/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 For each of the following iterat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 5

For each of the following iterated integrals, find the region \(R\) and write the integral in the other form (interchanging the order of integration): a) \(\int_{1 / 2}^{1} \int_{0}^{1-x} f(x, y) d y d x\) b) \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y d x\) c) \(\int_{0}^{1} \int_{y-1}^{0} f(x, y) d x d y\) d) \(\int_{0}^{1} \int_{1-x}^{1+x} f(x, y) d y d x\)

Short Answer

Expert verified
Answer: The region R can be described as: $$\frac{1}{2} \leq x \leq 1$$ $$0 \leq y \leq 1-x$$ The iterated integral in the other form is: $$\int_{0}^{1/2} \int_{1-y}^{1} f(x, y) dx dy$$

Step by step solution

01

Identify the limits of integration

For integral a) \(\int_{1 / 2}^{1} \int_{0}^{1-x} f(x, y) d y d x\), we have: - \(x\) ranges from \(1/2\) to \(1\) - \(y\) ranges from \(0\) to \(1-x\)
02

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$\frac{1}{2} \leq x \leq 1$$ $$0 \leq y \leq 1-x$$
03

Interchange the order of integration

To switch the order of integration, we will first solve for \(x\) in the inequality for \(y\): $$x \geq 1-y$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{0}^{1/2} \int_{1-y}^{1} f(x, y) d x d y$$ Answer for a): $$\int_{0}^{1/2} \int_{1-y}^{1} f(x, y) d x d y$$ #b) Find the region R and write the iterated integral in the other form#
04

Identify the limits of integration

For integral b) \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y d x\), we have: - \(x\) ranges from \(0\) to \(1\) - \(y\) ranges from \(0\) to \(\sqrt{1-x^2}\)
05

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$0 \leq x \leq 1$$ $$0 \leq y \leq \sqrt{1-x^2}$$
06

Interchange the order of integration

To switch the order of integration, we will first solve for \(x\) in the inequality for \(y\): $$x^2 \leq 1 - y^2$$ $$x \leq \sqrt{1 - y^2}$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x d y$$ Answer for b): $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x d y$$ #c) Find the region R and write the iterated integral in the other form#
07

Identify the limits of integration

For integral c) \(\int_{0}^{1} \int_{y-1}^{0} f(x, y) d x d y\), we have: - \(y\) ranges from \(0\) to \(1\) - \(x\) ranges from \(y-1\) to \(0\)
08

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$0 \leq y \leq 1$$ $$y-1 \leq x \leq 0$$
09

Interchange the order of integration

To switch the order of integration, we will first solve for \(y\) in the inequality for \(x\): $$y \geq x+1$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{-1}^{0} \int_{x+1}^{1} f(x, y) d y d x$$ Answer for c): $$\int_{-1}^{0} \int_{x+1}^{1} f(x, y) d y d x$$ #d) Find the region R and write the iterated integral in the other form#
10

Identify the limits of integration

For integral d) \(\int_{0}^{1} \int_{1-x}^{1+x} f(x, y) d y d x\), we have: - \(x\) ranges from \(0\) to \(1\) - \(y\) ranges from \(1-x\) to \(1+x\)
11

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$0 \leq x \leq 1$$ $$1-x \leq y \leq 1+x$$
12

