Problem 64 - Thomas Calculus

Question

Express the moment of inertia \(I_{z}\) of the solid hemisphere \(x^{2}+y^{2}+z^{2} \leq 1, z \geq 0,\) as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find \(I_{z}\) .

Step-by-Step Solution

Answer

The moment of inertia is \( \frac{2\pi^2}{15} \).

Step 1: Understanding the moment of inertia

The moment of inertia about the z-axis is given by \( I_z = \int_V \rho r^2 \, dV \), where \( \rho \) is the density and \( r^2 = x^2 + y^2 \). In this problem, the density is assumed to be constant, so you can focus on setting up the integral in the correct coordinate system.

Step 2: Expressing in Cylindrical Coordinates

In cylindrical coordinates, \( x = r\cos\theta \), \( y = r\sin\theta \), and \( z = z \). The differential volume element is \( dV = r \, dr \, d\theta \, dz \). Since \( r^2 = x^2 + y^2 \), the moment of inertia becomes:\[ I_z = \int_0^{2\pi} \int_0^1 \int_0^{\sqrt{1-r^2}} \rho r^3 \, dz \, dr \, d\theta \]

Step 3: Expressing in Spherical Coordinates

In spherical coordinates, \( x = \rho\sin\phi\cos\theta \), \( y = \rho\sin\phi\sin\theta \), and \( z = \rho\cos\phi \). The differential volume element is \( dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \). The limits for the hemisphere are \( \rho: [0, 1] \), \( \phi: [0, \frac{\pi}{2}] \), \( \theta: [0, 2\pi] \). Thus, the moment of inertia becomes:\[ I_z = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^1 \rho^4 \sin^3\phi \, d\rho \, d\phi \, d\theta \]

Step 4: Solving the Cylindrical Coordinates Integral

Solve the integral step by step.1. Integrate the innermost with respect to \(z\):\[ \int_0^{\sqrt{1-r^2}} \rho r^3 \, dz = \rho r^3 \sqrt{1-r^2} \]2. Integrate with respect to \( r \):\[ \int_0^1 \rho r^3 \sqrt{1-r^2} \, dr \]3. Finally, multiply by the integral with respect to \( \theta \):\[ \int_0^{2\pi} \rho 2r^3 \sqrt{1-r^2} \, d\theta \], leading to a factor multiplication with \( 2\pi \).

Step 5: Solving the Spherical Coordinates Integral

Integrate each component as follows:1. Start with \(\rho\):\[ \int_0^1 \rho^4 \, d\rho = \frac{1}{5} \]2. Integrate with respect to \( \phi \):\[ \int_0^{\frac{\pi}{2}} \sin^3\phi \, d\phi = \frac{4}{3} \times \frac{\pi}{4} \]3. Multiply by the \(\theta\) integral:\[ \int_0^{2\pi} \, d\theta = 2\pi \].Combine all results to find \( I_z = 2\pi \times \frac{1}{5} \times \frac{\pi}{3} \).

Step 6: Calculating the Final Result

Combine the results for each expression and solve. For example, in spherical coordinates:\[ I_z = \frac{2\pi \times \pi}{3 \times 5} = \frac{2\pi^2}{15} \] which is the result for the spherical coordinates. This value matches the evaluation of the cylindrical solution which should yield the same numerical result. The calculated moment of inertia captures the symmetry and properties of the solid hemisphere.

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional extension of polar coordinates, typically used in situations where symmetry around a central axis exists. You can think of them as a blend between the two-dimensional polar coordinates and the standard Cartesian coordinates.
To define a point in cylindrical coordinates, we use three values: \( r \), \( \theta \), and \( z \). Here:

• \( r \) is the perpendicular distance from the z-axis;
• \( \theta \) is the angle measured from the positive x-axis in the xy-plane;
• \( z \) is the same as in Cartesian coordinates, representing the height from the xy-plane.

In this exercise, you express the moment of inertia of a solid hemisphere using these coordinates. Doing so requires converting the volume element from Cartesian \( dV = dx \, dy \, dz \) to cylindrical, which becomes \( dV = r \, dr \, d\theta \, dz \).
This transformation makes it easier to calculate the moment of inertia for objects with circular symmetry like our hemisphere. Learning to switch between Cartesian and cylindrical coordinates will significantly enhance your ability to solve three-dimensional problems in physics and engineering.

Spherical Coordinates

Spherical coordinates offer another way to represent points in three-dimensional space. They are particularly useful when dealing with spherical symmetry, such as a solid hemisphere. In this coordinate system, we use three variables \( \rho \), \( \phi \), and \( \theta \):

• \( \rho \) is the radial distance from the origin;
• \( \phi \) is the polar angle measured from the positive z-axis;
• \( \theta \) is the azimuthal angle in the xy-plane from the positive x-axis, similar to cylindrical coordinates.

The spherical volume element is \( dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \), which differs from the rectangular and cylindrical volume elements.
Spherical coordinates are incredibly beneficial when calculating integrals over spheres or hemispheres, as they simplify the limits of integration to constant values. In the context of this exercise, using spherical coordinates helped simplify the integral required to express the moment of inertia \( I_z \). Moreover, these coordinates reduce the complexity of the calculations needed for objects with radial or spherical symmetry.

Iterated Integral

An iterated integral is used for calculating triple integrals by integrating one variable at a time, iteratively, rather than all at once. This concept simplifies complex mathematics by breaking down the integration process into more manageable steps.
When evaluating integrals representing volumes, masses, or moments of inertia, it is vital to understand how to correctly set up these iterated integrals based on the geometry of the problem.
Consider converting the moment of inertia formula \( I_z = \int_V \rho r^2 \, dV \) into an iterated integral. Both spherical and cylindrical coordinates can be used, but the choice depends on the symmetry of the problem:

• In cylindrical coordinates, the limits and integrals focus on \( z \), \( r \), and \( \theta \);
• In spherical coordinates, attention moves to \( \rho \), \( \phi \), and \( \theta \).

It is these limits of integration that dictate the bounds of the solid hemisphere. Mastering this process allows a seamless approach to compute physical properties like mass moments or centers of gravity in a wide array of practical applications.