Problem 23 - APEX Calculus

Question

Find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the disk centered at the origin with radius \(2 ; \delta(x, y)=\) \((x+y+4) \mathrm{kg} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

The center of mass is at (0, 0).

Step 1: Identify the Region and Density Function

We are given that the region \( R \) is a disk centered at the origin with a radius of 2. The density function for this region is \( \delta(x, y) = x + y + 4 \). First, we need to remember that the center of mass \( ( \bar{x}, \bar{y} ) \) is calculated using integrals of the form:\[ \bar{x} = \frac{1}{M} \iint_R x \delta(x, y) \, dA \]\[ \bar{y} = \frac{1}{M} \iint_R y \delta(x, y) \, dA \]\[ M = \iint_R \delta(x, y) \, dA \]

Step 2: Convert to Polar Coordinates

Since \( R \) is a circle, it's easier to use polar coordinates. In polar coordinates, \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The conversion of the disk is:- \( 0 \leq r \leq 2 \)- \( 0 \leq \theta \leq 2\pi \)The density function becomes \( \delta(r, \theta) = r \cos(\theta) + r \sin(\theta) + 4 \) and the area element \( dA = r \, dr \, d\theta \).

Step 3: Compute the Total Mass \( M \)

The total mass \( M \) is calculated by:\[ M = \int_0^{2\pi} \int_0^2 (r \cos(\theta) + r \sin(\theta) + 4) \, r \, dr \, d\theta \]Evaluating the inner integral first:\[ \int_0^2 (r^2 \cos(\theta) + r^2 \sin(\theta) + 4r) \, dr = \left[ \frac{r^3}{3} \cos(\theta) + \frac{r^3}{3} \sin(\theta) + 2r^2 \right]_0^2 \]Substituting the limits:\[ = \left(\frac{8}{3} \cos(\theta) + \frac{8}{3} \sin(\theta) + 8\right) \]

Step 4: Integrate over \( \theta \) to find \( M \)

Now integrate with respect to \( \theta \):\[ M = \int_0^{2\pi} \left(\frac{8}{3} \cos(\theta) + \frac{8}{3} \sin(\theta) + 8\right) \, d\theta \]Due to symmetry, the integrals of \( \cos(\theta) \) and \( \sin(\theta) \) over \( 0 \) to \( 2\pi \) are zero:\[ M = \int_0^{2\pi} 8 \, d\theta = 8[\theta]_0^{2\pi} = 16\pi \]

Step 5: Compute \( \bar{x} \) and \( \bar{y} \)

Calculate \( \bar{x} \):\[ \bar{x} = \frac{1}{16\pi} \int_0^{2\pi} \int_0^2 r \cos(\theta)(r \cos(\theta) + r \sin(\theta) + 4) \, r \, dr \, d\theta \]Similarly, calculate \( \bar{y} \):\[ \bar{y} = \frac{1}{16\pi} \int_0^{2\pi} \int_0^2 r \sin(\theta)(r \cos(\theta) + r \sin(\theta) + 4) \, r \, dr \, d\theta \]Both integrals over symmetric limits will yield 0 due to the symmetry of the trigonometric functions over complete cycles, thus finally giving:\[ \bar{x} = 0, \quad \bar{y} = 0 \]

Step 6: Conclusion: Find the Center of Mass

Since both \( \bar{x} \) and \( \bar{y} \) are 0, the center of mass of the lamina is at the origin (0,0). Given the symmetry of the region and the density function, this result is consistent.

Key Concepts

Density Function

A density function describes how mass is distributed across a region. In this case, the density function is given as \(\delta(x, y) = x + y + 4\text{ kg/m}^2\). This function tells us that mass varies at each point \((x, y)\) within the region.
The density depends on both the x and y coordinates and adds a constant 4. This variation play a crucial role when calculating properties like mass, center of mass, and moments.
To compute these properties, we integrate the density function over the entire region. In our exercise, we find the total mass \(M\) by integrating \(\delta(x, y)\) throughout the disk centered at the origin.

Polar Coordinates

When dealing with circular regions, such as disks, converting to polar coordinates simplifies the problem. In polar coordinates, a point is defined by \( (r, \theta) \), where \(r\) is the radius or distance from the origin, and \(\theta\) is the angle from the positive x-axis. These are related to Cartesian coordinates by:

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

The transformation of the region to polar coordinates is logical when dealing with circular symmetry.
So, the bounds for \(r\) range from 0 to the radius of the disk, here 2, and \(\theta\) ranges from 0 to \(2\pi\). This conversion transfers the problem to a more manageable framework for integration, making it easier to exploit the symmetry of the problem.

Integration

Integration is a fundamental tool for calculating areas, volumes, and mass, among other things. In this exercise, it allows us to compute the total mass and, subsequently, the center of mass of the lamina.
We integrated the density function over the entire disk to find the total mass \(M\). The expression: \[ M = \int_0^{2\pi} \int_0^2 (r \cos(\theta) + r \sin(\theta) + 4) \times r \, dr \, d\theta \] accounts for all the mass across the circular region.
The process involves

• Integrating first with respect to \(r\), the radial component
• Then integrating with respect to \(\theta\), the angular component

Executing these steps allows us to accumulate the mass over the entire region functionally.

Symmetry in Trigonometric Functions

Symmetry can drastically simplify calculations, especially when trigonometric functions are involved. In the exercise, we see that the integrals of \(\cos(\theta)\) and \(\sin(\theta)\) over \(0\) to \(2\pi\) are zero. This is a result of their periodic nature and symmetry over a full cycle.
Thus, using these properties allows us to claim that the center of mass is at (0,0) without needing to evaluate every term. This simplification shows that integrating these functions over symmetrical limits yields no contribution to the center of mass.
Recognizing and exploiting such symmetries in functions can save substantial time and effort in computations.