Chapter 16: Problem 57
Find the average value of \(f(x, y)=x^{2}+4 y\) on the rectangle \(0 \leq x \leq 3\) and \(0 \leq y \leq 6\)
Short Answer
Expert verified
The average value of the function is 15.
Step by step solution
01
Understand the Average Value Formula for a Function of Two Variables
To find the average value of a function \(f(x, y)\) over a region \(R\), we use the formula \[ \text{Average Value} = \frac{1}{\text{Area of } R} \iint_R f(x, y) \, dA \]. Here, the function is \(f(x, y) = x^2 + 4y\) and \(R\) is the rectangle defined by \(0 \leq x \leq 3\) and \(0 \leq y \leq 6\).
02
Calculate the Area of the Region R
The region \(R\) is a rectangle with dimensions \(3 \times 6\). Thus, the area of \(R\) is the product of its sides, i.e., \(3 \times 6 = 18\).
03
Set Up the Double Integral
Set up the double integral that will compute the total value of the function over the rectangle:\[ \iint_R x^2 + 4y \, dA = \int_0^6 \int_0^3 (x^2 + 4y) \, dx \, dy. \]
04
Evaluate the Inner Integral
Evaluate the inner integral with respect to \(x\):\[ \int_0^3 (x^2 + 4y) \, dx = \left[ \frac{x^3}{3} + 4yx \right]_0^3 = \left(\frac{3^3}{3} + 4y \times 3\right) - \left(\frac{0^3}{3} + 4y \times 0\right). \]This simplifies to \(9 + 12y\).
05
Evaluate the Outer Integral
Evaluate the outer integral with respect to \(y\):\[ \int_0^6 (9 + 12y) \, dy = \left[ 9y + 6y^2 \right]_0^6 = (9 \times 6 + 6 \times 6^2) - (9 \times 0 + 6 \times 0^2). \]This simplifies to \(54 + 216 = 270\).
06
Compute the Average Value
Finally, divide the result from the integral by the area of the region to find the average value:\[ \text{Average Value} = \frac{270}{18} = 15. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integration
Double integration is a fundamental concept in multivariable calculus, facilitating the process of finding the integral of a function with two variables, such as the given exercise requires. Instead of evaluating a single integral, you evaluate a nested or double integral over a specified region. This process is crucial for calculating things like area, volume, and average value under a surface represented by a function of two variables.
The double integral notation is written as \( \iint_R f(x, y) \, dA \), where \( R \) is the region over which you integrate. In a Cartesian coordinate system, this typically involves integrating with respect to \( x \) first and then \( y \), or vice versa. Each of these integrals is called an inner and an outer integral, respectively.
The double integral notation is written as \( \iint_R f(x, y) \, dA \), where \( R \) is the region over which you integrate. In a Cartesian coordinate system, this typically involves integrating with respect to \( x \) first and then \( y \), or vice versa. Each of these integrals is called an inner and an outer integral, respectively.
- Inner Integral: You integrate with respect to one variable, keeping the other fixed.
- Outer Integral: This is performed with respect to the second variable, using the result from the inner integral.
Area of a Region
The concept of the area of a region is fundamental when discussing the average value of a function over a designated space. For the exercise given, the region \( R \) is a rectangle in the first quadrant with boundaries \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq 6 \).
Calculating the area of such a region is straightforward since the area of a rectangle is simply the product of its length and width. Therefore, the area of \( R \) is calculated as \( 3 \times 6 = 18 \). This value is pivotal because it normalizes the result of the double integral, giving you an average rather than a total sum.
Calculating the area of such a region is straightforward since the area of a rectangle is simply the product of its length and width. Therefore, the area of \( R \) is calculated as \( 3 \times 6 = 18 \). This value is pivotal because it normalizes the result of the double integral, giving you an average rather than a total sum.
- The region's area enables the transformation from a total value (from integration) to an average value that reflects the function’s behavior per unit of the area.
- Utilizing the area as a divisor helps express how function values are distributed over the specified region.
Function of Two Variables
Functions of two variables, such as \( f(x, y) = x^2 + 4y \), are expressions where the output depends on two input variables. These are common in multivariable calculus and modeling scenarios where one dimension alone isn't enough to describe a situation.
Whenever dealing with two-variable functions, visualize them as surfaces in a three-dimensional space. In the exercise, the function \( f(x, y) = x^2 + 4y \) defines a surface over the rectangular region \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq 6 \). It gives every point on this rectangle a height, providing a more comprehensive view of the interactions between the two variables:
Whenever dealing with two-variable functions, visualize them as surfaces in a three-dimensional space. In the exercise, the function \( f(x, y) = x^2 + 4y \) defines a surface over the rectangular region \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq 6 \). It gives every point on this rectangle a height, providing a more comprehensive view of the interactions between the two variables:
- Height Interpretation: The function's output value can be imagined as a height above the \( xy \)-plane, illustrating how \( x \) and \( y \) control the surface's elevation.
- Behavior Analysis: The function's form helps predict how the outputs change across the region, offering insights into increasing or decreasing trends, peaks, and troughs.