Final answer:
The rotational inertia about the z-axis for the unit ball with density ρ(x,y,z) is found by setting up a triple integral in spherical coordinates, integrating the density function multiplied by the square of the perpendicular distance to the z-axis over the volume of the ball.
Step-by-step explanation:
To find the rotational inertia about the z-axis for the unit ball with density ρ(x,y,z), we need to set up an integral that considers mass distribution throughout the volume of the ball. The general formula for the moment of inertia of a volume is I = ∫∫∫ ρ(r) r^2 dV, where r is the distance from the axis of rotation. For a unit ball with a uniform density where x^2 + y^2 + z^2 < 1, the integral becomes more complex and is typically done using spherical coordinates because it respects the symmetry of the ball around the z-axis.
In spherical coordinates, a point is represented as (r, θ, φ), where r is the radial distance from the origin, θ is the azimuthal angle, and φ is the polar angle. The volume element in spherical coordinates is dV = r^2 sin(φ) dr dθ dφ. Since we are only interested in the component of r that is perpendicular to the z-axis (r sin(φ)), the rotational inertia integral is transformed into I = ∫∫∫ ρ(r) r^2 sin^2(φ) dV. This integral ranges from r = 0 to 1, θ = 0 to 2π, and φ = 0 to π.
Mathematically, the integral to find the rotational inertia about the z-axis for the unit ball with density ρ(x,y,z) is:
I = ∫∫∫ ρ(r) r^4 sin^3(φ) dr dθ dφ
where the limits for r are from 0 to 1, for θ are from 0 to 2π, and for φ are from 0 to π.