Question

Set up a triple integral that gives the moment of inertia about the \(z\) -axis of the solid region \(Q\) of density \(\rho .\) $$ \begin{array}{l} Q=\left\\{(x, y, z): x^{2}+y^{2} \leq 1,0 \leq z \leq 4-x^{2}-y^{2}\right\\} \\\ \rho=k x^{2} \end{array} $$

Short answer

The triple integral set up that gives the moment of inertia about the \(z\) -axis of the solid region \(Q\) of density \(\rho = k x^{2}\) is \[I_z = k \int_0^{2\pi} cos^2(\phi) d\phi \int_0^1 r^5 dr \int_0^{4-r^2} dz.\]

Step-by-step solution

Understanding the Problem

The moment of inertia about the \(z\) -axis of the solid region \(Q\) of density \(\rho\) with \(\rho = k x^{2}\) is given by the triple integral \(I_z = \int \int \int_Q r^2 \rho dV\), where \(r^2 = x^2+y^2\) is the radial coordinate squared in a cylindrical coordinate system (r, \(\phi\), z). The volume integral ranges over the region \(Q\) described by \(x^{2}+y^{2} \leq 1\) and \(0 \leq z \leq 4-x^{2}-y^{2}\) in cylindrical coordinates.

Set up the Integral in Cartesian Coordinates

To express the density \(\rho\) and the radial coordinate \(r\) squared, replace \(x\) in terms of cylindrical coordinates \(x = r cos(\phi)\) to get \(\rho = k r^2 cos^2(\phi)\) and \(r^2 = r^2\). Apply these replacements to the moment of inertia integral to obtain \(I_z = \int \int \int_Q r^2 (k r^2 cos^2(\phi)) dV\).

Convert the Volume Integral to Cylindrical Coordinates

The differential volume element in cylindrical coordinates over region \(Q\) is \(dV = r dz dr d\phi\). Therefore the limits of the integral will describe the region \(Q\) in cylindrical coordinates: \[0 \leq r \leq 1 \quad (circular cross section),\]\[0 \leq \phi \leq 2\pi \quad (full rotation around z axis), \]\[0 \leq z \leq 4 - r^2 \quad (conical shape).\]Applying these limits to the moment of inertia integral, \[I_z = \int_0^{2\pi} \int_0^1 \int_0^{4-r^2} r^2 * k r^2 cos^2(\phi) * r dz dr d\phi\]is the integral set up in cylindrical coordinates.

Simplify the Integral

The moment of inertia integral in cylindrical coordinates can be simplified to \[I_z = k \int_0^{2\pi} cos^2(\phi) d\phi \int_0^1 r^5 dr \int_0^{4-r^2} dz.\]

Key concepts

Cylindrical Coordinates

When dealing with problems in three dimensions, particularly those involving circular symmetry, cylindrical coordinates offer a more natural and convenient way to describe the location of points than Cartesian coordinates. In cylindrical coordinates, a point in 3D space is defined by three parameters: the radial distance from the z-axis (\r), the angle (\r\( \phi \)) around the z-axis, starting from the positive x-axis, and the height (\r\( z \)) above the xy-plane.

The transformation from Cartesian to cylindrical coordinates is given by:\r
\r\[ x = r\cos(\phi) \]\r
\r\[ y = r\sin(\phi) \]\r
\r\[ z = z \]\r
These coordinates are particularly helpful for solving integrals over volumes with circular or cylindrical symmetry because they simplify the boundaries of integration. In our exercise, where a region's boundaries are defined by an equation involving \r\( x^2 + y^2 \), converting to cylindrical coordinates neatly collapses this term into \r\( r^2 \), significantly simplifying the process of setting up the integral for the moment of inertia.

Solid Region

A solid region in mathematics is a three-dimensional space occupied by matter. The boundaries of a solid region can be described using inequalities in Cartesian, cylindrical, or spherical coordinates. The solid region \r\( Q \) in our problem is defined by 'all points within a unit cylinder's radius that extend up to a certain height determined by the cylinder’s radius'.

Understanding the shape is crucial for setting up the limits of integration for the moment of inertia. We use the description provided to visualize the region: \r\( r \) ranges from 0 to 1, making a full circle around the z-axis, which means \r\( \phi \) ranges from \r\( 0 \) to \r\( 2\pi \). The height \r\( z \) is variable, going from 0 all the way up to \r\( 4 - r^2 \), a conical shape tapering as we move up. In essence, our region \r\( Q \) can be thought of as a cone mounted on top of a cylinder, and the applied limits in the integration reflect this unique shape.

Density Function

The density function in a physical context is a mathematical description of how mass is distributed throughout a region of space. It can vary from point to point, depending on the specific density properties of the material.

For our exercise, the density function \r\( \rho \) is given as \r\( kx^2 \), which means density increases with the square of the distance along the x-axis from the origin. Since we are rotating around the z-axis, the density is not uniform; it varies with \r\( x \). This has implications for the moment of inertia, which relates to how mass is distributed in relation to the axis of rotation.

In order to integrate the density function over the solid region \r\( Q \), we convert \r\( x \) into cylindrical coordinates, yielding \r\( \rho = kr^2\cos^2(\phi) \). The density function affects the moment of inertia integral directly, as it is included within the integral expression. The triple integral accounts for the variable density and delivers the moment of inertia, reflecting how mass distribution influences rotational resistance around the z-axis.