91Ó°ÊÓ

When working with problems involving regions of circular symmetry, like our disk in this exercise, polar coordinates become invaluable. They allow us to simplify complex integrals by converting Cartesian coordinates (x, y) into polar coordinates (r, θ). This conversion uses the relationships:

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

This is particularly useful when our region, like the disk centered at (1, 2), is circular. By shifting this disk so its center aligns with the origin in polar coordinates, the limits for \( r \) simply range from 0 to the radius of the disk, and \( \theta \) from 0 to \( 2\pi \). This transformation simplifies the process of calculating integrals related to moments of inertia.

The density function, \( \rho(x, y) \), describes how mass is distributed over the lamina. In this exercise, it is given by the expression: \[ \rho(x, y) = x^2 + y^2 - 2x - 4y + 5 \] This quadratic form is not only a mathematical description but also represents the physical characteristic of the lamina. To use this function in polar coordinates, we replace \( x \) and \( y \) using their polar equivalents and then apply it to the calculations for moments of inertia and mass. Understanding this function is key, as it directly influences how the mass (and therefore inertia) calculations will behave across the region of interest.

Radii of gyration provide a measure of distribution of the mass of the lamina relative to a specific axis. For axes such as the x-axis, y-axis, and the origin, these radii can be calculated using:

• \( k_x = \sqrt{\frac{I_x}{M}} \)
• \( k_y = \sqrt{\frac{I_y}{M}} \)
• \( k_0 = \sqrt{\frac{I_0}{M}} \)

The process requires having the moments of inertia \( I_x, I_y, \text{and } I_0 \), and the total mass \( M \). These parameters show how far from the axis the mass appears to act as if concentrated, helping understand the dynamics associated with the structure's mass distribution.

Double integration is the mathematical procedure that we use to calculate quantities like moments of inertia and mass over a continuous area, such as the lamina in our problem. When expressed in polar coordinates, the double integral takes the form:\[\int_0^{2\pi} \int_0^2 f(r, \theta) \cdot r \, dr \, d\theta\]Here, \( f(r, \theta) \) represents the transformed function of interest (like our density function multiplied by \( y^2 \) for \( I_x \)). Double integration in polar coordinates considers both the radial and angular components, simplifying integration over circular regions. It allows us to sum our defined function over the area of the disk, yielding insights like total mass or inertia.

The mass of the lamina, often represented as \( M \), is a fundamental quantity required to compute moments of inertia and radii of gyration. It is obtained through the double integral of the density function over the region R, the disk in this instance:\[ M = \int_0^{2\pi} \int_0^2 \rho(r, \theta) \cdot r \, dr \, d\theta \]Using polar transformations, we calculate this integral, which accounts for the entire mass distributed across the lamina based on its density function. This mass not only indicates how heavy the lamina is but also plays a crucial role in further calculations of mechanical properties like the radii of gyration.