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Chapter 15: Problem 32

Sketch the region over which you are integrating, and then write down the integral with the order of integration reversed (changing the limits of integration as necessary). $$ \int_{-1}^{1} \int_{0}^{1+y} f(x, y) d x d y $$

Short Answer

Expert verified
The given integral is \(\int_{-1}^{1} \int_{0}^{1+y} f(x, y) d x d y\), defined over the region \(-1\leq y\leq 1\) and \(0 \leq x \leq 1+y\), which can be sketched on the xy-plane. After reversing the order of integration, the new limits of integration are found: \(y \geq x-1\) and \(-1 \leq y \leq 1\), with x-limits \(0 \leq x \leq 2\). Therefore, the integral with a reversed order of integration is: \(\int_{0}^{2} \int_{x-1}^{1} f(x, y) dy dx\).

Step by step solution

01

Sketch the region

The integral is defined over the rectangular region \(-1\leq y\leq 1\) and \(0 \leq x \leq 1+y\). We can represent this region using inequalities: 1. \(-1 \leq y \leq 1\) 2. \(0 \leq x \leq 1+y\) Now we can plot the region on the xy-plane by these inequalities: - For \(y=-1\), x-values vary from 0 to 0; - For \(y=1\), x-values vary from 0 to 2; - For \(x=0\), y-values vary from -1 to 1.
02

Reverse the order of integration

We need to change the order of integration from \(dx dy\) to \(dy dx\). To do this, let's find new limits of integration. First, we should express y from the second inequality: $$ 0\leq x \leq 1+y \Rightarrow y\geq x-1 $$ The equality holds when \(x=0\) and \(y=-1\). So, we have: 1. \(y \geq x-1\) 2. \(-1 \leq y \leq 1\) Now, we will find x-limits: - For \(y=-1\), the minimum x-value, \(x_{min} = -1+1 = 0\); - For \(y=1\), the maximum x-value, \(x_{max} = 1+1 = 2\); Thus, the new integral becomes: $$ \int_{0}^{2} \int_{x-1}^{1} f(x, y) dy dx $$

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Key Concepts

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

Region of Integration
Understanding the region of integration is crucial when tackling multivariable integrals. In our problem, the integration region was described by the inequalities
  • \(-1 \leq y \leq 1\)
  • \(0 \leq x \leq 1 + y\)
These inequalities define the bounds on the xy-plane. Imagine drawing vertical lines where y varies from -1 to 1 and horizontal lines where x ranges from 0 to 1+y. This creates a triangular region since as y increases from -1 to 1, the upper x limit grows from 0 to 2.
The sketch provides a visual cue of how these boundaries encapsulate the area of interest for integration. It's like drawing a border around the parts of the plane you will consider when integrating the function \(f(x, y)\). This helps in understanding which parts of the graph contribute to the integral's total value.
Order of Integration
The order of integration refers to the sequence in which you integrate multivariable functions. Initially, our integral required performing the operation with respect to x first and then y (notation \(dx dy\)). This means for each slice parallel to the x-axis, extend along the y direction. However, these problems often ask for reversing this sequence.
Reversing the order (to \(dy dx\)) means integrating with y first, followed by x. In practical terms, this shifts our focus from slices that move vertically to slices that extend horizontally, modifying the limits accordingly. The impact of such a change can simplify the calculations if the limits of one variable become easier to handle. This is an exercise in flexibility, helping us see the problem from a fresh angle, potentially revealing new ways to simplify the computations.
Limits of Integration
The limits of integration are boundaries within which integration is performed. Initially given by the inequalities:
  • \(-1 \leq y \leq 1\)
  • \(0 \leq x \leq 1 + y\)
These determine the chunk of the plane we include in our solution. When switching the order of integration, the challenge is to redefine these constraints so they fit our new interpretation (from \(dx dy\) to \(dy dx\)). This involves a bit of creative reasoning to express the same region under different terms.

For the new order \(dy dx\), the limits become:
  • \(x\) from \(0\) to \(2\)
  • \(y\) from \(x - 1\) to \(1\)
Adjusting these limits ensures the integration still spans the same region on the graph, capturing all possible y values for every x. This transformation is like describing a road trip with new landmarks; you're still covering the same journey, just looking at it from a different route.

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Most popular questions from this chapter

Package Dimensions: USPS The U.S. Postal Service (USPS) will accept only packages with a length plus girth no more than 108 inches. \({ }^{28}\) (See the figure.) What are the dimensions of the largest volume package that the USPS will accept? What is its volume? (This exercise is the same as Exercise 49 in the preceding section. This time, solve it using Lagrange multipliers.)

Use Lagrange multipliers to solve the given optimization problem. HINT [See Example 2.] Find the maximum value of \(f(x, y)=4 x y\) subject to \(x^{2}+y^{2}=8\). Also find the corresponding point(s) \((x, y)\).

Complete the following: The first step in calculating an integral of the form \(\int_{a}^{b} \int_{r(x)}^{s(x)} f(x, y) d y d x\) is to evaluate the integral ______ obtained by holding ______ constant and integrating with respect to ______.

Locate and classify all the critical points of the functions. HINT [See Example 2.] $$ t(x, y)=x^{4}+8 x y^{2}+2 y^{4} $$

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