Problem 45 - Thomas Calculus

Question

Thin shell of constant density Find the center of mass and the moment of inertia about the \(z\) -axis of a thin shell of constant density \(\delta\) cut from the cone \(x^{2}+y^{2}-z^{2}=0\) by the planes \(z=1\) and \(z=2\).

Step-by-Step Solution

Answer

The center of mass is (0, 0, 7/3), and the moment of inertia is \(\frac{15\pi\delta}{2}\).

Step 1: Parametrize the Cone

The cone can be parametrized using cylindrical coordinates where \( r = z \). For a point on the cone, \( x = z \cos\theta \) and \( y = z \sin\theta \). The range for \( z \) is from 1 to 2, based on the planes given.

Step 2: Express Area Element

The differential element for area in cylindrical coordinates is \( dA = z \, dz \, d\theta \). Therefore, the infinitesimal area element on the surface of the cone is \( dA = z \, dz \, d\theta \).

Step 3: Calculate Center of Mass

The center of mass \( (\bar{x}, \bar{y}, \bar{z}) \) uses the formulas \( \bar{x} = \frac{1}{A} \int x \, dA \), \( \bar{y} = \frac{1}{A} \int y \, dA \), and \( \bar{z} = \frac{1}{A} \int z \, dA \). For symmetry, \( \bar{x} = 0 \) and \( \bar{y} = 0 \). To find \( \bar{z} \), evaluate the integral: \( \bar{z} = \frac{1}{A} \int_0^{2\pi} \int_1^2 z^2 \, dz \, d\theta \).

Step 4: Solve for \( \bar{z} \)

First, calculate \( A = \oint dA = \int_0^{2\pi} \int_1^2 z \, dz \, d\theta = 2\pi [z^2/2]_1^2 = 3\pi \). Now, \( \bar{z} = \frac{1}{3\pi} \cdot 2\pi \cdot \left[\frac{z^3}{3}\right]_1^2 = \frac{7}{3} \).

Step 5: Moment of Inertia about the \(z\)-axis

The moment of inertia about the \(z\)-axis for a surface is \( I_z = \int (x^2 + y^2) \, \delta \, dA \). Here, \( x^2 + y^2 = z^2 \). Thus, \( I_z = \delta \int_0^{2\pi} \int_1^2 z^3 \, dz \, d\theta = \delta \cdot 2\pi [\frac{z^4}{4}]_1^2 = \frac{15\pi\delta}{2} \).

Key Concepts

Moment of Inertia

The moment of inertia is a measure of an object's resistance to rotational motion around an axis. In this exercise, we focus on the moment of inertia about the z-axis for a thin shell of a cone. To calculate it, we use the formula:- \( I_z = \int (x^2 + y^2) \delta \, dA \)Here, \( x^2 + y^2 = z^2 \), reflecting the geometry of the cone. The density \( \delta \) indicates how mass is distributed over the area. Rewriting the integral for our problem, we focus on the shell that lies between \( z = 1 \) and \( z = 2 \). The integral becomes:- \( I_z = \delta \int_0^{2\pi} \int_1^2 z^3 \, dz \, d\theta \)Executing this integral methodically, we find that the shell's moment of inertia about the z-axis is \( \frac{15\pi\delta}{2} \). This value describes how the shell's mass is distributed relative to the z-axis, impacting its rotational dynamics.

Cylindrical Coordinates

Cylindrical coordinates add a third dimension to polar coordinates, useful for expressing three-dimensional surfaces. We represent a point in terms of radial distance \( r \), angular coordinate \( \theta \), and height \( z \). For the cone in this exercise, the relation \( r = z \) provides a neat parametrization where:- \( x = z \cos\theta \)- \( y = z \sin\theta \)These expressions transform the surface into cylindrical coordinates. This allows for simpler integration over the cone surface, particularly when finding areas and volumes. The range of \( z \) given as between 1 and 2 helps to define the limits of integration.

Parametrization

Parametrization involves representing a geometric surface using equations that express surface points as functions of one or more variables. For the cone defined by \( x^2 + y^2 - z^2 = 0 \), the parametrization with cylindrical coordinates is highly effective. We chose:

• \( r = z \)
• \( x = z \cos\theta \)
• \( y = z \sin\theta \)

This choice helps describe every point on the cone and simplifies calculations. By translating the cone's equation into parameters, integration becomes manageable, aiding in determining physical properties like the center of mass or moment of inertia.

Integral Calculus

Integral calculus allows us to build cumulative quantities, like areas and volumes, by summing infinitesimal elements. In this exercise, integral calculus is essential for finding both the center of mass and moment of inertia.- **Center of Mass:** To find \( \bar{z} \), the weighted average along the z-axis, we evaluate: - \( \bar{z} = \frac{1}{A} \int_0^{2\pi} \int_1^2 z^2 \, dz \, d\theta \) - With \( A = \oint dA = 3\pi \), the computed integral reveals that \( \bar{z} = \frac{7}{3} \).- **Area Element:** Integral calculus sums the small area pieces \( dA = z \, dz \, d\theta \) to get the total surface area \( A \), critical for calculating average values like center of mass.These integrals transform geometric data into meaningful physical properties.