Problem 19 - Thomas Calculus

Question

Find the average value of \(f ( x , y ) = \sin ( x + y )\) over \(\begin{array} { l l } { \text { a. the rectangle } 0 \leq x \leq \pi , } & { 0 \leq y \leq \pi } \\ { \text { b. the rectangle } 0 \leq x \leq \pi , } & { 0 \leq y \leq \pi / 2 } \end{array}\)

Step-by-Step Solution

Answer

The average value over both rectangles is 0.

Step 1: Understand the Problem

We need to find the average value of the function \( f(x, y) = \sin(x + y) \) over two different rectangles in the \( xy \)-plane. The average value of a function \( f(x,y) \) over a region \( R \) can be calculated using the formula: \[ \text{Average value} = \frac{1}{\text{Area of } R} \int \int_{R} f(x, y) \, dA. \]

Step 2: Calculate the Area of Rectangle a

For rectangle (a), the boundaries are \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \pi \). The area is calculated as: \( \text{Area} = (\pi - 0)(\pi - 0) = \pi^2 \).

Step 3: Set Up the Double Integral for Rectangle a

For rectangle (a), we need to integrate \( \sin(x+y) \) from \( x = 0 \) to \( x = \pi \) and \( y = 0 \) to \( y = \pi \). The integral is: \[ \int_0^\pi \int_0^\pi \sin(x+y) \, dy \, dx. \]

Step 4: Solve the Double Integral for Rectangle a

First, solve the inner integral with respect to \( y \): \( \int \sin(x+y) \, dy = -\cos(x+y) + C \). Evaluate from \( y=0 \) to \( y=\pi \): \[ [-\cos(x+\pi) + \cos(x)] \]. Thus, the result becomes \( \int_0^\pi [\cos(x) + \cos(x) ] \, dx = \int_0^\pi 2 \cos(x) \, dx \). This simplifies to \( [2\sin(x)]_0^\pi = 0 \).

Step 5: Calculate the Average Value for Rectangle a

With the integral evaluated to zero and area \( \pi^2 \), the average value of the function over Region a is: \( \frac{0}{\pi^2} = 0 \).

Step 6: Calculate the Area of Rectangle b

For rectangle (b), the boundaries are \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \pi/2 \). The area is: \( \text{Area} = (\pi - 0)\left( \frac{\pi}{2} - 0 \right) = \frac{\pi^2}{2} \).

Step 7: Set Up and Solve the Double Integral for Rectangle b

For rectangle b, integrate \( \sin(x+y) \) from \( x = 0 \) to \( x = \pi \) and \( y = 0 \) to \( y = \pi/2 \). Solve: \( \int_0^\pi \int_0^{\pi/2} \sin(x+y) \, dy \, dx \). This results in evaluating: \( [-\cos(x+y)]_0^{\pi/2} = [\cos(x) - \cos(x + \pi/2)]= [\cos(x) - 0] \). The integral becomes \( \int_0^\pi \cos(x) \, dx \).

Step 8: Evaluate the Integral of Cosine for Rectangle b

The integral \( \int_0^\pi \cos(x) \, dx \) becomes \( [\sin(x)]_0^\pi = 0 \).

Step 9: Calculate the Average Value for Rectangle b

The integral evaluates to 0, and the area is \( \frac{\pi^2}{2} \). The average is \( \frac{0}{\frac{\pi^2}{2}} = 0 \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that focuses on the concept of integration. It deals with finding areas under curves, volumes of shapes, and other quantities that can be accumulated continuously. In this particular exercise, we use integral calculus to obtain the average value of a function over a defined region. The function given is \( f(x, y) = \sin(x + y) \), and we are tasked with finding its average over two specified rectangles.
To compute the average, we first determine the area of the region, which involves multiplying the lengths of the sides of the rectangle. For example, in Rectangle (a), the sides are \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \pi \), resulting in an area of \( \pi^2 \). Then, we use a double integral to integrate \( \sin(x + y) \) over this area. The general formula for the average value is given by:

• \( \text{Average value} = \frac{1}{\text{Area of } R} \int \int_{R} f(x, y) \, dA \)

This approach shows how integral calculus helps in determining cumulative measures over defined regions.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions involving more than one variable. In this exercise, we are working with a function \( f(x, y) \) that depends on both \( x \) and \( y \), which makes it necessary to apply multivariable techniques. Particularly, we employ double integration, a fundamental concept in multivariable calculus, to evaluate the average value of our function.
In double integration, we integrate consecutively over two variables, typically described as integrating over each limit of \( x \) first and then over \( y \), or vice versa. This involves setting up a region \( R \) over which the integration takes place. In our case:

• Rectangle (a) involves limits \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \pi \).
• Rectangle (b) involves limits \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \frac{\pi}{2} \).

These integrations over rectangles require solving an inner integral for one variable and then an outer integral for the other variable. This illustrates how multivariable calculus allows us to explore more complex shapes and functions.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that contain trigonometric expressions. In this exercise, we are dealing with the integral of \( \sin(x + y) \), which is a common trigonometric function. Such integrals often require special integration techniques, like substitution, perceived in how we handle the inner integration with respect to \( y \).
When integrating \( \sin(x + y) \) inside the given boundaries, we use an identity from calculus:

• \( \int \sin(x + y) \, dy = -\cos(x + y) + C \).

This identity helps us find definite integrals by evaluating it over the limits provided in the problem, allowing us to solve these trigonometric integrals systematically. In our calculations for both rectangles, despite the trigonometric nature, the integrals simplify to zero. This shows that even with potential complexity, trigonometric integrals can yield neat solutions when done methodically.