Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use spherical coordinates to find the indicated quantity. For any homogeneous solid \(S\), show that the average value of the linear function \(f(x, y, z)=a x+b y+c z+d\) on \(S\) is \(f(\bar{x}, \bar{y}, \bar{z})\), where \((\bar{x}, \bar{y}, \bar{z})\) is the center of mass.

Step-by-Step Solution

Answer

The average value of the function on the solid is \( f(\bar{x}, \bar{y}, \bar{z}) \), where \((\bar{x}, \bar{y}, \bar{z})\) is the center of mass.

Step 1: Understanding the Problem

We are asked to find the average value of a linear function on a homogeneous solid and show it equals the value of the function at the center of mass. The function is given by: \( f(x,y,z) = ax + by + cz + d \). Spherical coordinates are needed to express the volume integrals.

Step 2: Define Spherical Coordinates

Convert the Cartesian coordinates \((x, y, z)\) into spherical coordinates \((\rho, \theta, \phi)\): \( x = \rho \sin\phi \cos\theta \), \( y = \rho \sin\phi \sin\theta \), \( z = \rho \cos\phi \). The volume element in spherical coordinates is \( dV = \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \).

Step 3: Set Up the Integral for Average Value

The average value of \( f(x, y, z) \) over \( S \) is obtained by integrating the function over \( S \), divided by the volume of \( S \): \[ \text{Average} = \frac{1}{V} \int_S (ax + by + cz + d) \, dV \] where \( V \) is the total volume of \( S \).

Step 4: Express Integrals in Spherical Coordinates

Substitute the expressions for \( x, y, z \) into the integral: \[ \int_S ax + by + cz + d \, dV = \int_S \left[ a(\rho \sin\phi \cos\theta) + b(\rho \sin\phi \sin\theta) + c(\rho \cos\phi) + d \right] \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \]

Step 5: Simplify and Separate the Integrals

Separate the integrals for different parts: \[ a\int_S x \, dV + b\int_S y \, dV + c\int_S z \, dV + dV \] The first three integrals are related to the center of mass, \( \bar{x} \), \( \bar{y} \), and \( \bar{z} \).

Step 6: Use Center of Mass Definition

The center of mass \((\bar{x}, \bar{y}, \bar{z})\) of a homogeneous solid is \[ \bar{x} = \frac{1}{V} \int_S x \, dV \] \[ \bar{y} = \frac{1}{V} \int_S y \, dV \] \[ \bar{z} = \frac{1}{V} \int_S z \, dV \] Use these in the average to get \[ \frac{1}{V} (aV\bar{x} + bV\bar{y} + cV\bar{z} + dV) \]

Step 7: Conclude the Calculation

After the integration and substitution, the average value becomes \[ a\bar{x} + b\bar{y} + c\bar{z} + d = f(\bar{x}, \bar{y}, \bar{z}) \] Hence, the average value of \( f(x, y, z) \) on \( S \) is indeed the function evaluated at the center of mass.

Key Concepts

Average Value

The average value is a fundamental concept, often used to represent the central tendency or the typical value of a function over a specified domain. In the context of a function defined over a solid, like in our exercise, it describes how this function averages out throughout the volume of the solid.
For a homogeneous solid with volume \(V\), the average value \( \text{Average} \) of a function \(f(x, y, z)\) is obtained by dividing the integral of the function over the entire solid by the total volume \(V\) of the solid. The integral accounts for all the small increments of the function, weighted by their presence in each part of the solid.

• The formula for the average value is given by: \[ \text{Average} = \frac{1}{V} \int_S f(x, y, z) \, dV \]

This means you sum up the value of \(f\) across all parts of the solid, then normalize by the size of the solid. In our exercise, we showed that this average corresponds to the value of the function evaluated at the center of mass of the solid, which highlights the symmetry inherent in a homogeneous solid.

Center of Mass

The center of mass is a point that can be considered as the average location of the total mass of an object. For a homogeneous solid, where the density is uniform throughout, this point is particularly significant. It encapsulates the balance point of the solid.
For a solid \(S\), the coordinates of the center of mass \((\bar{x}, \bar{y}, \bar{z})\) are calculated by integrating the moments about each coordinate axis and dividing by the total mass or volume. Because the solid is homogeneous, we are essentially looking at the average positional vectors.

• The equations for finding the center of mass are given by: \[ \bar{x} = \frac{1}{V} \int_S x \, dV, \ \bar{y} = \frac{1}{V} \int_S y \, dV, \ \bar{z} = \frac{1}{V} \int_S z \, dV \]

Once you calculate these, you can determine the 'average' center point of the solid. This is particularly useful because it simplifies computations, as seen when equating the average value of the function over the solid to its value at the center of mass.

Volume Integrals

Volume integrals are a powerful mathematical tool used to sum over a three-dimensional region. They are used to calculate properties distributed within a volume, such as mass, charge, or in our case, the average value of a function within a solid.
When working with spherical coordinates, volume integrals transform Cartesian coordinates \((x, y, z)\) into spherical ones \((\rho, \theta, \phi)\), which may often simplify calculations due to the symmetry of the problem.

• The conversion is done through the relationships: \[ x = \rho \sin\phi \cos\theta, \ y = \rho \sin\phi \sin\theta, \ z = \rho \cos\phi \]
• The volume element becomes: \[\ dV = \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \]

This technique simplifies the integral setup and aids in solving complex problems by leveraging the natural symmetry in spherical shapes, making it easier to calculate integrals over spherical volumes or related shapes. The volume integral is crucial for determining the average value of a function in this exercise.

Homogeneous Solid

A homogeneous solid is one whose properties are uniform throughout its volume. This means the density, or the property being measured, is constant at every point within the solid. In mathematical problems, this uniform distribution simplifies calculations significantly.
When dealing with homogeneous solids in physics or math, the uniformity allows certain assumptions, such as considering the center of mass in simpler terms or leveraging symmetry to ease integrals and problem-solving.

• For example: Because the solid is homogeneous, the center of mass calculations simplifies to finding averages of the coordinate positions as there are no variations in density to consider across different points.
• This principle enables questions like our exercise, where we showed that certain integrals equate to these uniform center points.

The notion of homogeneity allows us to focus more on geometric and algebraic properties rather than complex density functions, making problem-solving more straightforward and intuitive.