Question

Find the average value of \(f\) over the region \(R .\) Notice how these functions and regions are related to the iterated integrals given in Exercises \(5-8\). \(f(x, y)=\sin x \cos y ; \quad R\) is bounded by \(x=0, x=\pi\) \(y=-\pi / 2\) and \(y=\pi / 2\).

Short answer

The average value of the function is \( \frac{4}{\pi^2} \).

Step-by-step solution

Identify the region boundaries

The region \( R \) is defined by \( x = 0 \), \( x = \pi \), \( y = -\pi/2 \), and \( y = \pi/2 \). This forms a rectangle in the \( xy \)-plane with width \( \pi \) and height \( \pi \).

Set up the double integral for the area of the region

To find the area of region \( R \), the double integral is set up as \( A = \int_{0}^{\pi} \int_{-\pi/2}^{\pi/2} 1 \; dy \; dx \). This integral measures the area across both dimensions.

Calculate the area of the region

Compute the area by integrating the function \( 1 \):\[A = \int_{0}^{\pi} \left[ y \right]_{-\pi/2}^{\pi/2} \; dx = \int_{0}^{\pi} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) \; dx = \int_{0}^{\pi} \pi \; dx = \pi^2.\]

Set up the double integral for the function over the region

To find the total integral of \( f(x, y) = \sin x \cos y \), set up the double integral: \( I = \int_{0}^{\pi} \int_{-\pi/2}^{\pi/2} \sin x \cos y \; dy \; dx \).

Evaluate the inner integral

Evaluate the inner integral with respect to \( y \):\[\int_{-\pi/2}^{\pi/2} \cos y \; dy = \left[ \sin y \right]_{-\pi/2}^{\pi/2} = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) = 1 + 1 = 2.\]

Evaluate the outer integral

Substitute the result from Step 5 into the outer integral: \[I = \int_{0}^{\pi} (\sin x)(2) \; dx = 2 \int_{0}^{\pi} \sin x \; dx = 2 [-\cos x]_{0}^{\pi} = 2[-(-1) - (1)] = 2[1 + 1] = 4.\]

Calculate the average value of the function

The average value of \( f(x, y) \) over the region \( R \) is:\[f_{\text{avg}} = \frac{I}{A} = \frac{4}{\pi^2}.\]

Key concepts

Average Value of a Function

The average value of a function over a specified region can offer insights into the general behavior of that function in the region. To understand this concept, think about it as the mean level of the function's output across the entire area. This is particularly useful when dealing with continuous functions over multidimensional spaces.

To find the average value of a function, say \( f(x, y) \), in a region \( R \), you divide the total integral of the function over the area by the area of the region itself:

• First, calculate the total integral \( I \) by extending the definite integration process to two dimensions, often referred to as performing a double integral.
• Next, determine the area \( A \) of the region \( R \). This is also found using integration but with the function set to \( 1 \).
• Finally, divide \( I \) by \( A \) to get the average value.

For example, in the original exercise, using the defined region boundaries and double integrals, the average value is found to be \( f_{\text{avg}} = \frac{4}{\pi^2} \). This means if you consider the fluctuations of the function \( \sin x \cos y \) across the rectangle, the mean value it takes throughout the region is \( \frac{4}{\pi^2} \).

Understanding this average helps in predicting the behavior of the function over the specific region.

Iterated Integrals

Iterated integrals extend the idea of integration from one dimension to two or more dimensions. In essence, a double integral calculates the volume under a surface described by a function over a given region. The process involves computing integrals sequentially.

Here’s a step-by-step guide to understanding iterated integrals:

• Inner Integral: First, identify and solve the inner integral. This involves holding one variable constant while integrating with respect to the others.
• Outer Integral: Use the result of the inner integral as the integrand for the outer integration and solve it with respect to the remaining variable.

In the provided solution, the inner integral of \( \sin x \cos y \) over \( y \) was solved first, resulting in \( 2\sin x \). Then, this result was used to compute the outer integral over \( x \). This sequential process is what defines iterated integrals.

The order of integration matters: switching the order could change the complexity or even the feasibility of solving the integral, though in current contexts, the final numerical result should remain the same.

Definite Integration

Definite integration involves finding the integral of a function within defined bounds. In simpler terms, it allows you to compute the accumulated "volume" under a curve between two limits.

For functions of two variables, the concept extends naturally to regions in the plane, where integration is done with respect to both variables across specific boundaries:

• Defining Limits: Set the limits for both the \( x \) and \( y \) directions based on the problem scope (e.g., \( x = 0 \) to \( x = \pi \)).
• Integration Process: Using the defined limits, perform integration firstly over one variable and then over the next (see iterated integrals).

In the example given, definite integration was applied twice: once to determine the area \( A \) of the region (using 1 as the function) and the other to determine the total integral value \( I \) of \( \sin x \cos y \).

This technique is crucial because it not only measures total quantities (like volume or area) but also sets the stage for understanding averages, changes, and accumulation within defined bounds, being integral to multi-variable calculus.