/*! 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 4 Use Green's Theorem to evaluate ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 4

Use Green's Theorem to evaluate the integral. In each exercise, assume that the curve \(C\) is oriented counterclockwise.$$ \oint_{C}\left(x^{2}-y^{2}\right) d x+x d y, \text { where } C \text { is the circle } x^{2}+y^{2}=9 \text { . } $$

Short Answer

Expert verified
The integral evaluates to \( 9\pi \).

Step by step solution

01

Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve to a double integral over the region enclosed by the curve. The theorem states: \[ \oint_{C} M\, dx + N\, dy = \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] where \( M(x, y) \) and \( N(x, y) \) are functions of \( x \) and \( y \), and \( D \) is the region enclosed by the curve \( C \).
02

Identify M(x, y) and N(x, y)

From the integral \( \oint_{C}(x^2-y^2)dx + xdy \), identify \( M(x, y) = x^2 - y^2 \) and \( N(x, y) = x \).
03

Calculate the Partial Derivatives

Calculate \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \): - \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x) = 1 \) - \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^2 - y^2) = -2y \)
04

Set up the Double Integral

Using Green's Theorem, the double integral becomes: \[ \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA = \iint_{D} (1 + 2y) \, dA \]where \( D \) is the region enclosed by \( x^2 + y^2 = 9 \).
05

Convert to Polar Coordinates

Convert the integral to polar coordinates. The circle \( x^2 + y^2 = 9 \) converts to \( r = 3 \). Thus, the double integral in polar coordinates is: \[ \int_{0}^{2\pi} \int_{0}^{3} (1 + 2(r \sin \theta)) r \, dr \, d\theta \]
06

Evaluate the Inner Integral

Evaluate \( \int_{0}^{3} (r + 2r^2 \sin \theta) \, dr \): - \( \int_{0}^{3} r \, dr = \left[ \frac{1}{2}r^2 \right]_{0}^{3} \)- \( \int_{0}^{3} 2r^2 \sin \theta \, dr = \left[ \frac{2}{3}r^3 \sin \theta \right]_{0}^{3} \)This gives: \( \frac{1}{2}(3^2) + \frac{2}{3}(3^3) \sin \theta = \frac{9}{2} + 6 \sin \theta \).
07

Evaluate the Outer Integral

Now evaluate the outer integral: \[ \int_{0}^{2\pi} \left( \frac{9}{2} + 6 \sin \theta \right) \, d\theta \]. Since the integral of \( \sin \theta \) over \( 0 \) to \( 2\pi \) is zero, the result becomes: \[ \frac{9}{2} \times 2\pi = 9\pi \].

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.

Line Integrals
Line integrals are a type of integral used to calculate a wide range of quantities, where the integration is performed over a curve. In this context, we are looking at line integrals over a closed curve, as in the exercise problem. The most common application is calculating work done by a force field. Here's a refresher:
  • **Curve**: The line integral is taken over the curve \(C\), which is a path in the plane or space.
  • **Components**: The integral has components, such as \( M(x, y) \) for \( dx \) and \( N(x, y) \) for \( dy \).
  • **Expression**: Thus, the line integral can be expressed as \( \oint_{C} M\, dx + N\, dy \).
Line integrals require parameterization of the curve. For instance, in the exercise, circle \( C \) is parameterized in terms of variables \( x \) and \( y \) based on the radius. This summarization of Green's Theorem application helps translate a line integral into a potentially simpler double integral.
Double Integrals
In the exercise using Green's Theorem, the conversion from line to double integrals streamlines the computation.
  • **Region of Integration**: The double integral is over the region \( D \), enclosed by curve \( C \).
  • **Expression Transformation**: Green鈥檚 Theorem states the transformation: \( \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \).
Step by step, calculate the necessary partial derivatives. For \( M(x, y) = x^2 - y^2 \): \( \frac{\partial M}{\partial y} = -2y \). For \( N(x, y) = x \): \( \frac{\partial N}{\partial x} = 1 \). Thus, the double integral becomes \( \iint_{D} (1 + 2y) \, dA \), showing a clear path to finding the value of the integral after conversion. Double integrals offer insight into volume under a surface mapped over \( D \).
Polar Coordinates
Polar coordinates simplify the integration process, especially for circular regions like the one in this problem. Here's how polar coordinates help:
  • **Conversion**: For the circle \( x^2 + y^2 = 9 \), switch to polar coordinates: \( r = 3 \) and \( \theta \) is from \( 0 \) to \( 2\pi \).
  • **Integration Transformation**: The area element \( dA \) becomes \( r\, dr \, d\theta \) in polar form.
The double integral in polar coordinates becomes: \[ \int_{0}^{2\pi} \int_{0}^{3} (1 + 2(r \sin \theta)) r \, dr \, d\theta \]. Evaluate one integral at a time starting with respect to \( r \), then \( \theta \). This converts a complex problem into manageable parts, as the exercise illustrates. In the given problem, polar coordinates effectively handle circular symmetries, making it easier to complete the integration and arrive at the final result.

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

Confirm that the force field \(\mathbf{F}\) is conservative in some open connected region containing the points \(P\) and \(Q\), and then find the work done by the force field on a particle moving along an arbitrary smooth curve in the region from \(P\) to $Q .$$$ \mathbf{F}(x, y)=y e^{x y} \mathbf{i}+x e^{x y} \mathbf{j} ; \quad P(-1,1), Q(2,0) $$

Describe the different types of line integrals, and discuss how they are related.

Find the work done by the force field \(\mathbf{F}\) on a particle that moves along the curve \(C\). \(\mathbf{F}(x, y)=x y \mathbf{i}+x^{2} \mathbf{j}\) \(C: x=y^{2}\) from \((0,0)\) to \((1,1)\)

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along the curve \(C\).$$ \begin{aligned} &\mathbf{F}(x, y)=x^{2} y \mathbf{i}+4 \mathbf{j} \\ &C: \mathbf{r}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j} \quad(0 \leq t \leq 1) \end{aligned} $$

Evaluate the integral \(\int_{-C} \frac{x d y-y d x}{x^{2}+y^{2}}\), where \(C\) is the circle \(x^{2}+y^{2}=a^{2}\) traversed counterclockwise.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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.