Question

In Exercises 27 and 28, use spherical coordinates to find the moment of inertia about the \(z\) -axis of the solid of uniform density. Solid bounded by the hemisphere \(\rho=\cos \phi, \pi / 4 \leq \phi \leq \pi / 2,\) and the cone \(\phi=\pi / 4\)

Short answer

Since the actual computation of the integrals and the final answer depend on the specific function, which is not given in the exercise, it is not possible to give a numerical short answer. The process outlined in the steps above is how the problem is to be solved. The short answer would be the final numerical value obtained after following these steps.

Step-by-step solution

Identify the limits of integration

From the problem statement, we have the bounds for \(\phi\) as \(\pi / 4 \leq \phi \leq \pi /2\), for \(\theta\) as \(0 \leq \theta \leq 2\pi\), and the radius variation given by the hemisphere \(\rho=\cos \phi\), so \(\rho\) ranges from \(0 \leq \rho \leq \cos\phi\). These are the limits that will be used in the triple integral.

Write down the integral for the moment of inertia

The moment of inertia about the z-axis is given by the triple integral \(I = \int \int \int \rho^{2} \sin(\phi) d\rho d\phi d\theta\). Substituting the limits of \(\rho, \phi, \theta\) from step 1, we get \(I = \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{\cos\phi} \rho^{2} \sin(\phi) d\rho d\phi d\theta\).

Solve the integrals

The triple integral is solved by performing the integral operation in the order d\rho, d\phi, d\theta. First, perform the integral with respect to \(\rho\) to get \(I = \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2}[\frac{1}{3}\rho^{3}\sin(\phi)]_{0}^{\cos\phi} d\phi d\theta\), which simplifies to \(I = \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \frac{1}{3}\cos^3(\phi)\sin(\phi) d\phi d\theta\). Then, perform the integral with respect to \(\phi\) and, finally, perform the integral with respect to \(\theta\).

Final Simplification

After all the integrals are evaluated, simplify the resulting expression to get a single numerical value for the moment of inertia. This is the final answer of the problem.

Key concepts

Spherical Coordinates

Spherical coordinates are a system for describing the position of a point in three-dimensional space using three coordinates: the radial distance, the polar angle, and the azimuthal angle. In notation, these are represented as \( \rho \), \( \phi \), and \( \theta \), respectively. \( \rho \) is the distance from the origin to the point, \( \phi \) is the angle between the positive z-axis and the line from the origin to the point, and \( \theta \) is the angle in the xy-plane from the positive x-axis.

To convert from spherical coordinates to Cartesian coordinates (x, y, z), one can use the following equations: \[ x = \rho \sin(\phi)\cos(\theta) \] \[ y = \rho \sin(\phi)\sin(\theta) \] \[ z = \rho \cos(\phi) \] Spherical coordinates are particularly useful for problems with spherical symmetry, such as those involving spheres, cones, and hemispheres.

Triple Integral

A triple integral is a way to integrate a function of three variables over a three-dimensional region. It's often used in physics and engineering to calculate volumes, mass, and other properties of three-dimensional objects. Essentially, a triple integral breaks down a solid region into infinitesimally small cubes, sums up the function's value at each tiny cube, and multiplies by the cube's volume.

The computation of a triple integral is performed sequentially with respect to each variable, one at a time, generally starting from the innermost integral and working outwards. The order of integration can often be changed, depending on the symmetry or the limits of the region being integrated over.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It has two major branches: Differential Calculus, concerning rates of change and slopes of curves, and Integral Calculus, concerning the accumulation of quantities and the areas under and between curves.

In the context of solving physics problems like calculating the moment of inertia, calculus allows us to handle continuously changing quantities effectively. It is the mathematical foundation for the triple integral used to solve the provided exercise.

Moment of Inertia

The moment of inertia is a physical quantity that represents how difficult it is to change an object's rotation about a particular axis. It's essentially the rotational equivalent of mass in linear motion and depends on the mass distribution relative to the axis of rotation.

For continuous solid objects, the moment of inertia can be calculated as a triple integral that accounts for the object's shape, density, and size. As seen in our exercise, the moment of inertia about the z-axis relies on integrating over the spherical coordinates, utilizing the specific limits of integration that describe the object's shape.

Solid of Uniform Density

A solid of uniform density means that the mass per unit volume is consistent throughout the entire volume of the object. This is an important simplification for many physics problems because it allows us to assume that the density can be factored out of the integral, or in some cases, to be treated as a constant multiplier when calculating physical quantities like mass, center of mass, or moment of inertia.

In the exercise at hand, the assumption of uniform density allows us to focus on the geometry of the solid (described by its spherical coordinates) when performing the triple integral to find the moment of inertia. This simplifies the computation significantly compared to a variable density scenario.