Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use triple iterated integrals to find the indicated quantities. Moment of inertia \(I_{x}\) about the \(x\) -axis of the solid bounded by the cylinder \(y^{2}+z^{2}=4\) and the planes \(x-y=0, x=0\), and \(z=0\) if the density \(\delta(x, y, z)=z .\) Hint: You will need to develop your own formula; slice, approximate, integrate.

Step-by-Step Solution

Answer

The moment of inertia about the x-axis is 0.

Step 1: Understand the Geometry of the Object

The solid is bounded by the vertical cylinder with axis along the x-axis: \( y^2 + z^2 = 4 \) and two planes: \( x = y \) and \( x = 0 \). Also, \( z = 0 \) provides the base. We need to find the moment of inertia around the x-axis.

Step 2: Set Up the Triple Integral

The moment of inertia \(I_{x}\) about the x-axis is given by the integral: \[ I_{x} = \int \int \int_{V} \delta(x, y, z) (y^2 + z^2) \, dv \] where \( \delta(x, y, z) = z \) is the density function. In cylindrical coordinates, the volume differential \( dv \) becomes \( r \, dr \, d\theta \, dx \).

Step 3: Convert the Integral into Cylindrical Coordinates

Convert to cylindrical coordinates where \( y = r \cos \theta \) and \( z = r \sin \theta \). Our cylinder bounds \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq \pi \), and \( 0 \leq x \leq y = r \cos \theta \). The integral becomes: \[ I_{x} = \int_{0}^{\pi} \int_{0}^{2} \int_{0}^{r \cos \theta} z (r^2) \cdot r \, dx \, dr \, d\theta \] where \( z = r \sin \theta \).

Step 4: Simplify the Integral Expression

Substitute \( z = r \sin \theta \) into the integral to obtain: \[ I_{x} = \int_{0}^{\pi} \int_{0}^{2} \int_{0}^{r \cos \theta} (r \sin \theta)(r^3) \, dx \, dr \, d\theta = \int_{0}^{\pi} \int_{0}^{2} r^4 (\sin \theta) (r \cos \theta) \, dr \, d\theta \] Cancel \( r^3 \cdot r \) to derive: \[ I_{x} = \int_{0}^{\pi} \int_{0}^{2} r^5 \sin \theta \cos \theta \, dr \, d\theta \]

Step 5: Evaluate the Iterated Integral

First, integrate the expression with respect to \( x \) (which ranges from 0 to \( r \cos \theta \)): \[ \int_{0}^{r \cos \theta} \, dx = r \cos \theta \]Substituting, we obtain: \[ I_x = \int_{0}^{\pi} \int_{0}^{2} r^4 \sin \theta \cos^2 \theta \, dr \, d\theta \]Then, integrate with respect to \( r \): \[ \int_{0}^{2} r^4 \, dr = \left. \frac{r^5}{5} \right|_{0}^{2} = \frac{32}{5} \]Now integrate with respect to \( \theta \): \[ I_x = \frac{32}{5} \int_{0}^{\pi} \sin \theta \cos^2 \theta \, d\theta \]

Step 6: Solve the Remaining Integral for \( \theta \)

To solve \( \int \sin \theta \cos^2 \theta \, d\theta \), use trigonometric identities: \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \).We reformulate the integral: \[ \int \sin \theta \left( \frac{1 + \cos 2\theta}{2} \right) \, d\theta \]The integral separates, forming two parts both of which give standard results, resulting in \[ \frac{1}{2} (-\frac{1}{2} \cos \theta \sin \theta + \frac{1}{4} \sin 2\theta) \] evaluated from 0 to \( \pi \). This evaluates to zero because the sine function is zero at these limits. Therefore,\( I_x = 0. \)

Step 7: Final Step: Summarize the Results

Evaluating the definite integral over the cylindrical bounds demonstrates that any contribution to the integral contributes to zero moment due to symmetry and boundary integration values of trigonometric evaluations, hence the moment of inertia about the x-axis is zero.

Key Concepts

Moment of Inertia

The moment of inertia is a physical quantity expressing an object's resistance to rotational motion around an axis. In our problem, the axis of interest is the x-axis. The formula involves integrating the mass distribution across the solid. In general, the moment of inertia, say about the x-axis, is: \[ I_{x} = \int \int \int_{V} \delta(x, y, z) (y^2 + z^2) \, dv \] where \( \delta(x, y, z) \) is the density function of the solid. Here, each small piece of the object, considered as a point mass, contributes to the total moment of inertia. The term \( y^2 + z^2 \) represents the distance squaring factor from the axis of rotation, reflecting the rotational effect. In our exercise, the solid is defined by a bounded region in 3D space, and we use this triple integral approach to account for the entire volume's impact. You need to carefully identify the boundaries and limits of integration to properly compute the integral.

Cylindrical Coordinates

Cylindrical coordinates are a system that extends polar coordinates by adding a height, making them very useful for integrating over cylindrical shapes. The coordinates \( (r, \theta, z) \) are defined as follows:

• \( r \): the radial distance from the origin to the projection of the point in the \( yz \)-plane.
• \( \theta \): the angle measured counterclockwise from the positive x-axis.
• \( z \): the height of the point above the \( xy \)-plane, equivalent to the z-coordinate in Cartesian coordinates.

For our given cylinder \( y^2 + z^2 = 4 \), the cylindrical radius \( r \) ranges from 0 to 2, covering the circle's radius. The angle \( \theta \) ranges from 0 to \( \pi \) because of the half-cylinder defined within the given bounds. Converting from Cartesian coordinates, we have \( y = r \cos \theta \) and \( z = r \sin \theta \), which translate into the elements of the volume differential \( dv = r \, dr \, d\theta \, dx \). This conversion simplifies the integration process for solids with cylindrical symmetry by aligning with the natural shape boundaries.

Density Function

The density function \( \delta(x, y, z) \) describes how mass is distributed throughout a solid. In physics, density is often mass per unit volume. In this exercise, the density is given as a function of \( z \), specifically \( \delta(x, y, z) = z \). This implies that the density, and hence mass, increases with height above the \( xy \)-plane. The impact of a density function on moment of inertia is substantial. Since the function influences the mass distribution, it changes the balance and impact of different parts of the object on its rotational inertia. Each \( z \) level adds more mass proportionally, weighted by \( z \), to contribute to inertia in a linear gradient fashion. During integration, this density affects the computation through the weight it gives each infinitesimally small volume element. The moment of inertia's final value reflects the overall influence of the density profile throughout the cylindrical volume.