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Find the average value of \(f(x, y)=1 /(x y)\) over the square \(\ln 2 \leq x \leq 2 \ln 2, \ln 2 \leq y \leq 2 \ln 2\)

Short Answer

Expert verified
The average value of the function over the given square is 1.

Step by step solution

01

Understand the Problem

We want to find the average value of the function \(f(x, y) = \frac{1}{xy}\) over a square defined by \(\ln 2 \leq x \leq 2 \ln 2\) and \(\ln 2 \leq y \leq 2 \ln 2\). This requires us to integrate the function over the given region and then divide by the area of the region.
02

Calculate the Area of the Region

The region is a square with side length \(2 \ln 2 - \ln 2 = \ln 2\). Thus, the area of the square is \((\ln 2)^2 = (\ln 2)^2\).
03

Set Up the Integral for Average Value

The average value of a function \(f(x, y)\) over a region \(R\) is given by \(\frac{1}{\text{Area of } R} \int \int_{R} f(x, y) \, dA\). Here, the integral becomes \[\frac{1}{(\ln 2)^2} \int_{\ln 2}^{2 \ln 2} \int_{\ln 2}^{2 \ln 2} \frac{1}{xy} \, dy \, dx.\]
04

Evaluate the Double Integral

First, integrate with respect to \(y\): \[\int_{\ln 2}^{2 \ln 2} \frac{1}{xy} \, dy = \frac{1}{x} \left[ \ln y \right]_{\ln 2}^{2 \ln 2} = \frac{1}{x} (\ln(2 \ln 2) - \ln(\ln 2)) = \frac{\ln 2}{x}\].
05

Integrate Over \(x\)

Next, integrate with respect to \(x\): \[\int_{\ln 2}^{2 \ln 2} \frac{\ln 2}{x} \, dx = \ln 2 \left[ \ln x \right]_{\ln 2}^{2 \ln 2} = \ln 2 (\ln(2 \ln 2) - \ln(\ln 2)) = \ln 2 \cdot \ln 2\].
06

Compute the Average Value

The average value is \[\frac{1}{(\ln 2)^2} \cdot \ln 2 \cdot \ln 2 = \frac{(\ln 2)^2}{(\ln 2)^2} = 1\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Integration
Double integration can be understood as the process of finding the integral of a function of two variables over a specified region. When dealing with double integrals, it is similar to how we integrate single-variable functions, but now we are working with two dimensions. This involves integrating two times, each with respect to one of the variables, often labeled as \(x\) and \(y\).

Think of double integration as calculating the volume under a surface defined by a function \(f(x, y)\) over a certain region \(R\). In the example problem, the function \(f(x, y) = \frac{1}{xy}\) is integrated over a square region. To achieve this, we first set up the integral based on the bounds of the square, and then proceed to carry out the integration process first with respect to \(y\) and then with respect to \(x\).
  • First integral: Integrate \(f(x, y)\) with respect to \(y\) while holding \(x\) constant.
  • Second integral: Integrate the result with respect to \(x\).
The ultimate goal of double integration is to accumulate all the tiny volumes described by the function and the area of the base region, yielding a total integral value.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to more than one variable at a time. This branch of mathematics allows us to study functions that depend on two or more variables, which is essential in fields like physics, engineering, and economics. In our example, we work with a function \(f(x, y) = \frac{1}{xy}\), which is defined over two variables \(x\) and \(y\).

This type of calculus is crucial because real-world applications often involve several changing quantities that depend on one another. Understanding these relationships gives insights that single-variable calculus cannot provide. Multivariable calculus includes several techniques such as partial derivatives, multiple integrals, and vector calculus. In double integration, which is part of multivariable calculus, each integral captures the variation concerning one of the variables.
  • Partial Derivatives: Measure how a function changes as individual variables change.
  • Multiple Integrals: Used to accumulate quantities over a region with more than one variable.
  • Applications: Can describe phenomena like fluid flow, electromagnetic fields, and much more.
Integration over a Region
Integration over a region involves calculating an integral across a specified area on the \(xy\)-plane. In double integrals, the region over which you are integrating is crucial because it directly impacts the calculation of the integral.

In our problem, the region is a square. This simplifies the integration process since the bounds of \(x\) and \(y\) are constants. The endpoints of the double integrals are determined by this region: \(\ln 2 \leq x \leq 2 \ln 2\) and \(\ln 2 \leq y \leq 2 \ln 2\).
  • The region is defined by the limits of \(x\) and \(y\).
  • The area of the region can often be found by analytical geometry methods, like multiplying the side lengths in case of a rectangle or square.
  • Integrating over the region accumulates function values across the defined area.
The integration bounds determine which portions of the function's 3-dimensional surface are included in the volume calculation.

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Most popular questions from this chapter

Let \(D_{\mathrm{u}} f\) denote the derivative of \(f(x, y)=\left(x^{2}+y^{2}\right) / 2\) in the direction of the unit vector \(\mathbf{u}=u_{1} \mathbf{i}+u_{2} \mathbf{j}\) a. Finding average value Find the average value of \(D_{\mathrm{u}} f\) over the triangular region cut from the first quadrant by the line \(x+y=1 .\) b. Average value and centroid Show in general that the average value of \(D_{\mathbf{u}} f\) over a region in the \(x y\) -plane is the value of \(D_{\mathbf{u}} f\) at the centroid of the region.

a. Suppose that a solid right circular cone \(C\) of base radius \(a\) and altitude \(h\) is constructed on the circular base of a solid hemisphere \(S\) of radius \(a\) so that the union of the two solids resembles an ice cream cone. The centroid of a solid cone lies one-fourth of the way from the base toward the vertex. The centroid of a solid hemisphere lies three-eighths of the way from the base to the top. What relation must hold between \(h\) and \(a\) to place the centroid of \(C \cup S\) in the common base of the two solids?

Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that $$ \int_{0}^{1} f(x)\left(\int_{0}^{x} g(x-y) f(y) d y\right) d x $$ $$ \begin{array}{l}{=\int_{0}^{1} f(y)\left(\int_{y}^{1} g(x-y) f(x) d x\right) d y} \\ {=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} g(|x-y|) f(x) f(y) d x d y}\end{array} $$

Moments of inertia of solid circular eylinder Find the moment of inertia of a solid circular cylinder of radius 1 and height 2 (a) about the axis of the cylinder and (b) about a line through the centroid perpendicular to the axis of the cylinder. (Take \(\delta=1 . )\)

In Exercises \(37-40,\) find the average value of \(F(x, y, z)\) over the given region. \(F(x, y, z)=x y z\) over the cube in the first octant bounded by the coordinate planes and the planes \(x=2, y=2,\) and \(z=2\)

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