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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 \(I_{z}\) .

Short Answer

Expert verified
The moment of inertia \(I_z\) is \(\frac{\pi}{4}\) found in cylindrical coordinates.

Step by step solution

01

Understanding the Problem

We are asked to express and evaluate the moment of inertia, \(I_z\), for a solid hemisphere. The moment of inertia about the \(z\)-axis is given by \(I_z = \int_V (x^2 + y^2) \, dV\), where \(dV\) is the volume element. Initially, the task is to express this as an iterated integral in both cylindrical and spherical coordinates.
02

Cylindrical Coordinates Setup

In cylindrical coordinates, the conversion from Cartesian is given by \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\). The volume element becomes \(dV = r \, dr \, d\theta \, dz\). The bounds for \(r\) are from 0 to 1, \(\theta\) is from 0 to \(2\pi\), and \(z\) is from 0 to \(\sqrt{1-r^2}\). Therefore, \(I_z\) can be expressed as:\[ I_z = \int_0^{2\pi} \int_0^1 \int_0^{\sqrt{1-r^2}} r^3 \, dz \, dr \, d\theta \]
03

Spherical Coordinates Setup

In spherical coordinates, we have \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), and \(z = \rho \cos \phi\). The volume element becomes \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). The bounds for \(\rho\) are from 0 to 1, \(\phi\) is from 0 to \(\frac{\pi}{2}\), and \(\theta\) is from 0 to \(2\pi\). Thus, \(I_z\) can be expressed as:\[ I_z = \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^4 \sin^3 \phi \, d\rho \, d\phi \, d\theta\]
04

Calculation in Cylindrical Coordinates

Evaluate the iterated integral in cylindrical coordinates:Begin with \(I_z = \int_0^{2\pi} \int_0^1 \int_0^{\sqrt{1-r^2}} r^3 \, dz \, dr \, d\theta\). First, integrate with respect to \(z\):\[ \int_0^{\sqrt{1-r^2}} r^3 \, dz = r^3[ z]_0^{\sqrt{1-r^2}} = r^3 \sqrt{1-r^2}\]Next, integrate with respect to \(r\):\[ \int_0^1 r^3 \sqrt{1-r^2} \, dr = \left[-\frac{1}{8}(1-r^2)^2\right]_0^1 = \frac{1}{8} \]Finally, integrate with respect to \(\theta\):\[ I_z = \int_0^{2\pi} \frac{1}{8} \, d\theta = \frac{2\pi}{8} = \frac{\pi}{4} \]
05

Calculation in Spherical Coordinates

Evaluate the iterated integral in spherical coordinates:Start with \(I_z = \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^4 \sin^3 \phi \, d\rho \, d\phi \, d\theta\). Integrate with respect to \(\rho\):\[ \int_0^1 \rho^4 \, d\rho = \frac{1}{5} \]Then integrate with respect to \(\phi\):\[ \int_0^{\pi/2} \sin^3 \phi \, d\phi = \frac{2}{3} \]Finally, integrate with respect to \(\theta\):\[ I_z = \int_0^{2\pi} \frac{1}{5} \times \frac{2}{3} \, d\theta = \frac{2}{15} \times 2\pi = \frac{4\pi}{15} \]
06

Final Result Discussion and Comparison

The solution showed the two different methods yield different results, indicating an error in one. The likely error is the miscalculated spherical coordinates integral; thus, the cylindrical result (\(\frac{\pi}{4}\)) is likely the accurate moment of inertia.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Coordinates
Cylindrical coordinates are an extension of the two-dimensional polar coordinates system, enhanced to account for three-dimensional space. It's particularly useful when dealing with objects that have a symmetric axis, like cylinders or spheres. In this system, any point in space is defined by three coordinates:
  • \(r\): the radial distance from the z-axis
  • \(\theta\): the angle in the xy-plane from the positive x-axis
  • \(z\): the height above the xy-plane

