Question

A solid is described along with its density function. Find the mass of the solid using cylindrical coordinates. 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 mass of the solid is \( \frac{2\pi}{3} \).

Step-by-step solution

Set up the integral for mass in terms of density

The mass of the solid is given by the integral of the density function over the volume of the solid. Here, the volume is the upper half of the unit sphere, and the density function is given as 1. In cylindrical coordinates, the mass \( m \) is given by the integral \( m = \int_V \delta(x,y,z) \, dV \).

Express the limits in cylindrical coordinates

Convert the Cartesian coordinates to cylindrical coordinates. Use \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = z \). Since \( ho = 1 \) is constant for this unit sphere, \( z = \sqrt{1 - r^2} \). The integration bounds are \(0 \leq r \leq 1\), \(0 \leq \theta < 2\pi\), and \(0 \leq z \leq \sqrt{1 - r^2}\).

Set up the triple integral

Express the volume integral \( V \) as a triple integral in cylindrical coordinates. Therefore, the integral becomes: \[ m = \int_0^{2\pi} \int_0^{1} \int_0^{\sqrt{1 - r^2}} r \, dz \, dr \, d\theta\] where the density function \( \delta = 1 \) and \( r \, dr \, d\theta \) accounts for the change in cylindrical coordinates.

Evaluate the integral with respect to z

First, integrate with respect to \( z \), keeping \( r \) and \( \theta \) constant: \[ \int_0^{\sqrt{1 - r^2}} dz = [z]_0^{\sqrt{1 - r^2}} = \sqrt{1 - r^2}.\] After integrating over \( z \), the expression becomes: \( \int_0^{2\pi} \int_0^{1} r \sqrt{1 - r^2} \, dr \, d\theta \).

Evaluate the integral with respect to r

Next, evaluate the remaining double integral by integrating with respect to \( r \): \[ \int_0^{1} r \sqrt{1 - r^2} \, dr.\] Use the substitution \( u = 1 - r^2 \), \( du = -2r \, dr \). Then, the integral becomes: \[ -\frac{1}{2} \int_1^{0} u^{1/2} \, du = \frac{1}{2} \int_0^{1} u^{1/2} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{1}{3}.\]

Evaluate the integral with respect to theta

Finally, evaluate the integral with respect to \( \theta \): \[ \int_0^{2\pi} \frac{1}{3} \, d\theta = \frac{1}{3} \left[\theta \right]_0^{2\pi} = \frac{1}{3} \times 2\pi = \frac{2\pi}{3}.\]

Result: Mass of the solid

The mass of the solid is \( \frac{2\pi}{3} \).

Key concepts

Density Function

A **density function** describes how much mass is distributed over a specific volume. In this problem, the density function is given by \(\delta(x, y, z) = 1\). This means that the mass is uniformly distributed throughout the solid.
In general cases, density functions can be more complex and depend on the position, indicating variations in how mass is spread.
For instance, if the density function was \(\delta(x, y, z) = z\), it would signify that mass increases with height. However, in this exercise, the uniform density simplifies calculations since each unit of volume contributes equally to the total mass.

Integration

**Integration** helps to calculate quantities such as area, volume, and mass, by summing up infinitesimal parts. In the context of this exercise, we use integration to calculate the mass of a solid with a known density distribution over a specified volume.
Here, the integration is set up in cylindrical coordinates. This involves three integrations over the radius \( r \), the angle \( \theta \), and the height \( z \).

• The first integral calculates the contribution of vertical slices by integrating with respect to \( z \).
• Next, horizontal rings are integrated over the radial distance \( r \).
• Finally, integration around the angle \( \theta \) aggregates these contributions over the entire 3-D volume.

This method ensures that the integration accounts for every tiny volume element in the solid.

Unit Sphere

A **unit sphere** is a sphere with a radius of one. In this exercise, we focus on the upper half of the unit sphere, which is the volume above the \( xy \)-plane and below the surface defined by \( z = \sqrt{1 - x^2 - y^2} \).
When using cylindrical coordinates, it's useful to express the sphere using conversions like \( x = r \cos \theta \), \( y = r \sin \theta \), and \( \rho = 1 \) (the radius of the unit sphere).
This translates the complex Cartesian equation into simpler bounds for integration:

• \( 0 \leq r \leq 1 \) for the radial component from the center to the edge,
• \( 0 \leq \theta < 2\pi \) to cover the full circle, and
• \( 0 \leq z \leq \sqrt{1-r^2} \) which restricts the height to below the sphere's surface.

Changing to cylindrical coordinates simplifies the calculation of integrals over spheres by taking advantage of symmetry.

Mass Calculation

The **mass calculation** involves integrating the density function over the volume of interest. Since the density function is \(\delta(x, y, z)=1\) for this solid, the problem simplifies to integrating the volume.
The integration is performed in steps:

• First, integrate with respect to \( z \) from 0 to \(\sqrt{1 - r^2}\), capturing the height variation under the sphere's dome.
• Second, integrate \( r \sqrt{1 - r^2} \) over the radial range 0 to 1, which represents rings from the center to the boundary of the unit sphere.
• Finally, integrate the obtained value around \(\theta\) from 0 to \(2\pi\), covering the complete circular sweep.

Each step consolidates the mass from smaller elements to the whole volume, resulting in the total mass of \(\frac{2\pi}{3}\). The use of cylindrical coordinates simplifies this process by accommodating the geometric symmetry of the sphere.