Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-22, use spherical coordinates to find the indicated quantity. Mass of the solid inside the sphere \(\rho=b\) and outside the sphere \(\rho=a(a

Step-by-Step Solution

Answer

The mass of the solid is \( \frac{k\pi}{2} (b^4 - a^4) \).

Step 1: Understand Spherical Coordinates

In spherical coordinates, a point is represented by three variables: the radial distance \( \rho \), the polar angle \( \theta \), and the azimuthal angle \( \phi \). Here, the radial distance \( \rho \) is given between two spheres, \( \rho = a \) and \( \rho = b \), with a density \( \delta = k \rho \) where \( k \) is a constant of proportionality.

Step 2: Set Up the Mass Integral

The mass of the solid within the given region can be expressed as the triple integral \( M = \iiint_V \delta(\rho) \, dV \). In spherical coordinates, the volume element \( dV = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi \). The integral for the mass becomes \[ M = \int_0^{2\pi} \int_0^{\pi} \int_a^b k \rho \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta. \]

Step 3: Simplify the Integral

The term \( k \rho \cdot \rho^2 \) simplifies to \( k \rho^3 \). Therefore, the integral is \[ M = k \int_0^{2\pi} \int_0^{\pi} \int_a^b \rho^3 \sin \phi \, d\rho \, d\phi \, d\theta. \]

Step 4: Integrate with Respect to \( \rho \)

Integrate the innermost integral with respect to \( \rho \): \[ \int_a^b \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_a^b = \frac{b^4}{4} - \frac{a^4}{4}. \]

Step 5: Integrate with Respect to \( \phi \)

Now integrate with respect to \( \phi \): \[ \int_0^{\pi} \sin \phi \, d\phi = [-\cos \phi]_0^{\pi} = 2. \]

Step 6: Integrate with Respect to \( \theta \)

Finally, integrate with respect to \( \theta \): \[ \int_0^{2\pi} d\theta = 2\pi. \]

Step 7: Compute the Mass

Substitute the results back to calculate the mass: \[ M = k \cdot (2\pi) \cdot 2 \cdot \left( \frac{b^4}{4} - \frac{a^4}{4} \right) = \frac{k\pi}{2} (b^4 - a^4). \]

Key Concepts

Mass Calculation

Mass calculation in spherical coordinates might seem complex at first, but we can break it down to make it understandable. In this scenario, we aim to find the mass of a solid region located between two concentric spheres. The spheres have radii \( \rho = a \) and \( \rho = b \). The density of this region isn't uniform; rather, it is proportional to the radial distance \( \rho \) from the origin.

• Density is given as \( \delta = k \rho \), where \( k \) is a constant.
• The volume element in spherical coordinates is \( dV = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi \).
• Hence, the mass, \( M \), can be expressed as the integral: \[ M = \int_0^{2\pi} \int_0^{\pi} \int_a^b k \rho^3 \sin \phi \, d\rho \, d\phi \, d\theta. \]

This setup is crucial for evaluating the mass through triple integration.

Density Proportionality

Understanding the concept of density proportionality is essential in this exercise. Here, the density function is not constant but varies with the distance from the origin. This means:

• The density \( \delta \) is directly proportional to the radial coordinate \( \rho \).
• This is expressed mathematically as \( \delta(\rho) = k \rho \), where \( k \) is a proportionality constant.
• Such a density function models the situation where denser regions are further away from the center, which can be common in real-world scenarios like planets or stars.

Knowing the density in this way helps us set the integrals correctly, making our calculations accurate and reflective of the actual physical properties.

Multiple Integrals

Multiple integrals are a way to compute quantities over regions in higher dimensions. In this context:

• We are using a triple integral because our region of interest is three-dimensional.
• The integral's role is to sum up contributions to the mass over the volume between the two spheres.
• Each layer of integration allows us to handle one dimension at a time: \( \rho \) is integrated from \( a \) to \( b \), \( \phi \) from 0 to \( \pi \), and \( \theta \) from 0 to \( 2\pi \).
• Breaking down the problem in this structured way simplifies tracking how each variable contributes to the overall mass.

The use of multiple integrals is key to solving complex geometry and density variations within regions like our concentric spheres.

Triple Integration

In spherical coordinates, triple integration serves as a powerful tool for calculating volumes and other quantities. Let's see how it applies here:

• Initially, we integrate with respect to \( \rho \) to account for the radial distance's varying impact on mass. The integration performed was \[ \int_a^b \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_a^b = \frac{b^4}{4} - \frac{a^4}{4}. \]
• The following step involves integrating with respect to the polar angle, \( \phi \). Here, we evaluated \[ \int_0^{\pi} \sin \phi \, d\phi = [-\cos \phi]_0^{\pi} = 2. \]
• Finally, integrating over the azimuthal angle, \( \theta \), gives us \[ \int_0^{2\pi} d\theta = 2\pi. \]
• Each of these integrations addresses a specific property of the space we're analyzing.

The combined result from these three integrals forms the final expression for the mass. Triple integration in spherical coordinates is essential for precise calculations in volumes with symmetry about a point, such as spherical regions.