Problem 38 - 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 cone bounded above \(z=\sqrt{x^{2}+y^{2}}\) and below the plane \(z=1\) with density function \(\delta(x, y, z)=z\).

Step-by-Step Solution

Answer

The center of mass is located at \((0, 0, \bar{z})\), where \(\bar{z}\) is found through simplified integral calculations due to symmetry.

Step 1: Convert the Given Surfaces to Spherical Coordinates

The given surfaces are the cone described by \( z = \sqrt{x^2 + y^2} \) and the plane \( z = 1 \). In spherical coordinates, \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), and \( z = \rho \cos \phi \). Therefore, the cone condition becomes \( \rho \cos \phi = \rho \sin \phi \), which simplifies to \( \tan \phi = 1 \) or \( \phi = \frac{\pi}{4} \). The plane \( z = 1 \) means \( \rho \cos \phi = 1 \).

Step 2: Determine the Limits of Integration

Given the conversion conditions, the limits of integration for the spherical coordinates will be:- \( \theta \) ranges from \( 0 \) to \( 2\pi \) (as the solid is symmetrical around the z-axis),- \( \phi \) ranges from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \) (since below the plane is the cone),- \( \rho \) ranges from \( \sec \phi \) to a value determined by the intersection of the cone and plane, which simplifies here since \( \rho = \cos^{-1}(\phi) \).

Step 3: Express the Density Function in Spherical Coordinates

The density function given is \( \delta(x, y, z) = z \). In spherical coordinates, this becomes \( \delta(\rho, \phi, \theta) = \rho \cos \phi \).

Step 4: Set Up the Triple Integral for Mass

The mass \( M \) of the solid is given by \[ M = \int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_{\sec \phi}^{1} \rho \cos \phi \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]Multiply \( \rho \cos \phi \) by \( \rho^2 \sin \phi \), and integrate to find the mass of the solid.

Step 5: Evaluate the Integral for Mass

We compute the integral:- Integrating with respect to \( \rho \) first: \[ \int_{\sec \phi}^{1} \rho^3 \cos \phi \sin \phi \, d\rho = \cos \phi \sin \phi \left[ \frac{1}{4}\rho^4 \right]_{\sec \phi}^{1} = \cos \phi \sin \phi \left( \frac{1}{4}(1 - \sec^4 \phi) \right) \]- Then integrate with respect to \( \phi \) and \( \theta \).

Step 6: Set Up and Evaluate Integrals for the Center of Mass Coordinates

The center of mass coordinates in the spherical system are:- \( \bar{x} = \frac{1}{M}\int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_{\sec \phi}^{1} \rho^3 \cos \phi \sin \phi \sin \phi \cos \theta \, d\rho \, d\phi \, d\theta \)- \( \bar{y} = \frac{1}{M}\int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_{\sec \phi}^{1} \rho^3 \cos \phi \sin \phi \sin \phi \sin \theta \, d\rho \, d\phi \, d\theta \)- \( \bar{z} = \frac{1}{M}\int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_{\sec \phi}^{1} \rho^3 \cos^2 \phi \sin \phi \, d\rho \, d\phi \, d\theta \)Evaluate these integrals.

Step 7: Compute the Center of Mass

Due to symmetry in this particular problem, \( \bar{x} = 0 \) and \( \bar{y} = 0 \). Therefore, we focus on finding \( \bar{z} \). The integral for \( \bar{z} \) simplifies due to symmetrical properties and bounds. Once evaluated, the center of mass coordinates will be \( (0, 0, \bar{z}) \).

Key Concepts

Spherical Coordinates

When you deal with three-dimensional problems, spherical coordinates come in handy. They are particularly useful when the object in question has a symmetry around a point, such as spherical or conical shapes. In spherical coordinates, any point in space is defined by three parameters:

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

The conversion from Cartesian coordinates \((x, y, z)\) to spherical coordinates \((\rho, \phi, \theta)\) is achieved with these formulas:

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

These formulas allow you to express complex surfaces, like cones or spheres, in a way that can simplify integration and other calculations.

Density Function

Density functions describe how mass is distributed throughout a solid. It tells you how much mass is present in a small volume around a particular point. In the problem given, the density function is specified as \( \delta(x, y, z) = z \), which implies that the density continues to change linearly with the z-coordinate.
When translating this function into spherical coordinates, you replace \( z \) with \( \rho \cos \phi \). Therefore, the density function becomes \( \delta(\rho, \phi, \theta) = \rho \cos \phi \). This modification is crucial when setting up integrals, as it maintains the proportional relationship between mass and position within the solid.
Interfaces with density become essential, especially when calculating mass, centers of mass, and other integral-related measures.

Triple Integral

A triple integral helps calculate a quantity over a three-dimensional region, such as volume or mass. In this context, it's used to determine the total mass of the solid and eventually the center of mass.
The triple integral of a density function over a region will yield the total mass of a solid:\[ M = \int \int \int \delta(\rho, \phi, \theta) \, dV \]In spherical coordinates, the volume element \( dV \) is expressed as:\[ dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]Thus, the integral for mass becomes:\[ M = \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_{\sec \phi}^{1} \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta \]Each part of this integrand accounts for different elements: \(\rho^3\) accounts for radial distance calculations, \(\cos \phi\) integrates the z-component of density, and \(\sin \phi\) adjusts the surface area measure akin to a Jacobian adjustment in a coordinate transformation. This step forms the backbone of physical calculations involving volume and mass.

Limits of Integration

Establishing correct limits of integration is vital for accurate results in multivariable calculus. For this problem in spherical coordinates:

• \( \theta \) ranges from \( 0 \) to \( 2\pi \), covering the entire circular range around the z-axis.
• \( \phi \) spans from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \), representing the angle of the cone intersecting a plane.
• \( \rho \) stretches from \( \sec \phi \) to 1, neatly specifying the radial starting and stopping points with respect to the surfaces that bound the solid.

These bounds encapsulate the entire geometry of the object — from the cone's slant surface defined by \( \tan \phi = 1 \) to the plane cap at \( z = 1 \). Incorrect limits result in portions of the solid being omitted or incorrectly computed, leading to errors in mass or center of mass measurements. By carefully setting these limits, you ensure that the entire volume is considered.