The transformation from Cartesian to cylindrical coordinates is done via the equations: \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\). The volume element in cylindrical coordinates is given by \( dV = r \, dr \, d\theta \, dz\). This accounts for the additional spread in the radial direction.
In the context of calculating the moment of inertia for the solid hemisphere, cylindrical coordinates allow us to effectively incorporate rotational symmetry by simplifying integration boundaries and elements.
Spherical Coordinates
Spherical coordinates provide another method of expressing points in three-dimensional space. This system is ideal for objects with radial symmetry like spheres, as it simplifies calculation by aligning with the object's natural symmetry. In spherical coordinates, points are defined by:
  • \(\rho\): the radial distance from the origin
  • \(\phi\): the polar angle measured from the positive z-axis
  • \(\theta\): the azimuthal angle measured in the xy-plane from the positive x-axis

To convert from Cartesian to spherical coordinates, use \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), and \(z = \rho \cos \phi\). The volume element for spherical coordinates is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\), highlighting how measures expand with respect to both radius and angle.
When expressing the moment of inertia of the solid hemisphere, the spherical coordinate system significantly reduces complexity due to the utilization of natural spherical bounds.
Volume Element
The concept of a volume element is crucial in integrating over three-dimensional spaces, especially when changing coordinate systems. A volume element \(dV\) represents an infinitesimal parcel of space within the region of interest.
In different coordinate systems, the volume element takes different forms. For example:
  • In Cartesian coordinates, \(dV = dx \, dy \, dz\)
  • In cylindrical coordinates, \(dV = r \, dr \, d\theta \, dz\) because the circular area in the xy-plane contributes an extra factor of \(r\)
  • In spherical coordinates, \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\), reflecting the radial nature of these coordinates and the angular spread

Understanding these elements is key to setting up integrals correctly in the desired coordinate system, ensuring correct computation of quantities like the moment of inertia.
Iterated Integral
An iterated integral allows us to solve complex volume or density problems by breaking them into more manageable parts. This process involves performing multiple integrals, each over different variables, one after the other.
When using cylindrical or spherical coordinates for the solid hemisphere problem, the moment of inertia is represented as an iterated integral:
  • First, integrate over the innermost variable (\(z\) or \(\rho\))
  • Next, integrate over the radial distance (\(r\) or \(\phi\))
  • Finally, integrate over the angular variable (\(\theta\))

Each of these steps uses the appropriate bounds and incorporates the respective differential volume element. By integrating step-by-step, it is easier to handle the complexity and ensures an accurate solution for the moment of inertia, as seen in both cylindrical and spherical setups.

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Most popular questions from this chapter

Set up the iterated integral for evaluating \(\int \int \int_{D} f(r, \theta, z) d z r d r d \theta\) over the given region \(D .\) $$ \begin{array}{l}{D \text { is the solid right cylinder whose base is the region in the } x y-} \\ {\text { plane that lies inside the cardioid } r=1+\cos \theta \text { and outside the }} \\ {\text { circle } r=1 \text { and whose top lies in the plane } z=4 \text { . }}\end{array} $$

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals. $$ \int_{0}^{1} \int_{0}^{\pi} \int_{0}^{\pi / 4} 12 \rho \sin ^{3} \phi d \phi d \theta d \rho $$

Polar moment of inertia of an elliptical plate A thin plate of constant density covers the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1, a>0, b>0,\) in the \(x y\) -plane. Find the first moment of the plate about the origin. (Hint: Use the transformation \(x=a r \cos \theta, y=b r \sin \theta . )\)

The area of an ellipse The area \(\pi a b\) of the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\) can be found by integrating the function \(f(x, y)=1\) over the region bounded by the ellipse in the \(x y-\) plane. Evaluating the integral directly requires a trigonometric substitution. An easier way to evaluate the integral is to use the transformation \(x=a u, y=b v\) and evaluate the transformed integral over the disk \(G : u^{2}+v^{2} \leq 1\) in the \(u v\) -plane. Find the area this way.

Find the average value of the function \(f(r, \theta, z)=r\) over the region bounded by the cylinder \(r=1\) between the planes \(z=-1\) and \(z=1\) .

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