Question

A lamina corresponding to a planar region \(R\) is given with a mass of 16 units. For each, compute \(I_{x}\) \(I_{y}\) and \(I_{0}\). \(R\) is the \(8 \times 2\) rectangle with corners at (-4,-1) and (4,1) with density \(\delta(x, y)=1\).

Short answer

\(I_x = \frac{16}{3}\), \(I_y = \frac{256}{3}\), \(I_0 = \frac{272}{3}\).

Step-by-step solution

Understand the Problem

We need to calculate the moments of inertia about the x-axis \(I_x\), y-axis \(I_y\), and the origin \(I_0\) for a rectangular lamina with the given density function \(\delta(x, y) = 1\). The mass of the rectangle is given as 16 units.

Set Up the Inertia Formulas

The formulas for moments of inertia for a rectangular region \(R\) with density \(\delta(x, y)\) are:\[I_x = \int \int_R y^2 \cdot \delta(x, y) \, dA, \quad I_y = \int \int_R x^2 \cdot \delta(x, y) \, dA, \quad I_0 = I_x + I_y.\]Here, \(dA\) is the differential area element.

Define the Region R

The region \(R\) is an \(8 \times 2\) rectangle with corners at \((-4,-1)\) and \((4,1)\). The limits of integration for \(x\) are from \(-4\) to \(4\) and for \(y\) are from \(-1\) to \(1\).

Compute \(I_x\)

Substitute \(\delta(x, y) = 1\) into the formula for \(I_x\):\[ I_x = \int_{-4}^{4} \int_{-1}^{1} y^2 \, dy \, dx. \]Perform the inner integration with respect to \(y\):\[ \int_{-1}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]Then, perform the outer integration with respect to \(x\):\[ I_x = \int_{-4}^{4} \frac{2}{3} \, dx = \frac{2}{3} \times 8 = \frac{16}{3}. \]

Compute \(I_y\)

Substitute \(\delta(x, y) = 1\) into the formula for \(I_y\):\[ I_y = \int_{-4}^{4} \int_{-1}^{1} x^2 \, dy \, dx. \]Perform the inner integration with respect to \(y\):\[ \int_{-1}^{1} \, dy = 2. \]Then, perform the outer integration with respect to \(x\):\[ I_y = \int_{-4}^{4} x^2 \, dx imes 2. \]Calculate\[ \int_{-4}^{4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-4}^{4} = \frac{64}{3} + \frac{64}{3} = \frac{128}{3}. \]Thus,\[ I_y = \frac{128}{3} \times 2 = \frac{256}{3}. \]

Compute \(I_0\)

Use the formula \(I_0 = I_x + I_y\), substitute the values:\[ I_0 = \frac{16}{3} + \frac{256}{3} = \frac{272}{3}. \]

Key concepts

Planar Region

In mathematics, a planar region refers to a flat, two-dimensional surface extending in the plane. It can be delimited by curves or straight lines that form a closed boundary. In this context, we're dealing with a planar region defined as a rectangle.
We consider a rectangle owing to its simplicity and straightforward boundary, which facilitates calculations involving integration.
For the given exercise, the rectangle stretches between corners at (-4, -1) and (4, 1), forming dimensions of 8 units in length and 2 units in width. The choice of rectangular planar regions often simplifies integration due to constant limits of integration.

Density Function

The density function, denoted as \(\delta(x, y)\), represents the density of a material distributed over a given region, such as a lamina. In this exercise, \(\delta(x, y) = 1\) signifies uniform density across the planar region, implying the material's mass is evenly spread.
This uniform density simplifies moment of inertia calculations, as the density function becomes a constant factor, rather than varying with position.
In more complex situations, density may vary with position, and would thus require integration to consider how mass distribution affects inertia.

Rectangular Lamina

A rectangular lamina is a flat sheet with a rectangular shape that has mass distributed evenly or according to a defined density function. It's a two-dimensional object, meaning it has no thickness.
This kind of lamina is perfect for study in mechanics, especially when analyzing concepts of inertia and center of mass.
In exercises involving moments of inertia, such as the given one, having a clear geometric shape like a rectangle facilitates computations, as both its boundaries and area are simple to define and integrate over.

Calculus

Calculus is the branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. Its methods are essential for analyzing changes and accumulations, like in the computation of moments of inertia.
In our problem, calculus, specifically integral calculus, is used to calculate the moments of inertia, which are cumulative properties that describe how mass is distributed in an object.
This branch enables precise determinations of quantities like area under curves and serves as a tool for making continuous approximations necessary in physical applications.

Integration

Integration is a fundamental operation in calculus, often used to calculate areas, volumes, and other quantities that represent accumulation. It involves summing infinitesimal pieces to find the whole quantity.
Here, double integration is used because it is necessary to consider the entire two-dimensional region – the rectangular lamina. The integrals are performed over the region's area, to determine the moments of inertia about different axes.
Specifically, \(I_x\) and \(I_y\) are calculated by integrating over the coordinates \(x\) and \(y\). These integrations take into account the region's geometry, demonstrating how integration facilitates computations of physical properties like inertia.