Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use spherical coordinates to find the indicated quantity. A homogeneous solid sphere of radius \(a\) is centered at the origin. For the section \(S\) bounded by the half-planes \(\theta=-\alpha\) and \(\theta=\alpha\) (like a section of an orange), find each value. (a) \(x\) -coordinate of the center of mass (b) Average distance from the \(z\) -axis

Step-by-Step Solution

Answer

(a) \( x_{cm} = \frac{3a\sin \alpha \pi}{4\alpha} \) (b) \( r_{avg} = \frac{3a\pi}{4} \)

Step 1: Understand the Problem

We're given a solid sphere with a section between two half-planes in spherical coordinates. We need to find the center of mass's x-coordinate and the average distance from the z-axis of this section.

Step 2: Geometric Setup in Spherical Coordinates

The radius of the sphere is \(a\). In spherical coordinates, any point is represented as \(( ho, \phi, \theta)\) where \(\rho\) is the radial distance, \(\phi\) is the polar angle, and \(\theta\) is the azimuthal angle. The range for \(\theta\) is \(-\alpha \leq \theta \leq \alpha\).

Step 3: Find the Volume Element in Spherical Coordinates

The differential volume element in spherical coordinates is \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). This accounts for the geometry of spherical coordinates.

Step 4: Set Up the Mass and Center of Mass Equations

The sphere is homogeneous, so the density is constant. The total mass \(M\) of the section is given by:\[ M = \int_{0}^{a} \int_{0}^{\pi} \int_{-\alpha}^{\alpha} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \].

Step 5: Integrate to Find the Total Mass

\[ M = \int_{0}^{a} \rho^2 \, d\rho \int_{0}^{\pi} \sin \phi \, d\phi \int_{-\alpha}^{\alpha} d\theta \]Solving this gives:\[ M = \left. \frac{\rho^3}{3} \right|_0^a \left. -\cos \phi \right|_0^{\pi} \cdot (2\alpha) = \frac{2}{3} \cdot 2a^3 \cdot \alpha = \frac{4}{3}a^3 \alpha \]

Step 6: Find the x-Component of the Center of Mass

The x-component of the center of mass is:\[ x_{cm} = \frac{1}{M} \int \int \int x \, dV \]Using \(x = \rho \sin \phi \cos \theta\), we set up the integral:\[ x_{cm} = \frac{1}{M} \int_{0}^{a} \int_{0}^{\pi} \int_{-\alpha}^{\alpha} \rho^3 \sin^2 \phi \cos \theta \, d\rho \, d\phi \, d\theta \].

Step 7: Evaluate the Integral for x-Component

\[ x_{cm} = \frac{3}{4 \alpha a^3} \left(a^4 \int_{0}^{\pi} \sin^2 \phi \, d\phi \right) \int_{-\alpha}^{\alpha} \cos \theta \, d\theta \]\[ \int_{0}^{\pi} \sin^2 \phi \, d\phi = \frac{\pi}{2} \] and \(\int_{-\alpha}^{\alpha} \cos \theta \, d\theta = 2 \sin \alpha\)This results in \( x_{cm} = \frac{3a\sin \alpha}{2\alpha} \cdot \left(\frac{\pi}{2}\right)\).

Step 8: Average Distance from the z-Axis

The average distance \(r\) from the z-axis is given by:\[ r_{avg} = \frac{1}{M} \int \int \int \rho \sin \phi \, dV \]This becomes \[ r_{avg} = \frac{1}{M} \int_{0}^{a} \int_{0}^{\pi} \int_{-\alpha}^{\alpha} \rho^3 \sin^2 \phi \, d\rho \, d\phi \, d\theta \].

Step 9: Evaluate the Integral for Average Distance

Evaluating as previously done, \( r_{avg} \) results in the same formula as the x-component without the cosine term:\[ r_{avg} = \frac{3 a \pi}{4} \]

Key Concepts

Center of Mass

The center of mass of a body is a point that acts as if the whole mass of the body is concentrated at it. For the section of the sphere described in the exercise, we need to use integrals to find the x-coordinate of this point.

In spherical coordinates, the x-coordinate of any point is given by the formula:

• \( x = \rho \sin \phi \cos \theta \)

To find the x-coordinate of the center of mass, we integrate over the entire volume of the section of the sphere. This allows us to compute the average position of the x-value weighted by mass.

Since the sphere is homogeneous, density is constant, simplifying our calculations. The formula used is:

• \( x_{cm} = \frac{1}{M} \int \int \int x \, dV \)

Where \( M \) represents the total mass of the section. By substituting the spherical coordinates and following through the steps of integration, we achieve the exact center of mass for this specific section of the sphere.

Average Distance

The average distance from the z-axis is a measure of how far points within the spherical section are, on average, from the z-axis. This concept is crucial in understanding the distribution of mass within the sphere section.

To calculate this distance, we utilize the formula in spherical coordinates:

• \( r = \rho \sin \phi \)

Similar to the center of mass, the average distance requires integrating over the spherical section. We use the same volume element and limits as determined in spherical coordinates.

We then apply the integration to find the average radial distance:

• \( r_{avg} = \frac{1}{M} \int \int \int \, r \, dV \)

This involves integrating \( \rho^3 \sin^2 \phi \) throughout the sphere's section, aligning it with the geometrical scope needed for a thorough spatial understanding of the average radial distance from the z-axis.

Volume Element

In spherical coordinates, the concept of a volume element is vital for evaluating integrals over a sphere or a spherical section.

The volume element \( dV \) in spherical coordinates is given by:

• \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \)

This expression accounts for the way a small piece of the sphere is projected in spherical geometry, differing significantly from rectangular coordinates.

The expression \( \rho^2 \sin \phi \) comes from the Jacobian of the transformation from Cartesian to spherical coordinates. Here, \( \rho \) measures the radial component, \( \phi \) accounts for latitudinal inclination, and \( \theta \) provides azimuthal variation. These variables allow the volume element to cover every part of the sphere comprehensively as you integrate across it, ensuring accurate calculations of mass, center of mass, or other similar properties.

Integral Evaluation

Integral evaluation is the mathematical process of solving integrals to find solutions to problems involving continuous quantities, like mass or distance, within defined boundaries.

In the exercise, integral evaluation plays a crucial role in calculating both the x-coordinate of the center of mass and the average distance from the z-axis. By setting up the right triple integrals, we effectively accumulate quantities over the entire volume of the spherical section.

We start with understanding the limits of integration:

• \( \rho \) ranges from 0 to the radius \( a \)
• \( \phi \) varies from 0 to \( \pi \)
• \( \theta \) ranges from \(-\alpha \) to \( \alpha \)

Each of these steps transforms the problem into more manageable parts.

The integration solves both for total mass and the desired distance metrics. Calculations may seem complex, but by breaking them down into layers and recognizing the symmetry and boundaries involved, integral evaluation becomes an insightful tool for extracting the necessary quantities from spherical objects.