crucial for integrating over the sphere to calculate various properties like total volume and probabilities.

The **density function**, often denoted as \(f(r)\), represents how the probability density varies with radius within the sphere. Given a specific distribution function \(F(r)\), the density function can be obtained by differentiating \(F(r)\) with respect to \(r\). For a uniform distribution within a sphere of radius 1, we derived the distribution function as \(F(r) = r^3\).
Thus, the density function is:
\[ f(r) = \frac{dF(r)}{dr} = \frac{d}{dr} (r^3) = 3r^2 \]
This tells us how the probability density changes as we move from the center of the sphere to the edge. The density function essentially provides a more detailed picture of the particle distribution within the sphere.

The **distribution function**, denoted as \(F(r)\), gives the probability that the particle is within a certain radius \(r\) from the center of the sphere. Since the probability distribution is uniform, the function is proportional to the volume of the sphere with radius \(r\).
We start by calculating the volume of a sphere with radius \(r\):
\[ V_{sphere \, of \, radius \, r} = \frac{4}{3} \pi r^3 \]
The total volume of the sphere (with radius 1) is:
\[ V_{total} = \frac{4}{3} \pi \]
Therefore, the distribution function is:
\[ F(r) = \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi} = r^3 \]
This means \(F(r)\) grows with the cube of the radius, indicating that larger radii correspond to larger volumes and thus higher cumulative probability.