Problem 15 - APEX Calculus

Question

Find the mass/weight of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\). \(R\) 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 mass of the lamina is \(16\pi \; \text{kg}.\)

Step 1: Define the Problem

We have a lamina (a flat surface with mass) described by a region \( R \), which is a disk centered at the origin with radius 2. The density function of the lamina is given by \( \delta(x, y) = x + y + 4 \; \text{kg/m}^2 \). The task is to find the total mass of this lamina.

Step 2: Determine the Integral Setup

The mass \( M \) of the lamina is found by integrating the density function over the region \( R \). The integral that gives the mass is given by:\[M = \int \int_R \delta(x, y) \, dA\]Since \( R \) is a circular region, it is advantageous to convert to polar coordinates where \( x = r\cos\theta \), \( y = r\sin\theta \), and \( dA = r\,dr\,d\theta \).

Step 3: Convert the Density Function to Polar Coordinates

Substituting \( x = r\cos\theta \) and \( y = r\sin\theta \) into the density function gives:\[\delta(r, \theta) = r\cos\theta + r\sin\theta + 4\]

Step 4: Set Limits for the Integral in Polar Coordinates

The region \( R \) is a disk centered at the origin with radius 2. In polar coordinates, this corresponds to:- \( 0 \leq r \leq 2 \)- \( 0 \leq \theta \leq 2\pi \)

Step 5: Calculate the Double Integral

Set up and evaluate the double integral in polar coordinates:\[M = \int_0^{2\pi} \int_0^2 (r \cos\theta + r \sin\theta + 4) r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 (r^2 \cos\theta + r^2 \sin\theta + 4r) \, dr \, d\theta\]First, evaluate the integral with respect to \( r \):\[\int_0^2 (r^2 \cos\theta + r^2 \sin\theta + 4r) \, dr = \left[\frac{r^3}{3}\cos\theta + \frac{r^3}{3}\sin\theta + 2r^2\right]_0^2\]Substitute the limits:\[= \left(\frac{8}{3}\cos\theta + \frac{8}{3}\sin\theta + 8\right) - 0\]Evaluate the \( \theta \) integral:\[M = \int_0^{2\pi} \left(\frac{8}{3}\cos\theta + \frac{8}{3}\sin\theta + 8\right) \, d\theta\]

Step 6: Complete the Evaluation of the Theta Integral

The definite integrals of \( \cos\theta \) and \( \sin\theta \) over \( [0, 2\pi] \) are zero, so:\[\int_0^{2\pi} \left(\frac{8}{3}\cos\theta + \frac{8}{3}\sin\theta\right) \, d\theta = 0\]Thus, the mass \( M \) simplifies to:\[M = \int_0^{2\pi} 8 \, d\theta = 8[\theta]_0^{2\pi} = 8 \times 2\pi = 16\pi\]

Step 7: Provide the Final Answer

The mass of the lamina is given by the expression \( 16\pi \). In terms of specific units, this would be mass in kilograms, given the density was in \( \text{kg/m}^2 \).

Key Concepts

Double Integral

A double integral is a way to compute the accumulation of a quantity over a two-dimensional area. It's like taking slices through an area and adding up all the values. Imagine you are calculating how much paint you need to cover a wall, but instead, you're covering a specific region with a density of material. In our problem, the double integral helps us find the total mass by summing the density function over the entire region. To set up a double integral, we first define the limits for the region over which we want to integrate. In Cartesian coordinates, this is usually straightforward with known boundaries for each variable, like rectangles. However, for circular regions or other non-rectangular areas, polar coordinates are often more suitable. This is because polar coordinates naturally describe circular shapes, and integrating in polar coordinates can simplify the calculation.

Polar Coordinates

Polar coordinates are an alternative coordinate system to Cartesian coordinates, which are based on distance from a central point (the origin) and the angle from a reference direction, usually the positive x-axis. This system is particularly useful when dealing with circular regions, as it aligns more naturally with circles than the typical x-y axis system does.To convert from Cartesian to polar coordinates, we use the relationships:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• The area element \( dA = r \, dr \, d\theta \)

In our problem, the region is a disk centered at the origin with radius 2, making polar coordinates a great fit. These relationships help us transform the coordinates of our density function to make integration over the circular region more straightforward.

Density Function

A density function gives a way to describe how much mass is distributed over an area. It's the amount of mass per unit area, reported in units like \( ext{kg/m}^2 \). Interpreting density functions correctly is crucial, as they allow us to model real-world scenarios where materials are distributed unevenly over space.The problem gives a density function of \( \delta(x, y) = x + y + 4 \). This implies the density varies linearly with a little bump from the constant 4. When converting to polar coordinates, it becomes \( \delta(r, \theta) = r \cos \theta + r \sin \theta + 4 \). Here, the change of variables does not alter the composition of the function but adapts it to fit the polar system, easing the integration process for the mass of our region.

Circular Region

Circular regions in mathematics are areas defined by a circle. For calculations involving mass or area inside a circular region, using polar coordinates simplifies the computation because circular geometry aligns naturally with the radial measures employed in polar coordinates.For our region, defined as a disk centered at the origin with radius 2, using polar coordinates means the radial distance \( r \) varies from 0 to 2, and the angle \( \theta \) spans from 0 to \( 2\pi \). This description allows the integral to account for every point within the circular boundary efficiently.Handling circular regions in this way translates the integration process into manageable mathematical limits, allowing us to compute the total mass of the lamina accurately.