/*! 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 197 Use Green's Theorem to evaluate ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use Green's Theorem to evaluate integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}, \quad\) where \(\mathbf{F}(x, y)=\left(x y^{2}\right) \mathbf{i}+x \mathbf{j}, \quad\) and \(C\) is a unit circle oriented in the counterclockwise direction.

Short Answer

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

Step by step solution

01

Understand Green's Theorem

Green's Theorem relates a line integral around a closed curve \( C \) to a double integral over the region \( D \) bounded by \( C \). The theorem states: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] for \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \).
02

Identify P and Q from F

Given \( \mathbf{F}(x, y) = (x y^2) \mathbf{i} + x \mathbf{j} \), we identify \( P(x, y) = x y^2 \) and \( Q(x, y) = x \).
03

Calculate Partial Derivatives

Calculate the partial derivatives \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). We find \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = 2xy \).
04

Set Up the Double Integral

Substitute the partial derivatives into Green's Theorem, we get: \[ \iint_{D} \left( 1 - 2xy \right) \, dA \]. Here, \( D \) is the unit circle \( x^2 + y^2 \leq 1 \).
05

Convert to Polar Coordinates

Because \( D \) is a circle, it's convenient to use polar coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \), \( dA = r \, dr \, d\theta \), and \( r \) ranges from 0 to 1 and \( \theta \) from 0 to \( 2\pi \).
06

Evaluate the Double Integral

Substitute the polar expressions into the integral: \[ \int_{0}^{2\pi} \int_{0}^{1} \left( 1 - 2r^2 \cos \theta \sin \theta \right) r \, dr \, d\theta \]. Simplify and integrate: \[ \int_{0}^{2\pi} \left. \left( \frac{r^2}{2} - \frac{r^4}{2} \sin 2\theta \right) \right|_{0}^{1} \, d\theta = \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \pi \].
07

Conclusion

The value of the integral \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \) using Green's Theorem is \( \pi \).

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 91Ó°ÊÓ!

Key Concepts

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

Line Integral
A line integral is a fundamental concept in vector calculus that helps measure the degree of circulation of a vector field along a given path. In simpler terms, it calculates the integral of a vector field along a specified curve. You imagine it as a way to add up tiny segments of work done by a force or field along a path. Here's why it's interesting:
  • It can be used to evaluate how a field like force acts over a path.
  • It can provide insights into physical scenarios, such as work done by a force field on a particle traveling along a path.
For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \), the line integral along a curve \( C \) is:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \]Green's Theorem gives us a toolkit where, instead of evaluating this possibly complex line integral directly, we convert it into an equivalent double integral over the region \( D \) enclosed by the curve \( C \). This is often simpler to compute.
Double Integral
A double integral extends the concept of a single integral to compute the volume under a surface in a given region. In the context of Green's Theorem, it transforms a line integral around a closed curve into a double integral over the plane region it encircles. When evaluating the double integral in Green's Theorem, we compute:\[\iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\]Here’s what happens in a practical sense:
  • Once partial derivatives are calculated, they are plugged into the equation.
  • Now, instead of analyzing the path itself, the focus shifts to examining the region bounded by the path.
This process simplifies calculations, especially in symmetric regions or when the vector field setup is complicated. It also allows transforming real-world physical problems into a more treatable mathematical framework.
Polar Coordinates
Polar coordinates offer an advantageous system, especially when dealing with circular or symmetric regions like the unit circle. By translating Cartesian coordinates \( (x, y) \) into polar coordinates using \( x = r \cos \theta \) and \( y = r \sin \theta \), we can handle circular domains more gracefully. Here's why polar coordinates are effective:
  • They simplify integrations over circular areas.
  • They inherently account for the radial symmetry of the circle.
  • Changing from Cartesian differential area \( dA \) to \( r \, dr \, d\theta \) matches the circle's geometry better.
In the solution to the Green's Theorem problem, polar coordinates were introduced to convert the domain from the Cartesian plane into the circular region, where \( r \) ranges between 0 and 1, while \( \theta \) extends from 0 to \( 2\pi \). This simplification is crucial not just for the ease of integration but also for accurately capturing the essence of circular trajectories.

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

Let \(E\) be the solid unit cube with diagonally opposite comers at the origin and \((1,1,1),\) and faces parallel to the coordinate planes. Let \(S\) be the surface of \(E\), oriented with the outward-pointing normal. Use a CAS to find \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\) using the divergence theorem \(\mathbf{F}(x, y, z)=2 x y \mathbf{i}+3 y e^{z} \mathbf{j}+x \sin z \mathbf{k}\)

Use Stokes' theorem for vector field \(\mathbf{F}(x, y, z)=-\frac{3}{2} y^{2} \mathbf{i}-2 x y \mathbf{j}+y z \mathbf{k},\) where \(S\) is that part of the surface of plane \(x+y+z=1\) contained within triangle \(C\) with vertices \((1,0,0),(0,1,0),\) and \((0,0,1),\) traversed counterclockwise as viewed from above.

Use Stokes' theorem for vector field \(\mathbf{F}(x, y, z)=z \mathbf{i}+3 x \mathbf{j}+2 z \mathbf{k} \quad\) where \(\quad S \quad\) is \(\quad\) surface \(z=1-x^{2}-2 y^{2}, z \geq 0, \quad C \quad\) is \(\quad\) boundary circle \(x^{2}+y^{2}=1,\) and \(S\) is oriented in the positive \(z\) -direction.

Find the line integral of \(\mathbf{F}(x, y, z)=12 x^{2} \mathbf{i}-5 x y \mathbf{j}+x z \mathbf{k}\) over path \(C\) defined by \(y=x^{2}, \quad z=x^{3}\) from point (0,0,0) to point \((2,4,8) .\)

Let \(\mathbf{F}(x, y, z)=x y \mathbf{i}+2 z \mathbf{j}-2 y \mathbf{k}\) and let \(C\) be the intersection of plane \(x+z=5\) and cylinde \(x^{2}+y^{2}=9, \quad\) which is oriented counterclockwise when viewed from the top. Compute the line integral of \(\mathbf{F}\) over \(C\) using Stokes' theorem.

See all solutions

Recommended explanations on Math Textbooks

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.