Problem 82 - Advanced Engineering Mathematics

Question

Find the moment of inertia about the \(z\) -axis of the solid that is bounded by the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) if the density at a point \(P\) is directly proportional to the distance from the origin.

Step-by-Step Solution

Answer

\(I_z = \frac{2}{9} \pi k a^6.\)

Step 1: Understand the Problem

We need to find the moment of inertia about the z-axis for a solid bounded by a sphere centered at the origin with radius \(a\). The density is given to be directly proportional to the distance from the origin, implying \(\rho = k \cdot r\) where \(r = \sqrt{x^2+y^2+z^2}\) and \(k\) is a constant.

Step 2: Set up the Integral for Moment of Inertia

The moment of inertia about the \(z\)-axis, denoted as \(I_z\), is obtained by integrating \(\rho \cdot r_{ot}^2 \) over the volume of the sphere, where \(r_{\bot}^2 = x^2 + y^2\) (distance in the xy-plane). The integral becomes \(I_z = \int_V k \sqrt{x^2+y^2+z^2} (x^2+y^2)\, dV\).

Step 3: Convert to Spherical Coordinates

In spherical coordinates: \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), and \(z = \rho\cos\phi\). The volume element \(dV = \rho^2\sin\phi\, d\rho\, d\phi\, d\theta\) and the limits are \(0 \leq \rho \leq a\), \(0 \leq \phi \leq \pi\), \(0 \leq \theta \leq 2\pi\).

Step 4: Substitute and Simplify the Integral

Substitute into the integral: \(I_z = \int_0^{2\pi} \int_0^{\pi} \int_0^{a} k \rho \rho^2\sin^2\phi\ \rho^2\sin\phi \cos^2\theta\, d\rho\, d\phi\, d\theta\). This simplifies to \( I_z = k \int_0^{2\pi} \int_0^{\pi} \int_0^{a} \rho^5 \sin^3\phi \cos^2\theta\, d\rho\, d\phi\, d\theta\).

Step 5: Evaluate the Integral

Separate the integral: \(I_z = k \left( \int_0^{2\pi} \cos^2\theta\, d\theta \right) \left( \int_0^{\pi} \sin^3\phi\, d\phi \right) \left( \int_0^{a} \rho^5\, d\rho \right)\). Calculate each part: 1. \(\int_0^{2\pi} \cos^2\theta\, d\theta = \pi\)2. \(\int_0^{\pi} \sin^3\phi\, d\phi\) can be calculated using substitution yielding \(\frac{4}{3}\).3. \(\int_0^{a} \rho^5\, d\rho = \frac{a^6}{6}\).

Step 6: Calculate Final Moment of Inertia

After simplifying and evaluating the integrals, substitute back to find: \[ I_z = k \cdot \pi \cdot \frac{4}{3} \cdot \frac{a^6}{6} = \frac{2}{9} \pi k a^6. \] Therefore, the moment of inertia is directly related to\(k\), the proportionality constant of density.

Key Concepts

Spherical Coordinates

When we are dealing with three-dimensional problems involving symmetry, like a sphere, spherical coordinates come in handy. They allow us to work with the geometry in a more natural and simplified way. In spherical coordinates, any point in space is defined by three parameters: \(\rho\), \(\phi\), and \(\theta\).

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

For a sphere of radius \(a\), \(\rho\) ranges from 0 to \(a\), \(\phi\) varies from 0 to \(\pi\), and \(\theta\) ranges from 0 to \(2\pi\). This sets up our coordinate system to match the symmetries of the problem. During the transformation from Cartesian to spherical coordinates, we also change the volume element. The volume element \(dV\) in spherical coordinates becomes \(\rho^2\sin\phi \, d\rho \, d\phi \, d\theta\). This new form of \(dV\) accounts for how space becomes denser as you move further from the axis of rotation.

Density Distribution

In the context of the problem, the density of the sphere is not uniform but rather varies with position. Specifically, it is directly proportional to the distance from the origin. This means that the density function is given by \(\rho = k \cdot r\), where \(k\) is a constant of proportionality, and \(r = \sqrt{x^2 + y^2 + z^2}\) is the radial distance from the origin.

This kind of density distribution means that points further from the origin are denser (have more mass) than those closer to the origin. Such a property significantly affects the computation of moment of inertia because heavier points further from the axis contribute more to the integral. This is due to the nature of the moment of inertia formula, which involves multiplying mass by the square of the distance to the axis. Understanding how density varies allows us to correctly set up the integral to find the moment of inertia.

Integral Calculus

Integral calculus is essential in computing the moment of inertia, especially for objects with continuous mass distribution like a sphere. The moment of inertia \(I_z\) about the z-axis in this problem is calculated using a triple integral. The expression \(I_z = \int_V \rho \cdot r_{\bot}^2 \, dV\) is complex, which is why converting to spherical coordinates simplifies the process.

After swapping to spherical coordinates, the integral becomes \[ I_z = k \int_0^{2\pi} \int_0^{\pi} \int_0^{a} \rho^5 \sin^3\phi \cos^2\theta \, d\rho \, d\phi \, d\theta \].

• We integrate with respect to \(\theta\), which accounts for the symmetry around the z-axis.
• The integration with respect to \(\phi\) considers the variation in mass along different polar angles.
• The radius \(\rho\) integration covers the change over the radial distance.

Each part of the integral represents a different element in the calculation, and together, they culminate in calculating \(I_z\) which represents the sphere's resistance to rotational acceleration about the z-axis. This method reflects the power of integral calculus in solving such complex three-dimensional geometry problems.