Problem 78 - University Calculus: Early Transcendentals

Question

A solid ball is bounded by the sphere \(\rho=a\) Find the moment of inertia about the \(z\) -axis if the density is a. \(\delta(\rho, \phi, \theta)=\rho^{2}\) b. \(\delta(\rho, \phi, \theta)=r=\rho \sin \phi\)

Step-by-Step Solution

Answer

The moment of inertia for part a is \(\frac{16 \pi}{15} a^7\) and for part b is \(\frac{8 \pi}{15} a^7\).

Step 1: Analyze the problem

Identify the type of problem and the appropriate approach.

Step 2: Solve

The moment of inertia for part a is \(\frac{16 \pi}{15} a^7\) and for part b is \(\frac{8 \pi}{15} a^7\)..

Step 3: Verify

Check the solution for correctness.

Key Concepts

Understanding Spherical Coordinates

When dealing with three-dimensional shapes like a sphere, using spherical coordinates can simplify calculations. Unlike Cartesian coordinates, which use x, y, and z, spherical coordinates use three parameters:

• \( \rho \): The radial distance from the origin to the point.
• \( \phi \): The polar angle, measured from the positive z-axis.
• \( \theta \): The azimuthal angle, in the x-y plane from the positive x-axis.

These coordinates make problems involving spheres more manageable because they align with the spherical shape. For a solid ball bounded by a sphere \( \rho = a \), \( \rho \) varies from 0 to \( a \), \( \phi \) from 0 to \( \pi \), and \( \theta \) from 0 to \( 2\pi \). The volume element in spherical coordinates is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \), which matches the natural symmetries of the sphere.

Exploring Density Functions

In physics, the density function \( \delta(\rho, \phi, \theta) \) describes how mass is distributed in space. It can vary depending on the specific problem. In our case, the exercise considers two different density functions:

• \( \delta(\rho, \phi, \theta) = \rho^{2} \): Here, the density increases with the square of the radial distance. This function suggests that the outer parts of the sphere are denser than the center.
• \( \delta(\rho, \phi, \theta) = r = \rho \sin \phi \): This considers the cylindrical distance from the z-axis, meaning that density depends on how far a point is from this axis within the sphere.

Choosing a density function affects how the moment of inertia is calculated. A higher density further from the axis increases the moment of inertia, since mass at a greater distance contributes more to rotational dynamics.

Introduction to Integral Calculus

Integral calculus is a fundamental mathematical tool used to compute areas, volumes, and other quantities under curves or surfaces. In our context, we use integral calculus to determine the moment of inertia for a sphere. The moment of inertia about the z-axis, \( I_z \), is calculated by the triple integral:\[I_z = \int_{V} r^2 \delta(\rho, \phi, \theta) \, dV\]where \( r = \rho \sin \phi \) is the perpendicular distance from the axis of rotation, and \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \) is the volume element in spherical coordinates. This integral accumulates contributions from each infinitesimal volume within the sphere, accounting for both the distance from the axis and the local density. For each density function in the exercise, the integration limits correspond to the bounded region of the solid ball and are integral to obtaining the correct moment of inertia result. Understanding these concepts can help explain not just this problem, but also similar physical calculations involving rotational dynamics.