Problem 27 - APEX Calculus

Question

In Exercises \(27-30,\) 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 \(4 \times 4\) square with corners at (-2,-2) and (2,2) with density \(\delta(x, y)=1\)

Step-by-Step Solution

Answer

The moments of inertia are: \( I_x = \frac{64}{3} \), \( I_y = \frac{64}{3} \), \( I_0 = \frac{128}{3} \).

Step 1: Identify the Mass and Shape

The mass of the lamina is 16 units, and the region \( R \) is a \( 4 \times 4 \) square with corners at (-2, -2) and (2, 2). The density of the lamina is constant, \( \delta(x, y) = 1 \).

Step 2: Define the Planar Region

The region \( R \) can be expressed as \( -2 \leq x \leq 2 \) and \( -2 \leq y \leq 2 \). This is the range for our integrals when calculating the moments of inertia.

Step 3: Calculate Moment of Inertia with Respect to x-axis, \( I_x \)

The formula for \( I_x \) is \( \iint_R y^2 \, \delta(x,y) \, dA \). Since \( \delta(x, y) = 1\), the integral becomes \( \int_{-2}^2 \int_{-2}^2 y^2 \, dy \, dx \). Evaluating this computation, it becomes 1. Integrate \( y^2 \) from \(-2\) to \(2\), resulting in \( \frac{16}{3} \).2. Integrate a constant \( \frac{16}{3} \) with respect to \( x \) over \(-2\) to \(2\), resulting in \( \frac{64}{3} \).

Step 4: Calculate Moment of Inertia with Respect to y-axis, \( I_y \)

The formula for \( I_y \) is \( \iint_R x^2 \, \delta(x,y) \, dA \). Replace and evaluate the integral \( \int_{-2}^2 \int_{-2}^2 x^2 \, dx \, dy \):1. Integrate \( x^2 \) from \(-2\) to \(2\), resulting in \( \frac{16}{3} \).2. Integrate this constant \( \frac{16}{3} \) with respect to \( y \) from \(-2\) to \(2\) resulting in \( \frac{64}{3} \).

Step 5: Calculate Moment of Inertia about the Origin, \( I_0 \)

The formula for \( I_0 \) is \( \iint_R (x^2 + y^2) \, \delta(x,y) \, dA \). This simplifies to \( \int_{-2}^2 \int_{-2}^2 (x^2 + y^2) \, dA \):1. Split this into two separate integrals: \( \int_{-2}^2 \int_{-2}^2 x^2 \, dy \, dx + \int_{-2}^2 \int_{-2}^2 y^2 \, dx \, dy \).2. Both integrals were computed previously to get \( \frac{64}{3} \) each.3. Therefore, \( I_0 = \frac{64}{3} + \frac{64}{3} = \frac{128}{3} \).

Key Concepts

Planar Region

In this exercise, we are working with a planar region represented by a square lamina. The lamina is situated on the Cartesian plane. This square covers an area defined with the corners at (-2, -2) and (2, 2). When dealing with planar regions in such mathematical problems, it is crucial to understand that they are essentially flat, two-dimensional surfaces. These surfaces are described by their bounds along the x-axis and y-axis.
In our example, the square, which is our planar region, extends from

• -2 to 2 along the x-axis
• -2 to 2 along the y-axis.

This understanding of the planar region is fundamental for setting the limits when calculating integrals over the region, which we will address in the following sections.

Constant Density

The concept of constant density is a simplification often used in physics and engineering problems. In our exercise, the density of the lamina is given as a constant value,

• \( \delta(x, y) = 1 \)

This means that the mass distribution across the lamina is uniform. Every point within the square carries the same amount of mass per unit area.
Having a constant density simplifies many calculations, including the determination of mass and moments of inertia. Instead of integrating a more complex density function over the region, we simply multiply by the area integral, as the density does not vary.
In practical terms, constant density implies that the object behaves consistently throughout, making calculations predictable and manageable.

Double Integral

Double integrals are integral calculus tools used to calculate quantities over two-dimensional regions. In the problem, we use double integrals to find moments of inertia with respect to the x-axis, y-axis, and the origin. A double integral

• First, integrates a function with respect to one variable while treating the other variable as a constant.
  
• Second, integrates the result with respect to the second variable allowing the manipulation of the whole surface area.

For the square lamina, these integrals take the form

• \( I_x = \int_{-2}^2 \int_{-2}^2 y^2 \, dy \, dx \)
• \( I_y = \int_{-2}^2 \int_{-2}^2 x^2 \, dx \, dy \)
• \( I_0 = \int_{-2}^2 \int_{-2}^2 (x^2 + y^2) \, dA \)

Each of these requires breaking the problem into one-dimensional integration steps, which we have provided in the original solution with values simplifying direct computation of the moments of inertia.

Lamina Mass Calculation

The mass calculation for a lamina involves integrating the density over the entire region. Fortunately, our problem is simplified by the fact that the density is constant and equal to 1. Therefore, the mass of the lamina is simply the area covered by the planar region multiplied by the density.
In our square, this calculation worked out to be straightforward because:
The area of the region is

• Width = 4 (from -2 to 2)
• Height = 4 (from -2 to 2),

producing a total area of:

• 16 square units.

Multiply this by the density (which is 1), to find the mass is 16 units, aligning perfectly with the given mass in the exercise. This simplifies the consideration in computation involving moments, as knowing the mass as the integral simplifies the moment calculations precisely.