Problem 89 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

EXPOSURE TO DISEASE The likelihood that a person with a contagious disease will infect others in a social situation may be assumed to be a function \(f(s)\) of the distance \(s\) between individuals. Suppose contagious individuals are uniformly distributed throughout a rectangular region \(R\) in the \(x y\) plane. Then the likelihood of infection for someone at the origin \((0,0)\) is proportional to the exposure index \(E\), given by the double integral $$ E=\iint_{R} f(s) d A $$ where \(s=\sqrt{x^{2}+y^{2}}\) is the distance between \((0,0)\) and \((x, y)\). Find \(E\) for the case where $$ f(s)=1-\frac{s^{2}}{9} $$ and \(R\) is the square $$ R:-2 \leq x \leq 2,-2 \leq y \leq 2 $$

Step-by-Step Solution

Answer

E = 11.26

Step 1: Define the Function and Region

Given the function describing the likelihood of infection is \( f(s) = 1 - \frac{s^2}{9} \), and the region \( R \) is the square with boundaries \( -2 \leq x \leq 2 \) and \( -2 \leq y \leq 2 \).

Step 2: Set up the Double Integral

The exposure index \( E \) is given by the double integral \[ E = \iint_{R} f(s) \, dA \]. Substitute \( f(s) = 1 - \frac{s^2}{9} \) and \( s = \sqrt{x^2 + y^2} \) to get \[ E = \iint_{R} \left( 1 - \frac{x^2 + y^2}{9} \right) dA \].

Step 3: Evaluate the Double Integral

The region \( R \) is \( -2 \leq x \leq 2 \) and \( -2 \leq y \leq 2 \). Evaluate the double integral by setting up and computing the integrals:\[ E = \int_{-2}^{2} \int_{-2}^{2} \left( 1 - \frac{x^2 + y^2}{9} \right) dy \, dx \].

Step 4: Integrate with Respect to y

First, integrate \( \left( 1 - \frac{x^2 + y^2}{9} \right) \) with respect to \( y \) over \( -2 \leq y \leq 2 \):\[ \int_{-2}^{2} \left( 1 - \frac{x^2}{9} - \frac{y^2}{9} \right) dy = \ \left[ y - \frac{yx^2}{9} - \frac{y^3}{27} \right]_{-2}^{2} \].

Step 5: Simplify the Result of Integration

Evaluate the expression by substituting the limits of integration:\[ \left[ 2 - \frac{2x^2}{9} - \frac{8}{27} \right] - \left[ -2 + \frac{2x^2}{9} + \frac{8}{27} \right] = 2 - \frac{2x^2}{9} - \frac{8}{27} + 2 - \frac{2x^2}{9} - \frac{8}{27} = 4 - \frac{4x^2}{9} - \frac{16}{27} \].

Step 6: Integrate with Respect to x

Now integrate the resultant function with respect to \( x \) over \( -2 \leq x \leq 2 \):\[ E = \int_{-2}^{2} \left( 4 - \frac{4x^2}{9} - \frac{16}{27} \right) dx \].

Step 7: Compute the Final Integral

Evaluate the remaining integral:\[ \int_{-2}^{2} 4 dx - \int_{-2}^{2} \frac{4x^2}{9} dx - \int_{-2}^{2} \frac{16}{27} dx \]. \ The first part results in \[ 4 \int_{-2}^{2} dx = 4[x]_{-2}^{2} = 4(2 - (-2)) = 4 \times 4 = 16 \]. \ The second part results in \[ \frac{4}{9} \int_{-2}^{2} x^2 dx = \frac{4}{9} \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{4}{9} \left[ \frac{8}{3} - \left( -\frac{8}{3}\right) \right] = \frac{4}{9} \times \frac{16}{3} = \frac{64}{27} \]. \ The third part results in \[ \frac{16}{27} \int_{-2}^{2} dx = \frac{16}{27}[x]_{-2}^{2} = \frac{16}{27} \times 4 = \frac{64}{27} \]. \ Therefore, \[ E = 16 - \frac{64}{27} - \frac{64}{27} = 16 - \frac{128}{27} = 16 - 4.74 = 11.26 \].

Key Concepts

Double Integrals

Double integrals are a way to integrate a function over a two-dimensional region. They are very useful in calculus, especially for finding areas, volumes, and average values of functions over a given region. In this exercise, we use a double integral to find the exposure index, which measures the likelihood of infection in a rectangular region. The double integral in this case is written as: \[E = \iint_{R} f(s) \, dA\]. Here, \(dA\) represents the area element in the region \(R\), and \(f(s)\) is the function describing the likelihood of infection. We need to integrate \(f(s)\) over the entire region defined by the given boundaries. Use double integrals when dealing with problems in two dimensions that involve accumulating quantities over an area.

Distance Functions

A distance function gives the distance between two points. In this problem, we need the distance from the origin \(0, 0\) to any point \(x, y\) in the rectangular region. The distance function is given by \(s = \sqrt{x^2 + y^2}\). This distance is then used to calculate the likelihood of infection using the function: \[f(s) = 1 - \frac{s^2}{9}\]. The function decreases as the distance \(s\) increases, representing a reduced likelihood of infection farther from the origin. The distance function is a crucial part of setting up and solving the double integral. Understanding how to calculate distances in a plane using the Pythagorean theorem is essential for problems like these.

Rectangular Regions in Plane

Rectangular regions in the plane are defined by their boundaries. In this problem, the region \( R \) is defined as \[-2 \leq x \leq 2, -2 \leq y \leq 2\]. Such regions are common in double integrals because they allow for straightforward integration limits. When dealing with rectangular regions, you typically integrate first in one direction (e.g., along the \(y\) axis) and then in the perpendicular direction (along the \(x\) axis). Defining the limits clearly is crucial for setting up and solving double integrals correctly. In this exercise, the integration limits help us break down the problem into manageable integration steps, ensuring an accurate solution.