Problem 42 - University Calculus: Early Transcendentals

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 \(\frac{4\pi}{15}\).

Step 1: Define the Moment of Inertia

The moment of inertia about the z-axis is given by the integral \( I_{z} = \iiint_{V} (x^2 + y^2) ho \, dV \), where \(\rho\) is the density, which we'll assume to be constant and equals to 1 for simplicity.

Step 2: Express the Integral in Cylindrical Coordinates

In cylindrical coordinates, the relation between Cartesian and cylindrical coordinates are \(x = r \cos(\theta), y = r \sin(\theta), z = z\), and \(dV = r \, dr \, d\theta \, dz\). Thus, \(x^2 + y^2 = r^2\). The limits for \(r\) are from 0 to 1, for \(\theta\) are from 0 to \(2\pi\), and for \(z\) are from 0 to \(\sqrt{1 - r^2}\). Then the integral becomes: \[I_{z} = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-r^2}} r^3 \, dz \, dr \, d\theta.\]

Step 3: Express the Integral in Spherical Coordinates

In spherical coordinates, \(x = \rho \sin\phi \cos\theta\), \(y = \rho \sin\phi \sin\theta\), and \(z = \rho \cos\phi\) with \(dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\). For a hemisphere, \(\phi\) ranges from 0 to \(\pi/2\) and \(\theta\) ranges from 0 to \(2\pi\), \(\rho\) ranges from 0 to 1. Substitute \(x^2 + y^2 = \rho^2 \sin^2 \phi\). The integral becomes: \[ I_{z} = \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{1} \rho^4 \sin^3\phi \, d\rho \, d\phi \, d\theta.\]

Step 4: Calculate the Moment of Inertia

We'll solve the integral in spherical coordinates since it simplifies computations. The integral becomes: \[I_{z} = \int_{0}^{2\pi} d\theta \int_{0}^{\pi/2} \sin^3\phi \, d\phi \int_{0}^{1} \rho^4 \, d\rho.\] First, compute the innermost integral: \[\int_{0}^{1} \rho^4 \, d\rho = \left[ \frac{\rho^5}{5} \right]_{0}^{1} = \frac{1}{5}.\] Then, the \(\phi\) integral: \[\int_{0}^{\pi/2} \sin^3\phi \, d\phi = \int_{0}^{\pi/2} (1 - \cos^2\phi) \sin\phi \, d\phi.\] Let \(u = \cos\phi\) resulting in \(du = -\sin\phi \ \, d\phi\): \[\int_{1}^{0} (1 - u^2)(-du) = \int_{0}^{1} (1-u^2) \, du = \left[ u - \frac{u^3}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}.\] Finally, compute the \(\theta\) integral: \[\int_{0}^{2\pi} d\theta = 2\pi.\]Putting all together: \[I_{z} = \left(\frac{1}{5}\right)\left(\frac{2}{3}\right)(2\pi) = \frac{4\pi}{15}.\]

Step 5: Conclusion

The moment of inertia about the z-axis of the solid hemisphere is \(\frac{4\pi}{15}\) given that the density is constant and equals 1.

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are extremely useful when dealing with symmetrical bodies aligned along an axis. They are defined using three parameters: radius \(r\), angle \(\theta\), and height \(z\). This system is particularly suitable for geometries involving circles or cylinders. In these coordinates, any point in space is represented by:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)
• \(z = z\)

Using cylindrical coordinates to express volume elements is another key feature and is given by \(dV = r \, dr \, d\theta \, dz\). It's perfect for our hemisphere which has rotational symmetry around the z-axis. In the exercise, for the solid hemisphere, the limits for \(r\) vary from 0 to 1, while \(\theta\) spans from 0 to \(2\pi\), and \(z\) changes from 0 to \(\sqrt{1-r^2}\). This setup will let us rewrite the moment of inertia as an integral that respects the natural shape and boundaries of the solid hemisphere. Cylindrical coordinates simplify problems involving such conditions, making calculations more intuitive.

Spherical Coordinates

Spherical coordinates are another way to handle three-dimensional space but are particularly handy for spheres or hemispheres. Here, each point is determined by three parameters: radial distance \(\rho\), polar angle \(\phi\), and azimuthal angle \(\theta\). This system is beneficial when dealing with problems of spherical symmetry. A point in space is expressed as:

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

The volume element in spherical coordinates is \(dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\). For a hemisphere, \(\phi\) ranges from 0 to \(\pi/2\), \(\theta\) from 0 to \(2\pi\), and \(\rho\) from 0 to 1. These boundaries align perfectly with the natural geometry of the hemisphere. Transforming the moment of inertia using these boundaries integrates across the entire hemisphere easily and efficiently, reducing the complexity of calculations. This method shows its strength in expressing problems of perfect radial geometry.

Solid Hemisphere

A solid hemisphere is exactly half of a perfect 3D circle (sphere), characterized by a flat circular base and a curved cap. The task of calculating the moment of inertia for such a shape involves evaluating rotational dynamics. The moment of inertia, \(I_z\), measures how resistant the hemisphere is to rotational acceleration about the z-axis, analogous to mass in linear dynamics.
Given this shape's symmetry and constraints such as \(x^2 + y^2 + z^2 \leq 1\) and \(z \geq 0\), calculating its moment of inertia requires integrating across its entire volume. Considering constant density simplifies this by focusing only on geometric barriers. Two coordinate systems, cylindrical and spherical, convert this 3D integral into manageable, calculable terms tapping into the hemisphere's natural symmetry with different emphases. Both systems foster understanding by each offering distinctive paths to the final solution, thus illustrating the power and flexibility inherent in mathematical tools for physics.