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 \(\frac{\pi k}{2} (b^4 - a^4)\).

Step 1: Understand the Sphere and Density Function

We have a solid bounded by two spheres: one with radius \(a\) and another with radius \(b\). The density is given to be proportional to the distance \(\rho\) from the origin, so the density function can be expressed as \(\delta(\rho) = k \rho\), where \(k\) is the proportionality constant.

Step 2: Set Up the Mass Integral

The mass \(M\) of the solid can be calculated by integrating the density over the volume of the solid in spherical coordinates. Therefore, the integral is set up as: \[ M = \int_{\phi = 0}^{\pi} \int_{\theta = 0}^{2\pi} \int_{\rho = a}^{b} k \rho \cdot \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \]This integral accounts for the entire volume of the solid between the two spheres.

Step 3: Integrate with Respect to \(\rho\)

Focus on the integration with respect to \(\rho\): \[ \int_{a}^{b} k \rho^3 \, d\rho \]Evaluating this integral: \[= k \left[ \frac{\rho^4}{4} \right]_{a}^{b} = k \left( \frac{b^4}{4} - \frac{a^4}{4} \right) \]

Step 4: Integrate with Respect to \(\theta\) and \(\phi\)

Now perform the integration over \(\theta\):\[ \int_{0}^{2\pi} \, d\theta = 2\pi \]Then perform the integration over \(\phi\):\[ \int_{0}^{\pi} \sin\phi \, d\phi = 2 \]

Step 5: Compute the Final Mass

Substituting the results into the overall integral:\[ M = 2 \pi \cdot 2 \cdot \frac{k}{4} (b^4 - a^4) \]\[= \frac{\pi k}{2} (b^4 - a^4) \]This is the mass of the solid between the two spheres, accounting for all variables.

Key Concepts

Mass Integral

When dealing with physical bodies, the mass integral allows us to calculate the total mass of a solid object by integrating its density over a given volume. Here, the mass of a solid bounded by two spherical surfaces—one of radius \( a \) and another of radius \( b \) (where \( a < b \))—can be computed.

• In spherical coordinates, you describe any point within these boundaries using \( \rho \) (radial distance), \( \theta \) (azimuthal angle), and \( \phi \) (polar angle).
• These coordinates simplify setting up integrals that involve spherical symmetry, such as mass integrals within spherical shells.

Mass, \( M \), can be expressed through the integral: \[M = \int_{\text{Volume}} \delta(\rho) \, dV,\] where \( \delta(\rho) \) is the density function and \( dV \) is the differential volume element. In spherical coordinates, the differential volume element used is \( \rho^2 \sin\phi \, d\rho \, d\theta \, d\phi \). This integral setup effectively computes the total mass by accumulating all infinitesimal mass contributions over the defined volume.

Density Function

The density function is a critical aspect in calculating the mass of a solid because it describes how mass is distributed throughout the volume.

• In this scenario, density is proportional to the distance from the origin, represented mathematically as \( \delta(\rho) = k \rho \).
• Here, \( k \) is a constant of proportionality, meaning the density increases linearly with distance from the center.

Understanding the density function helps us express the mass in terms of \( \rho \). By setting up the integral with this specific density function, we can capture how mass changes as a function of position in the solid. This linear relationship with \( \rho \) is often chosen in problems with spherical symmetry to simplify the model.

Calculus Problem Solving

Calculus provides us with tools to solve problems involving continuous change and accumulation. In scenarios like these, we leverage integration to evaluate total mass derived from a given density function.

• First, a detailed understanding of the objects in question—two concentric spheres in this case—is crucial.
• Next, setting up the proper integral, using the correct limits for \( \rho \), \( \theta \), and \( \phi \), allows us to model the problem accurately.
• Integration, a core calculus concept, helps us find quantities like mass, volume, or charge by summing continuous distributions.

By meticulously integrating across each coordinate: \( \rho \), then \( \theta \), and finally \( \phi \), we can solve for the total mass. Each step involves evaluating an integral within specified limits until you achieve a complete solution.

Triple Integral in Spherical Coordinates

A triple integral in spherical coordinates is an extension of the double integral but includes an additional dimension. It is particularly useful for objects with symmetry in three dimensions like spheres.

• The triple integral has three sequential integrations—one for each of the spherical variables: \( \rho \), \( \theta \), and \( \phi \).
• The integration limits vary based on the geometry or constraints of the physical scenario.

For our given problem, the setup is: \[\int_{\phi = 0}^{\pi} \int_{\theta = 0}^{2\pi} \int_{\rho = a}^{b} k \rho^3 \sin\phi \, d\rho \, d\theta \, d\phi,\] where \( \rho^3 \sin\phi \) accounts for contributions to each volume element. This setup helps us achieve an accurate model for the total mass of the solid within the given spherical boundaries. By executing these integrals sequentially, starting from \( \rho \), to \( \theta \), and finally to \( \phi \), we successfully compute the mass in a methodical and mathematically effective manner.