Problem 36 - APEX Calculus

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 spherical shell bounded between \(x^{2}+y^{2}+z^{2}=16\) and \(x^{2}+y^{2}+z^{2}=25\) with density function \(\delta(x, y, z)=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\)

Step-by-Step Solution

Answer

The center of mass is at (0, 0, 0).

Step 1: Convert to Spherical Coordinates

In spherical coordinates, we have the transformations: \[ x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi \]where \( \rho \) is the radial distance, \( \phi \) is the polar angle (angle from the positive \( z \)-axis), and \( \theta \) is the azimuthal angle (angle in the \( x-y \) plane from the positive \( x \)-axis). For our spherical shell, \( 16 \leq \rho^2 \leq 25 \) or \( 4 \leq \rho \leq 5 \). The angles range \( 0 \leq \phi \leq \pi \) and \( 0 \leq \theta < 2\pi \).

Step 2: Express Density Function

The density function \( \delta(x, y, z) = \sqrt{x^2 + y^2 + z^2} \) in spherical coordinates is just \( \rho \) because \(\sqrt{(\rho \sin\phi \cos\theta)^2 + (\rho \sin\phi \sin\theta)^2 + (\rho \cos\phi)^2} = \rho \).

Step 3: Formula for Mass and Integrals

The mass \( M \) of the solid can be computed using the integral:\[M = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\] which simplifies to:\[ M = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^3 \sin\phi \, d\rho \, d\phi \, d\theta \]

Step 4: Compute Mass Integral

Compute the integral step-by-step:- Integrate with respect to \( \rho \):\[ \int_4^5 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_4^5 = \frac{5^4}{4} - \frac{4^4}{4} = \frac{625}{4} - \frac{256}{4} = \frac{369}{4} \]- Integrate with respect to \( \phi \):\[ \int_0^{\pi} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi} = -(-1 + 1) = 2 \]- Integrate with respect to \( \theta \):\[ \int_0^{2\pi} 1 \, d\theta = [\theta]_0^{2\pi} = 2\pi \]Thus, the total mass \( M \) is:\[ M = \frac{369}{4} \times 2 \times 2\pi = \frac{369\pi}{2} \]

Step 5: Compute Moments for Center of Mass

Now, compute moments to find the center of mass:- The moment about the \( yz \)-plane is \( Mx = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^4 \sin\phi \cos\theta \, d\rho \, d\phi \, d\theta = 0 \) because integrating \( \cos\theta \) from \( 0 \) to \( 2\pi \) results in zero.- Similarly, \( My = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^4 \sin\phi \sin\theta \, d\rho \, d\phi \, d\theta = 0 \) for the same reason.- \( Mz = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^4 \sin\phi \cos\phi \, d\rho \, d\phi \, d\theta \) becomes:\[\left( \frac{369}{4} \right) \times \int_0^{\pi} \sin\phi\cos\phi \, d\phi \times \int_0^{2\pi} 1 \, d\theta = 0\],as integrating \( \sin(2\phi)/2 \) from \( 0 \) to \( \pi \) yields zero.

Step 6: Center of Mass Calculation

The center of mass, \( \bar{x}, \bar{y}, \bar{z} \) is given by\( \bar{x} = \frac{Mx}{M}, \bar{y} = \frac{My}{M}, \bar{z} = \frac{Mz}{M} \).Since all moments were 0, \( \bar{x} = \bar{y} = \bar{z} = \frac{0}{\frac{369\pi}{2}} = 0 \). Therefore, the center of mass is at the origin (0, 0, 0).

Key Concepts

Spherical Coordinates

When discussing the center of mass, spherical coordinates provide an incredibly useful way to deal with problems involving objects with circular symmetry. Spherical coordinates transform 3D points from Cartesian coordinates using three parameters: \( \rho \), \( \phi \), and \( \theta \).

Here's a quick breakdown of these parameters:

• \( \rho \) is the radial distance from the origin to the point.
• \( \phi \) (or polar angle) is the angle from the positive z-axis.
• \( \theta \) (or azimuthal angle) is the angle in the x-y plane from the positive x-axis.

Using the formulas, a point (x, y, z) converts to spherical coordinates as follows:

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

Remember, it is often easier to calculate integrals and other properties of spherical objects using spherical coordinates rather than Cartesian coordinates. By setting boundaries in \( \rho \), \( \phi \), and \( \theta \), you can precisely define any spherical region or shell.

Density Function

The density function in physics describes how mass is distributed within a particular object. For the given spherical shell, the density is set by the function \( \delta(x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}} \).

In spherical coordinates, this function simplifies beautifully to \( \rho \). This simplification occurs because:

• The Cartesian expression \( \sqrt{x^{2} + y^{2} + z^{2}} \) directly corresponds to the distance from the origin, precisely what \( \rho \) measures.

Thus, when working with problems involving density and spherical coordinates, properties like mass, volume, and center of mass become convenient to calculate. With \( \rho \) serving as the density function, the computations conclude smoothly when integrated over the defined limits.

Mass Integral

Calculating the mass of an object with varying density requires performing volume integration. For spherical coordinates, mass can be found with a triple integral that incorporates radius \( \rho \), density \( \rho \), and the geometric terms for spherical coordinates.

The formula for the mass \( M \) is given by:\[ M = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^3 \sin\phi \, d\rho \, d\phi \, d\theta \]
This formula is derived from:

• \( \rho^2 \) as a substitution for the volume element of a sphere in spherical coordinates.
• Multiplying by \( \rho \) because it serves as the density function.

Each step involves integrating over a different variable in sequence \( \rho \), \( \phi \), and finally \( \theta \). Successfully completing these integrations yields the total mass M, which is crucial for computing the center of mass.

Moments Calculation

Moments are critical for determining the position of the center of mass in any solid. In three-dimensional space, the center of mass is calculated using moments about the coordinate planes.

Here's a simple overview of moments:

• Moment about the yz-plane is denoted \( Mx \).
• Moment about the xz-plane is denoted \( My \).
• Moment about the xy-plane is denoted \( Mz \).

The calculations involve evaluating integrals:\[ Mx = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^4 \sin\phi \cos\theta \, d\rho \, d\phi \, d\theta \]\[ My = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^4 \sin\phi \sin\theta \, d\rho \, d\phi \, d\theta \]\[ Mz = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^4 \sin\phi \cos\phi \, d\rho \, d\phi \, d\theta \]
Because of symmetry for a full spherical shell, these moments often simplify. The integrals yield zero for the x and y components due to their corresponding cosine and sine functions integrating to zero over a complete periodic interval. Thus, these computations ultimately provide coordinates for the center of mass.