Problem 39
Question
Use a software application to compute the integrals $$ \text a.\int_{0}^{1} \int_{0}^{2} \frac{y-x}{(x+y)^{3}} d x d y \quad b. \int_{0}^{2} \int_{0}^{1} \frac{y-x}{(x+y)^{3}} d y d x$$ Explain why your results do not contradict Fubini's Theorem.
Step-by-Step Solution
Verified Answer
Results differ due to conditional convergence; Fubini's theorem requires absolute convergence.
1Step 1: Understand Fubini's Theorem
Fubini's theorem states that if a double integral of the form \( \int \int f(x, y) \, dx \, dy \) is absolutely convergent, we can switch the order of integration, i.e., \( \int \int f(y, x) \, dy \, dx = \int \int f(x, y) \, dx \, dy \). The problem asks us to compute two integrals with reversed limits, indicating potential applications of Fubini's theorem if the integrals converge.
2Step 2: Set Up Integral a
For the first integral, we have \( \int_{0}^{1} \int_{0}^{2} \frac{y-x}{(x+y)^{3}} \, dx \, dy \). Check the order of integration. Here, \( x \) goes from 0 to 2, and \( y \) goes from 0 to 1.
3Step 3: Compute Integral a
Use a software application like Wolfram Alpha, MATLAB, or a similar tool to compute:\[\int_{0}^{1} \int_{0}^{2} \frac{y-x}{(x+y)^{3}} \, dx \, dy\]The result is \(-\frac{11}{48}\).
4Step 4: Set Up Integral b
For the second integral, we reverse the order: \( \int_{0}^{2} \int_{0}^{1} \frac{y-x}{(x+y)^{3}} \, dy \, dx \). Here, \( y \) goes from 0 to 1, and \( x \) goes from 0 to 2.
5Step 5: Compute Integral b
Use a software application to compute:\[\int_{0}^{2} \int_{0}^{1} \frac{y-x}{(x+y)^{3}} \, dy \, dx\]The result is \(\frac{11}{48}\).
6Step 6: Explain Differences and Fubini's Theorem
The results are \(-\frac{11}{48}\) and \(\frac{11}{48}\), which are additive inverses. This does not contradict Fubini's theorem; instead, it highlights the function \( \frac{y-x}{(x+y)^{3}} \) is not absolutely integrable over the region \([0, 1] \times [0, 2]\). This means the behavior changes with the order due to conditional convergence.
Key Concepts
Double IntegrationConditional ConvergenceOrder of IntegrationAbsolute Convergence
Double Integration
Double integration is a method used in calculus to compute the volume under a surface defined by a function of two variables. When performing double integration, we integrate a function first with respect to one variable, usually denoted as the inner integral, and then integrate the result with respect to the other variable, known as the outer integral.
This process requires specifying the limits of integration for each variable. In our case, integrals are performed over specified regions: one where
This process requires specifying the limits of integration for each variable. In our case, integrals are performed over specified regions: one where
- The variable \( x \) ranges from 0 to 2 and \( y \) from 0 to 1, and
- Another where \( x \) ranges from 0 to 1 and \( y \) from 0 to 2.
Conditional Convergence
Conditional convergence occurs when the integral or series converges in a specific order or condition, but it does not converge absolutely. In the context of our exercise, the function \( \frac{y-x}{(x+y)^3} \) behaves in such a way that switching the order of integration affects its value.
Even though integrals of the form
Even though integrals of the form
- \( \int_0^1 \int_0^2 \frac{y-x}{(x+y)^{3}} \, dx \, dy \) and
- \( \int_0^2 \int_0^1 \frac{y-x}{(x+y)^{3}} \, dy \, dx \)
Order of Integration
The order of integration in a double integral refers to the sequence in which we perform the integration of variables. While the two integrals in our problem are mathematically equivalent in function expression, the results \(-\frac{11}{48}\) and \(\frac{11}{48}\) show a sensitivity to the order of integration.
This is due to how the boundaries are set and the varying contribution of the function \( \frac{y-x}{(x+y)^3} \) within those bounds. Switching from
This is due to how the boundaries are set and the varying contribution of the function \( \frac{y-x}{(x+y)^3} \) within those bounds. Switching from
- \( dx \) first then \( dy \)
- To \( dy \) first and then \( dx \)
Absolute Convergence
For a double integral to be absolutely convergent, the absolute value of the function being integrated must also converge. This means that if we took the absolute value of the function and then integrated it over the entire region, the result would still converge.
In our scenario, the function \( \frac{y-x}{(x+y)^3} \) does not meet this requirement. Thus, the integrals are not absolutely convergent. This lack of absolute convergence leads to differences in results when reversing the order of integration, as each scenario reflects a unique pathway to convergence.
Understanding absolute convergence helps clarify why Fubini's theorem doesn't apply here to allow an interchange of integrals without affecting the result due to the path-dependent nature of the integral in this context.
In our scenario, the function \( \frac{y-x}{(x+y)^3} \) does not meet this requirement. Thus, the integrals are not absolutely convergent. This lack of absolute convergence leads to differences in results when reversing the order of integration, as each scenario reflects a unique pathway to convergence.
Understanding absolute convergence helps clarify why Fubini's theorem doesn't apply here to allow an interchange of integrals without affecting the result due to the path-dependent nature of the integral in this context.
Other exercises in this chapter
Problem 38
In Exercises \(37-40,\) find the average value of \(F(x, y, z)\) over the given region. $$ \begin{array}{l}{F(x, y, z)=x+y-z \text { over the rectangular solid
View solution Problem 39
Verify that \(f\) gives a joint probability density function. Then find the expected values \(\mu_{X}\) and \(\mu_{Y}\) . $$ f(x, y)=\left\\{\begin{array}{ll}{x
View solution Problem 39
Sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{3 / 2} \int_{0}^{9-4
View solution Problem 39
The region that lies inside the cardioid \(r=1+\cos \theta\) and outside the circle \(r=1\) is the base of a solid right cylinder. The top of the cylinder lies
View solution