Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(R=\\{(x, y): 1 \leq x \leq 4,0 \leq y \leq 2\\}\). Evaluate \(\iint_{R} f(x, y) d A\), where \(f\) is the given function. \(f(x, y)= \begin{cases}2 & 1 \leq x<3,0 \leq y<1 \\ 1 & 1 \leq x<3,1 \leq y \leq 2 \\ 3 & 3 \leq x \leq 4,0 \leq y \leq 2\end{cases}\)

Step-by-Step Solution

Answer

The integral over the region \( R \) is 12.

Step 1: Understand the Region R

The region \( R \) is defined by the Cartesian product \( \{(x, y): 1 \leq x \leq 4, 0 \leq y \leq 2\} \). This defines a rectangle with the corners (1,0), (4,0), (1,2), and (4,2) on the xy-plane.

Step 2: Break Down Function f(x, y)

The function \( f(x,y) \) is piecewise defined as follows: \( f(x, y) = 2 \) for \( 1 \leq x < 3, 0 \leq y < 1 \), \( f(x, y) = 1 \) for \( 1 \leq x < 3, 1 \leq y \leq 2 \), and \( f(x, y) = 3 \) for \( 3 \leq x \leq 4, 0 \leq y \leq 2 \). Each condition refers to specific sub-regions of \( R \).

Step 3: Define Sub-regions for Integration

The region can be divided into three sub-regions based on the function definition: 1. \( R_1: 1 \leq x < 3, 0 \leq y < 1 \)2. \( R_2: 1 \leq x < 3, 1 \leq y \leq 2 \)3. \( R_3: 3 \leq x \leq 4, 0 \leq y \leq 2 \)

Step 4: Integrate Over Sub-region R1

For \( R_1 \), where \( f(x, y) = 2 \), the integral is \[ \iint_{R_1} f(x, y) \, dA = \int_{x=1}^{3} \int_{y=0}^{1} 2 \, dy \, dx \].Calculate the inner integral first: \[ \int_{0}^{1} 2 \, dy = 2 \times 1 = 2 \].Then, evaluate the outer integral: \[ \int_{1}^{3} 2 \, dx = 2 \times 2 = 4 \].Thus, \( \iint_{R_1} f(x, y) \, dA = 4 \).

Step 5: Integrate Over Sub-region R2

For \( R_2 \), where \( f(x, y) = 1 \), the integral is \[ \iint_{R_2} f(x, y) \, dA = \int_{x=1}^{3} \int_{y=1}^{2} 1 \, dy \, dx \].Calculate the inner integral: \[ \int_{1}^{2} 1 \, dy = 1 \times 1 = 1 \].Then, evaluate the outer integral: \[ \int_{1}^{3} 1 \, dx = 1 \times 2 = 2 \].Thus, \( \iint_{R_2} f(x, y) \, dA = 2 \).

Step 6: Integrate Over Sub-region R3

For \( R_3 \), where \( f(x, y) = 3 \), the integral is \[ \iint_{R_3} f(x, y) \, dA = \int_{x=3}^{4} \int_{y=0}^{2} 3 \, dy \, dx \].Calculate the inner integral: \[ \int_{0}^{2} 3 \, dy = 3 \times 2 = 6 \].Then, evaluate the outer integral: \[ \int_{3}^{4} 6 \, dx = 6 \times 1 = 6 \].Thus, \( \iint_{R_3} f(x, y) \, dA = 6 \).

Step 7: Sum of the Integrals Over All Sub-regions

Add the results from all the sub-regions: \[ \iint_R f(x, y) \, dA = \iint_{R_1} f(x, y) \, dA + \iint_{R_2} f(x, y) \, dA + \iint_{R_3} f(x, y) \, dA \]\[ = 4 + 2 + 6 = 12 \].This gives the total integral over the entire region \( R \).

Key Concepts

Piecewise Functions

Piecewise functions are a special type of function that have different expressions depending on the input value. Imagine switching gears to adapt to changes; this is precisely what piecewise functions do in the world of math. Their domains are broken into various intervals, each associated with a different sub-function.

The provided piecewise function, \( f(x, y) \), has three distinct conditions. Each condition applies to a separate sub-region within the overall region of integration. For example:

• \( f(x, y) = 2 \) when \( 1 \leq x < 3 \) and \(0 \leq y < 1 \)
• \( f(x, y) = 1 \) when \( 1 \leq x < 3 \) and \( 1 \leq y \leq 2 \)
• \( f(x, y) = 3 \) when \( 3 \leq x \leq 4 \) and \( 0 \leq y \leq 2 \)

Understanding which specific part to use is crucial when evaluating integrals. It ensures that each portion behaves according to its own defined rules within those sub-regions.

Region of Integration

In integral calculus, the region of integration plays a critical role as it describes the area over which we are integrating. For our exercise, this region is denoted by \( R=\{(x, y): 1 \leq x \leq 4, 0 \leq y \leq 2\} \), essentially forming a rectangle in the Cartesian plane.
Identifying this region is the starting point in solving multi-variable integration problems. We need to fully understand the limits for both \( x \) and \( y \) axes to apply integration properly.

• \( 1 \leq x \leq 4 \) defines the horizontal stretch of the rectangle.
• \( 0 \leq y \leq 2 \) determines its vertical stretch.

Integrating over this region means calculating the area under the curve defined by the function within these limits. By evaluating the integral over the specified intervals, we can find how the function behaves throughout the entire rectangle.

Cartesian Coordinates

Cartesian coordinates are a way of representing points on a plane by specifying their horizontal and vertical positions. Think of this system as providing an address for each point in a two-dimensional space. Each point is expressed as \((x, y)\), where \(x\) is the distance along the horizontal (x-axis) and \(y\) is the distance along the vertical (y-axis).

In our exercise, these coordinates define the boundaries of the region of integration, with rectangle corners at points like (1, 0) and (4, 2). Being in Cartesian coordinates allows us to easily visualize the region and understand how the piecewise function is applied.

• The x-coordinates \( 1 \) and \( 4 \) specify where the region begins and ends horizontally.
• The y-coordinates \( 0 \) and \( 2 \) highlight the vertical boundaries.

Understanding where every point lies in Cartesian space is fundamental, helping simplify complex calculations as we set up and solve integrals.

Integral Calculus

Integral calculus is a branch of calculus focused on accumulation functions, which helps calculate areas under curves or the cumulative sum of these areas. At its heart, it deals with integrals, which collect and sum up continuous data points.

In this task, we are evaluating an integral over a defined region \( R \). By performing double integration, we account for area accumulation both horizontally and vertically under the piecewise function.

• First, we integrate with respect to \( y \), gathering area slices vertically.
• Then, we integrate with respect to \( x \), combining these slices horizontally.

The result of this process is the total area under the curve across the specified region. Importantly, integrating over the subdivided regions accounts for each condition of the piecewise function. Thus, ensuring the integration reflects all variations in the function's behavior accurately.