Interchange the order of integration

To switch the order of integration, we will first solve for \(x\) in the inequality for \(y\): $$x \leq y-1$$ $$x \geq -y+1$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{0}^{1} \int_{-y+1}^{y-1} f(x, y) d x d y$$ Answer for d): $$\int_{0}^{1} \int_{-y+1}^{y-1} f(x, y) d x d y$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Order of Integration
The order of integration refers to the sequence in which we integrate a function of multiple variables. Consider iterated integrals where we perform one integral at a time, typically starting with the innermost integral. When we change the order of integration, it's crucial to take into account the limits and ensure they correctly describe the intended region of integration. For example, in the exercise a), the original order \( dy dx \) suggests that we integrate with respect to \( y \) first, holding \( x \) constant within its bounds. To interchange this order, we re-configure the limits so that \( x \) can be integrated first. This process must be done carefully to maintain the integrity of the region over which we are integrating. The skill of altering the order of integration efficiently is valuable, particularly when one order presents a simpler integral to solve than the other.
Region of Integration
The region of integration, denoted as \( R \), defines the subset of the plane or space over which the function is being integrated. Specifying the correct region is a fundamental step in setting up iterated integrals. It's visualized by the inequalities that are provided by the limits of integration. When changing the order of integration, as we saw in the solution for exercise a), we must ensure that the new limits still represent the same physical region. Graphing these regions on a coordinate system can help demonstrate the bounds on each variable and can be instrumental in understanding the relationships between them. The region of integration for each part of the given exercise is illustrated by inequalities, and the interchangeable order impacts how we describe this region but not its geometry.
Double Integrals
Double integrals extend the concept of a single integral to two dimensions. They are represented as iterated integrals with two different variables. The double integral of a function \( f(x,y) \) over a region \( R \) sums up the values of \( f \) at all points within that region, effectively calculating the 'volume' under the surface described by \( f(x,y) \) over \( R \). The solution steps illustrate how to compute iterated integrals by integrating one variable at a time and showing that the choice of which variable to integrate first can be swapped. For example, in exercise b), we integrate with respect to \( y \) first. However, if we find it more convenient or viable, we can choose to integrate with respect to \( x \) first, as long as we identify the proper limits that correspond to our region \( R \). This flexibility can simplify calculations, especially in cases where one order of integration might be easier to evaluate than the other.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate with the aid of the substitution indicated: a) \(\int_{0}^{1}\left(1-x^{2}\right)^{3 / 2} d x, x=\sin \theta\) b) \(\int_{0}^{1} \frac{1}{1+\sqrt{1+x}} d x, x=u^{2}-1\) c) \(\int_{0}^{\pi / 2} \frac{1}{\sin x+\cos x+2} d x, t=\tan (x / 2)\) d) \(\int_{0}^{\pi / 4} \frac{x \cos x(x \sin x-\cos x)}{1+x \cos x} d x, t=1+x \cos x\)

The error function \(y=\operatorname{erf}(x)\) is defined by the equation $$ y=\operatorname{erf}(x)=\int_{0}^{x} e^{-t^{2}} d t $$ This function is of great importance in probability and statistics and is tabulated in the books mentioned after (4.24). Establish the following properties: a) \(\operatorname{erf}(x)\) is defined and continuous for all \(x\); b) \(\operatorname{erf}(-x)=-\operatorname{erf}(x)\); c) \(-1<\operatorname{erf}(x)<1\) for all \(x\).

a) Prove that (4.60) remains valid for improper integrals, that is, if \(f(x)\) is continuous for \(x_{1} \leq x

Let an ellipse be given by parametric equations: \(x=a \cos \phi, y=b \sin \phi, b>a>0\). Show that the length of arc from \(\phi=0\) to \(\phi=\alpha\) is given by $$ s=b \int_{0}^{\alpha} \sqrt{1-k^{2} \sin ^{2} \phi} d \phi, \quad k^{2}=\frac{b^{2}-a^{2}}{b^{2}} . $$ 5\. Show that the function \(F(x)\) defined by (4.24) has the following properties: a) \(F(x)\) is defined and continuous for all \(x\); b) as \(x\) increases, \(F(x)\) increases; c) \(F(x+\pi)-F(x)=2 K\), where \(K\) is a constant; d) \(\lim _{x \rightarrow \infty} F(x)=\infty, \lim _{x \rightarrow-\infty} F(x)=-\infty\).

Let \(R_{u v}\) be the square \(0 \leq u \leq 1,0 \leq v \leq 1\). Show that the given equations define a one-to-one mapping of \(R_{u v}\) onto \(R_{x y}\) and graph \(R_{x y}\) : a) \(x=u+u^{2}, y=e^{v}\) b) \(x=u e^{v}, y=e^{v}\) c) \(x=2 u-v^{2}, y=v+u v\), d) \(x=5 u-u^{2}+v^{2}, y=5 v+10 u v\) z d y d x$.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.