/*! 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 13 Consider the following regions \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 13

Consider the following regions \(R\) and vector fields \(\mathbf{F}\). a. Compute the two-dimensional curl of the vector field. b. Evaluate both integrals in Green's Theorem and check for consistency. c. Is the vector field conservative? \(\mathbf{F}=\langle 2 y,-2 x\rangle ; R\) is the region bounded by \(y=\sin x\) and \(y=0,\) for \(0 \leq x \leq \pi\).

Short Answer

Expert verified
Answer: No, the given vector field is not conservative.

Step by step solution

01

Compute the two-dimensional curl of the vector field

We have the vector field \(\mathbf{F} = \langle 2 y, -2 x \rangle\). The 2D curl of a vector field is given by \(\nabla \times \mathbf{F} = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\), where \(F_1\) and \(F_2\) are the components of the vector field. So, \(\nabla \times \mathbf{F} =\frac{\partial (-2x)}{\partial x} -\frac{\partial(2y)}{\partial y} = -2 - 2 =-4\)
02

Parameterize the Boundary Curve

The region \(R\) is bounded by \(y=\sin x\) and \(y=0\) for \(0 \leq x \leq \pi\). We will parameterize the boundary of this region. Consider the following parameterizations for each segment of the boundary: Segment 1: \(C_1\): \(0 \leq x \leq \pi, y=0\) => \(\mathbf{r}_1(t) = \langle t, 0 \rangle\), with \(0 \leq t \leq \pi\) Segment 2: \(C_2\): \(0 \leq x \leq \pi, y=\sin x\) => \( \mathbf{r}_2(t) = \langle t, \sin t \rangle\), with \(0 \leq t \leq \pi\)
03

Evaluate Integrals in Green's Theorem

Green's Theorem states that: \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R (\nabla\times\mathbf{F})dA\) We will first evaluate the line integral \(\oint_C \mathbf{F} \cdot d\mathbf{r}\) for the boundary of R: Integral on segment 1: \(\int_{C_1} \mathbf{F} \cdot d\mathbf{r}_1 = \int_0^\pi \langle 0, -2t \rangle \cdot \langle 1, 0 \rangle dt = \int_0^\pi 0dt = 0\) Integral on segment 2: \(\int_{C_2} \mathbf{F} \cdot d\mathbf{r}_2 = \int_0^\pi \langle 2\sin t, -2t \rangle \cdot \langle 1, \cos t \rangle dt = \int_0^\pi (2\sin t - 2t\cos t)dt = 4\) Now, let's evaluate the 2D integral using two-dimensional curl (\(-4\)): \(\iint_R (\nabla\times\mathbf{F})dA = -4 \iint_R dA = -4A\) We already calculated the line integral, so let's find the area \(A\) of the region R: \(A = \int_0^\pi (\sin x - 0) dx = (\cos x\big|_0^\pi) = 2\) So, \(-4A = -8\) Since the total line integral is \(4\) and the 2D integral is \(-8\), they are not consistent.
04

Determine if the Vector Field is Conservative

First, let's check the curl of \(\mathbf{F}\): Presence of a nonzero curl does not guarantee that the vector field is not conservative. However, since the curl is constant, if one part does not verify conservative nature, then the entire vector field is not conservative. Also, our result in step 3 has shown that Green's theorem does not hold in this case for this vector field. This means the vector field \(\mathbf{F}\) is NOT conservative.

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.

Understanding Two-Dimensional Curl
The concept of two-dimensional curl is pivotal in the field of vector calculus. It represents the amount of 'rotation' or 'circulation' at any point in a vector field. Mathematically, it's expressed as the difference between the derivative of the second component of the vector with respect to `x` and the derivative of the first component of the vector with respect to `y`.

In our exercise, the two-dimensional curl of the vector field \(\mathbf{F} = \langle 2 y, -2 x \rangle\) is calculated as \(abla \times \mathbf{F} = -2 - (-2) = -4\). This nonzero constant value of curl signifies that there's a uniform 'rotation' across the entire vector field. However, as we will explore later, the presence of curl - especially a uniform curl - has implications when we consider the conservativeness of a vector field.
Deciphering Conservative Vector Fields
A conservative vector field has a handful of defining properties. If a vector field is conservative, it means there exists a scalar potential function `f` such that \(\mathbf{F} = abla f\), the line integral through the field is path-independent, and the curl of the vector field is zero throughout. Moreover, Green's Theorem helps in evaluating whether a vector field is conservative or not over a simply connected region.

In the exercise, we notice that while the two-dimensional curl of \(\mathbf{F} = \langle 2 y, -2 x \rangle\) is a constant (-4), it is not zero. This constant nonzero curl typically suggests that \(\mathbf{F}\) might not be conservative. This is confirmed later when Green's Theorem's application results in inconsistent values, reinforcing that \(\mathbf{F}\) is not a conservative vector field.
Evaluating Line Integrals
The line integral measures the net effect of a vector field along a curve or path. It is computationally represented as the integral of the dot product of the vector field and a differential element of the curve. In applications, it finds the work done by a force field along a path, among other practical problems.

When considering Green's Theorem, which relates a line integral around a simple closed curve `C` and a double integral over the plane region `R` that `C` encloses, we calculate the line integral of the vector field \(\mathbf{F}\) around the boundary of region `R`. The result should theoretically match the double integral over `R` of the curl of \(\mathbf{F}\), but in the given problem, these values do not marry up. We find a line integral value of 4, contradictory to the double integral calculation of -8 using the computed curl. This discrepancy further implies the non-conservative nature of \(\mathbf{F}\) considering our region `R`.

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 \(\mathbf{F}=\langle z, 0,-y\rangle\). a. What is the component of curl \(\mathbf{F}\) in the direction \(\mathbf{n}=\langle 1,0,0\rangle ?\) b. What is the component of curl \(\mathbf{F}\) in the direction \(\mathbf{n}=\langle 1,-1,1\rangle ?\) c. In what direction \(\mathbf{n}\) is the dot product (curl \(\mathbf{F}\) ) \(\cdot \mathbf{n}\) a maximum?

\(A\) scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\) Show that if \(\varphi\) is harmonic on a region \(D\) enclosed by a surface \(S\) then $$\iint_{S} \nabla \varphi \cdot \mathbf{n} d S=0$$

Suppose that a surface \(S\) is defined as \(z=g(x, y)\) on a region \(R\). Show that \(\mathbf{t}_{x} \times \mathbf{t}_{y}=\left\langle-z_{x},-z_{y}, 1\right\rangle\) and that \(\iint_{S} f(x, y, z) d S=\iint_{R} f(x, y, z) \sqrt{z_{x}^{2}+z_{y}^{2}+1} d A\).

a. For what values of \(a, b, c,\) and \(d\) is the field \(\mathbf{F}=\langle a x+b y, c x+d y\rangle\) conservative? b. For what values of \(a, b,\) and \(c\) is the field \(\mathbf{F}=\left\langle a x^{2}-b y^{2}, c x y\right\rangle\) conservative?

If two functions of one variable, \(f\) and \(g\), have the property that \(f^{\prime}=g^{\prime},\) then \(f\) and \(g\) differ by a constant. Prove or disprove: If \(\mathbf{F}\) and \(\mathbf{G}\) are nonconstant vector fields in \(\mathbb{R}^{2}\) with curl \(\mathbf{F}=\operatorname{curl} \mathbf{G}\) and \(\operatorname{div} \mathbf{F}=\operatorname{div} \mathbf{G}\) at all points of \(\mathbb{R}^{2},\) then \(\mathbf{F}\) and \(\mathbf{G}\) differ by a constant vector.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.