Problem 33 - APEX Calculus

Question

A solid is described along with its density function. Find the mass of the solid using spherical coordinates. The conical region bounded above \(z=\sqrt{x^{2}+y^{2}}\) and below the sphere \(x^{2}+y^{2}+z^{2}=1\) with density function \(\delta(x, y, z)=z\).

Step-by-Step Solution

Answer

The mass of the solid is \( \frac{\pi}{8} \).

Step 1: Identify the Boundaries of the Solid Region

The solid lies between the cone defined by \( z = \sqrt{x^2 + y^2} \) and the sphere \( x^2 + y^2 + z^2 = 1 \). In spherical coordinates, the cone becomes \( \phi = \frac{\pi}{4} \) and the sphere becomes \( \rho = 1 \). Thus, the boundaries in spherical coordinates are \( \phi \leq \frac{\pi}{4} \) and \( \rho \leq 1 \).

Step 2: Set Up the Integral in Spherical Coordinates

Convert the density function \( \delta(x, y, z) = z \) into spherical coordinates: \( z = \rho \cos \phi \). The volume element in spherical coordinates is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). Therefore, the mass integral becomes \( M = \int_0^{2\pi} \int_0^{\pi/4} \int_0^1 \rho \cos \phi \, \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).

Step 3: Integrate Over \( \rho \)

Integrate \( \int_0^1 \rho^3 \cos \phi \sin \phi \, d\rho \). As \( \, \cos \phi \sin \phi \, \) does not depend on \( \rho \), factor it out: \( \cos \phi \sin \phi \int_0^1 \rho^3 \, d\rho = \cos \phi \sin \phi \left[ \frac{\rho^4}{4} \right]_0^1 = \cos \phi \sin \phi \cdot \frac{1}{4} \).

Step 4: Integrate Over \( \phi \)

Integrate \( \int_0^{\pi/4} \frac{1}{4} \cos \phi \sin \phi \, d\phi \). This integral can be simplified using the identity \( \sin 2\phi = 2\sin \phi \cos \phi \), so: \( \frac{1}{8} \int_0^{\pi/4} \sin 2\phi \, d2\phi = \frac{1}{16} \int_0^{\pi/2} \sin u \, du = \frac{1}{16} \left[-\cos u\right]_0^{\pi/2} = \frac{1}{16} (0 + 1) = \frac{1}{16} \).

Step 5: Integrate Over \( \theta \)

Finally, integrate \( \int_0^{2\pi} \frac{1}{16} \, d\theta = \frac{1}{16} [\theta]_0^{2\pi} = \frac{1}{16} (2\pi - 0) = \frac{\pi}{8} \).

Step 6: Conclusion: Calculate the Mass

The total mass of the solid is \( \frac{\pi}{8} \).

Key Concepts

Density Function

The density function, denoted as \( \delta(x, y, z) \), is a mathematical representation of how mass is distributed throughout a solid object. In this scenario, the density function is given as \( \delta(x, y, z) = z \), which means that the density changes with the z-coordinate. Higher values of \( z \) lead to higher densities.
This concept is essential because it affects how the mass integral is set up, as variations in density will affect the total mass of the solid. Understanding the density function allows us to determine how density is distributed across different parts of a solid.

Mass Integral

The mass integral is a mathematical calculation used to determine the total mass of a solid with a given density distribution. For solids described in spherical coordinates, the mass integral incorporates the density function along with the volume element.
The integral setup in our case is: \[ M = \int_0^{2\pi} \int_0^{\pi/4} \int_0^1 \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta \]
This integral uses the density function \( \delta(x, y, z) = z \), transformed to spherical coordinates. The result of this integration process will yield the total mass by combining all the contributions at various points in the solid.

Volume Element

A volume element is a small "piece" of volume used in integration to sum up infinitely small sections of a solid. In spherical coordinates, the volume element \( dV \) is defined as \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).
This specific element accounts for changes in volume based on the radial distance \( \rho \), inclination angle \( \phi \), and azimuthal angle \( \theta \). Usually, these volume elements are multiplied by the density function to determine the contribution of each small section to the total mass.

• The term \( \rho^2 \) accounts for the change in volume as we move away from the origin.
• \( \sin \phi \) adjusts for the position relative to the horizontal plane.

By using this volume element, we faithfully capture the structure and volume of the solid during integration.

Triple Integration

Triple integration is the process of integrating a function of three variables, such as \( x, y, z \) or in spherical coordinates \( \rho, \phi, \theta \). It is used to calculate the integral over three-dimensional spaces.
In this exercise, the triple integral is used to find the mass of a solid by integrating over its entire volume. We perform integration sequentially over \( \rho \), \( \phi \), and \( \theta \). Each integration affects the next, step-by-step reducing the dimensionality until we achieve a singular, scalar value representing total mass.
The progression is:
1. Integrate with respect to \( \rho \).2. Integrate with respect to \( \phi \).3. Finally, integrate with respect to \( \theta \).
This structured approach is fundamental in calculus for handling complex three-dimensional calculations, ensuring that all subtleties of the geometry of the solid are taken into account.