Question

A solid is described along with its density function. Find the center of mass of the solid using spherical coordinates. (Note: these are the same solids and density functions as found in Exercises 31 through \(34 .)\) The upper half of the unit ball, bounded between \(z=0\) and \(z=\sqrt{1-x^{2}-y^{2}},\) with density function \(\delta(x, y, z)=1\)

Short answer

The center of mass is at \((0, 0, \frac{3}{8})\)."}

Step-by-step solution

Define the Solid in Spherical Coordinates

The unit ball in spherical coordinates is defined with the ranges: - Radius: \(0 \leq \rho \leq 1\) - Azimuthal Angle: \(0 \leq \phi \leq \pi/2\) (since it's the upper half)- Polar Angle: \(0 \leq \theta < 2\pi\).

Express Variables in Spherical Coordinates

In spherical coordinates, the relationships are: - \(x = \rho \sin(\phi) \cos(\theta)\)- \(y = \rho \sin(\phi) \sin(\theta)\)- \(z = \rho \cos(\phi)\)

Set Up the Integral for Mass of the Solid

Since the density function is constant \(\delta(x, y, z) = 1\), the mass \(M\) of the solid is given by the integral: \[ M = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \]

Calculate the Mass Integral

Calculate the integrals separately:1. \(\int_{0}^{1} \rho^2 \, d\rho = \frac{1}{3}\)2. \(\int_{0}^{\pi/2} \sin(\phi) \, d\phi = 1\)3. \(\int_{0}^{2\pi} d\theta = 2\pi\)Combine to find the mass:\[ M = \frac{1}{3} \times 1 \times 2\pi = \frac{2\pi}{3} \]

Set Up the Integrals for the Center of Mass

Calculate the moments about each axis (\(M_x\), \(M_y\), \(M_z\)) using the integral formulas:- \(M_x = \int \int \int x \delta(x,y,z) \cdot dV\)- \(M_y = \int \int \int y \delta(x,y,z) \cdot dV\)- \(M_z = \int \int \int z \delta(x,y,z) \cdot dV\)

Solve for the Center of Mass Coordinates

The center of mass coordinates \((\overline{x}, \overline{y}, \overline{z})\) are given by:\[ \overline{x} = \frac{1}{M} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} (\rho \sin(\phi) \cos(\theta)) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \]\[ \overline{y} = \frac{1}{M} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} (\rho \sin(\phi) \sin(\theta)) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \]\[ \overline{z} = \frac{1}{M} \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} (\rho \cos(\phi)) \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \]

Calculation of Integrals for Center of Mass

Due to symmetry, \(\overline{x} = 0\) and \(\overline{y} = 0\). For \(\overline{z}\), calculate:\[ \int_{0}^{2\pi} d\theta = 2\pi \]\[ \int_{0}^{\pi/2} \cos(\phi) \sin(\phi) \, d\phi = \frac{1}{2} \]\[ \int_{0}^{1} \rho^3 \, d\rho = \frac{1}{4} \]Combine:\[ \overline{z} = \frac{1}{M} (2\pi) \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) = \frac{1}{M} \times \frac{\pi}{4} = \frac{3}{8} \]

Conclusion on the Center of Mass

The center of mass of the upper half of the unit ball is located at \((0, 0, \frac{3}{8})\).

Key concepts

Spherical Coordinates

Spherical coordinates are a way to represent points in three-dimensional space using three parameters: radius \(\rho\), azimuthal angle \(\phi\), and polar angle \(\theta\). These coordinates are particularly useful for problems involving spheres or spherical symmetry. Here’s how they work in detail:

• Radius \(\rho\): This is the distance from the origin to the point. In the case of a unit sphere, this ranges from 0 to 1.
• Azimuthal Angle \(\phi\): This angle is measured from the positive z-axis down to the point. For the top half of a sphere, this ranges from 0 to \(\pi/2\).
• Polar Angle \(\theta\): This is the angle in the xy-plane, measured from the positive x-axis. It ranges from 0 to \(2\pi\).

Together, these coordinates allow us to easily describe the upper half of a unit ball in mathematical terms and simplify calculations like finding the center of mass. The transformation from Cartesian coordinates is provided as:

• \(x = \rho \sin(\phi) \cos(\theta)\)
• \(y = \rho \sin(\phi) \sin(\theta)\)
• \(z = \rho \cos(\phi)\)

Density Function

A density function in the context of calculating mass and center of mass of a solid is a mathematical function that describes how mass is distributed in space. For this specific problem, the density function is given as \(\delta(x, y, z) = 1\). Here’s what that implies in simple terms:

• Uniform Density: Since the density is constant (1), mass is uniformly spread throughout the volume of the solid. This simplifies calculations as the density won't change at different points within the solid.
• Physical Interpretation: The density function could represent various physical properties, like matter or energy per unit volume. In this case, it means that every tiny volume of our solid has the same amount of mass.

Mass Integral

To find the total mass of a solid using spherical coordinates, we must integrate the density over the entire volume of the solid.
This involves setting up a triple integral that accounts for all possible positions within our defined boundaries. For our problem, the mass \(M\) is calculated with the integral:
\[ M = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \]
This equation can be broken down into several parts:

• The integral over \(\rho\) considers the change in radius, from the center of the sphere to its surface.
• The integral over \(\phi\) accounts for the vertical angle from the z-axis to the xy-plane.
• The integral over \(\theta\) considers the rotation around the z-axis.

Individually solving these integrals step by step gives the total mass:

• \(\int_{0}^{1} \rho^2 \, d\rho = \frac{1}{3}\)
• \(\int_{0}^{\pi/2} \sin(\phi) \, d\phi = 1\)
• \(\int_{0}^{2\pi} d\theta = 2\pi\)

When combined, these integrals give the total mass \(M = \frac{2\pi}{3}\). This mass forms the basis for further calculations of the center of mass.

Center of Mass Calculation

Finding the center of mass of a solid is about determining where the average position of mass lies. In spherical coordinates, this involves calculating three separate moments for each axis and using these to find the center of mass coordinates. Let’s explore this a bit deeper:

• Moments: These are integrals that give us weighted average positions along the x, y, and z axes, considering both density and volume.
• Integrals for Center of Mass: Each coordinate of the center of mass \((\overline{x}, \overline{y}, \overline{z})\) is calculated using its specific integral:
  - \(\overline{x} = \frac{1}{M} \int x \, dV\), due to symmetry, \(\overline{x} = 0\)
  - \(\overline{y} = \frac{1}{M} \int y \, dV\), similarly, \(\overline{y} = 0\)
  - \(\overline{z} = \frac{1}{M} \int z \, dV\)

For the z-coordinate, we use:
\[ \int_{0}^{2\pi} d\theta = 2\pi \]
\[ \int_{0}^{\pi/2} \cos(\phi) \sin(\phi) \, d\phi = \frac{1}{2} \]
\[ \int_{0}^{1} \rho^3 \, d\rho = \frac{1}{4} \]
Combing these results, you find \(\overline{z} = \frac{3}{8}\). Thus, the center of mass of the solid lies at the point \((0, 0, \frac{3}{8})\), reflecting the symmetry present in this upper half of the sphere.