/*! 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 Change the order of integration ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 5

Change the order of integration in the integral \(\int_{0}^{1} \int_{y^{2}}^{\sqrt{y}} f(x, y) d x d y\)

Short Answer

Expert verified
Question: Swap the order of integration in the given integral: \(\int_{0}^{1} \int_{y^{2}}^{\sqrt{y}} f(x, y) d x d y\). Answer: The swapped-order integral is \(\int_{1}^{\infty} \int_{x^{4}}^{x^{2}} f(x, y) dy dx\).

Step by step solution

01

Analyze the limits of integration for x

Observe the given limits of integration for x. The inner integral has x ranging from \(y^2\) to \(\sqrt{y}\). This implies that for a fixed value of y, x varies between these two limits. Write down this inequality as follows: \(y^2 \leq x \leq \sqrt{y}\).
02

Analyze the limits of integration for y

Observe the given limits of integration for y. The outer integral has y ranging from 0 to 1. This means y is bound between 0 and 1. Write down these limits as an inequality: \(0 \leq y \leq 1\).
03

Determine the new limits of integration for y

We need to determine the boundaries for y in terms of x so that we can use these values when swapping the order of integration. From the inequality derived in Step 1, solve for y: \(y^2 \leq x \leq \sqrt{y} \Rightarrow y \leq x^2 \And y \geq x^4\) The inequalities we have now are: 1. \(0 \leq y \leq 1\) 2. \(y \leq x^2\) 3. \(y \geq x^4\) We need to find the limits of integration for y when integrating with respect to y first.
04

Determine the new limits of integration for x

From the inequality derived in Step 1, we notice that we still need the limits of integration for x when integrating with respect to x later. To find these limits, we can use inequality #2 and #3. For all possible values of x, #2 can be written as: \(0 \leq x^4 \leq x^2\). Let us find the range of x to satisfy this inequality. We have \(x^4 \leq x^2\). Divide both sides by \(x^2\) and rearrange to find the roots: \(x^2 - 1 \geq 0 \Rightarrow x \leq -1 \Or x \geq 1\) Since y varies between 0 and 1, we choose \(x \geq 1\)
05

Rewrite the integral with swapped order of integration

Now that we have the new limits of integration for x and y, we can rewrite the integral with a swapped order of integration. The given integral becomes: \(\int_{1}^{\infty} \int_{x^{4}}^{x^{2}} f(x, y) dy dx\) The final answer is the swapped-order integral: \(\int_{1}^{\infty} \int_{x^{4}}^{x^{2}} f(x, y) dy dx\). Note that although the limits for x make it look improper, the original function's domain makes it a well-defined integral.

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!

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 the following integrals using polar coordinates. Assume \((r, \theta)\) are polar coordinates. A sketch is helpful. $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. Let \(R\) be the unit disk centered at \((0,0) .\) Then $$\iint_{R}\left(x^{2}+y^{2}\right) d A=\int_{0}^{2 \pi} \int_{0}^{1} r^{2} d r d \theta$$ b. The average distance between the points of the hemisphere \(z=\sqrt{4-x^{2}-y^{2}}\) and the origin is 2 (calculus not required). c. The integral \(\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y\) is easier to evaluate in polar coordinates than in Cartesian coordinates.

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by one leaf of the rose \(r=\sin 2 \theta,\) for \(0 \leq \theta \leq \pi / 2\)

The solid bounded by the paraboloids \(z=2 x^{2}+y^{2}\) and \(z=27-x^{2}-2 y^{2}\).

Solids bounded by hyperboloids Find the volume of the solid below the hyperboloid \(z=5-\sqrt{1+x^{2}+y^{2}}\) and above the following polar rectangles. $$R=\\{(r, \theta): \sqrt{3} \leq r \leq 2 \sqrt{2}, 0 \leq \theta \leq 2 \pi\\}$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.