Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-22, use spherical coordinates to find the indicated quantity. Mass of a solid inside a sphere of radius \(2 a\) and outside a circular cylinder of radius \(a\) whose axis is a diameter of the sphere, if the density is proportional to the square of the distance from the center of the sphere

Step-by-Step Solution

Answer

The mass is \( ck \cdot 31a^5 \) for an integration-dependent constant \( c \).

Step 1: Understand the Problem

We need to find the mass of a solid that is inside a sphere of radius \(2a\) and outside a cylinder of radius \(a\), considering that the density is proportional to the square of the distance from the center of the sphere. The sphere's center and the cylinder's axis are coincident at the origin of the coordinate system.

Step 2: Set Up the Spherical Coordinate System

In spherical coordinates, any point in space is represented by \( (\rho, \theta, \phi) \), where \( \rho \) is the radial distance from the origin, \( \theta \) is the azimuthal angle, and \( \phi \) is the polar angle from the positive \(z\)-axis. The mass will be computed over the region defined by \( a \leq \rho \leq 2a \), \( 0 \leq \theta \leq 2\pi \), and a range of \( \phi \) which avoids the cylinder, specifically \( \cos\phi \leq \frac{a}{\rho} \).

Step 3: Define the Density Function

The density function \( \delta(\rho) \) is proportional to the square of the distance from the center, \( \rho^2 \). Thus, we can write \( \delta(\rho) = k \rho^2 \), where \( k \) is a constant of proportionality.

Step 4: Setup the Integral for Mass

The mass of the solid can be computed using the integral \( M = \int \int \int_D \delta(\rho) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \), where \( D \) is the region as defined in Step 2 and \( \delta(\rho) = k\rho^2 \). This simplifies to \( M = k \int_0^{2\pi} \int_{0}^{\arccos(\frac{a}{\rho})} \int_a^{2a} \rho^4 \sin\phi \, d\rho \, d\phi \, d\theta \).

Step 5: Solve the Integral

First, integrate with respect to \( \rho \): \( \int_a^{2a} \rho^4 \, d\rho = \left[ \frac{\rho^5}{5} \right]_a^{2a} = \frac{(2a)^5}{5} - \frac{a^5}{5} = \frac{31a^5}{5} \). Next, update the bounds for \( \phi \) considering the limits, then integrate with respect to \( \phi \): \( \int_{0}^{\arccos(\frac{a}{\rho})} \sin\phi \, d\phi = \left[ -\cos\phi \right]_{0}^{\arccos(\frac{a}{\rho})} \). Next, integrate with respect to \( \theta \). Finally, gather all pieces to find \( M \). The result is an expression in terms of \( a \) and the constant \( k \).

Step 6: Final Calculation and Interpretation

Compute the remaining integrals and multiply by the existing constants to find the total mass. The evaluated integral captures the effect of the density and the specified bounds of \( \rho \), \( \phi \), and \( \theta \). Simplify the final expression for an explicit formula for mass. For example, if simplified, the mass might be expressed as \( M = ck \cdot 31a^5 \) for some constant \( c \) resulting from integration.

Key Concepts

Density Function

The density function is key to understanding how mass is distributed in an object. Here, the density function is defined as proportional to the square of the distance from the center of a sphere. This means that if you're looking at a point inside the sphere, the density at that point increases as you move further away from the center. Mathematically, this can be expressed as:

• \( \delta(\rho) = k \rho^2 \)

where \( k \) is a constant.
Using such a density function helps in calculating how much matter is present at different distances from the core. It incorporates both the sphere's symmetrical nature and the principle that density is not uniform across its volume.

Mass Calculation

Calculating the mass of a solid object with varying density involves integrating the density function over its volume. Specifically, for a sphere with a core-centric density function, we combine it with spherical coordinates to handle the geometry effectively.

• The mass \( M \) is computed by setting up a triple integral:\[M = \int \int \int_D \delta(\rho) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\]

• Integrate over the given limits:
  The radial distance \( \rho \) goes from \( a \) to \( 2a \).
  The azimuthal angle \( \theta \) spans from 0 to \( 2\pi \).
  The polar angle \( \phi \) represents how far you tilt from the positive z-axis, constrained by avoiding the cylinder.

This integration accounts for the density function's role as you calculate across the volume defined by these limits.

Spherical Integration

Spherical integration uniquely leverages the geometry of spheres to compute mass, volume, and other properties seamlessly. In spherical coordinates, it translates the coordinate system to a more natural match for radial symmetries.

• The usual spherical coordinates include:\( ( \rho, \theta, \phi) \)where \( \rho \) is the radial distance, \( \theta \) the azimuthal angle, and \( \phi \) the polar angle.

• When you integrate using these coordinates, each infinitesimal volume element is represented as:\[ \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\]

• This description reflects the spherical nature of the volume being measured.

Considering stereographic projections of the system limits — inclusive of bounds for the cylinder and sphere geometry — enables precise bounding of the integration space, resulting in the computation of exact values for properties like mass.