/*! 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 Why does a two-dimensional vecto... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Why does a two-dimensional vector field with zero curl on a region have zero circulation on a closed curve that bounds the region?

Short Answer

Expert verified
Answer: A two-dimensional vector field with zero curl is irrotational, meaning there is no rotation at any point in the field. According to Green's theorem, the line integral of the vector field along the boundary of the region is equal to the double integral of the curl of the field over the region. Since the curl of the vector field is zero throughout the region, the double integral on the right side of the equation will also be zero, resulting in zero circulation along the closed curve that bounds the region.

Step by step solution

01

Understand Curl and Circulation

Curl is a measure of the rotation of a vector field, and circulation is a measure of the net amount of rotation along a closed curve. In a two-dimensional vector field with zero curl, the field is irrotational, meaning there is no rotation at any point.
02

Understand the Relationship between Curl and Circulation

In order for a vector field to have zero circulation on a closed curve, the field's curl must be zero everywhere inside the enclosed region. If its curl were non-zero anywhere inside the region, there would be a component of rotation, which would cause a non-zero circulation on any closed curve.
03

Explain the Green's Theorem Connection

According to Green's theorem, for a two-dimensional simply connected region, the line integral of a vector field F along the boundary of the region is equal to the double integral of the curl of F over the region, i.e., \begin{gather} \oint_{\partial R} \mathbf{F} \cdot d\mathbf{r} = \iint_R (\nabla \times \mathbf{F}) \cdot d\mathbf{A} \end{gather} Where R is the region enclosed by the boundary curve and \(\partial R\) is the closed curve. In our case, the curl of the vector field is zero throughout the region. This means the double integral on the right side of the equation will also be zero.
04

Conclude the Result

By Green's theorem and the fact that the curl of the vector field is zero in the region, we have: \begin{gather} \oint_{\partial R} \mathbf{F} \cdot d\mathbf{r} = 0 \end{gather} This shows that the circulation along the closed curve (\(\partial R\)) that bounds the region with zero curl is indeed zero.

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.

Curl
The concept of "curl" is essentially about rotation or swirling of a vector field. Imagine you're watching water flow in a pond, and you're curious about whether the water is swirling around specific spots. The "curl" at a point tells you if the water is rotating at that spot and in which direction it might be swirling.

Mathematically, for a two-dimensional vector field \( \mathbf{F} = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the curl is calculated as a scalar value, given by \[ abla \times \mathbf{F} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\] If the curl is zero, it means there's no swirling or rotation at that point or, in simpler terms, the vector field is "irrotational."

Generally, if a vector field's curl is everywhere zero within a region, it indicates that the whole field has no local rotations there. This property is crucial when linking it to circulation and Green's Theorem.
Circulation
"Circulation" in a vector field refers to the total amount of rotation or flow along a closed curve. Imagine tracing a path around the boundary of a region and accounting for how much the field flows along your path. It's like summing up tiny segments of swirling currents you would encounter if traveling along the curve.

The circulation is given by the line integral formula \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r}\] where \( C \) is the closed path.

If you have a two-dimensional vector field with zero curl, any closed path within that field will experience zero circulation. This happens because in areas without any local rotations, moving around a closed loop will cancel out any net swirling effect. Therefore, fields with zero curl lead directly to zero circulation for any path within the boundary, making the field irrotational across the closed curves.
Green's Theorem
"Green's Theorem" connects line integrals around a simple closed curve to a double integral over the region it encloses. It gives us a powerful tool to relate circulation (line integral) to the curl (a type of surface integral).

For any vector field \( \mathbf{F} \), if you want to check the circulation around a boundary \( \partial R \), you can calculate it using the formula: \[ \oint_{\partial R} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} (abla \times \mathbf{F}) \cdot d\mathbf{A} \] Green's Theorem tells us that the circulation around the curve is equal to the sum of all the little swirls (curl) in the area \( R \) enclosed by the curve.

In scenarios when the vector field is irrotational within a region (meaning the curl is zero), the right-hand side of the equation becomes zero. As a result, this leads to zero circulation around the boundary. Green's Theorem elegantly confirms that if there's no internal rotation, there's no net circulation on the enclosing path. Such understanding simplifies problems dealing with path independence in vector fields.

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

Prove the following identities. Assume \(\varphi\) is a differentiable scalar- valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \times(\mathbf{F} \times \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}-\mathbf{G}(\nabla \cdot \mathbf{F})-(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F}(\nabla \cdot \mathbf{G})$$

The area of a region R in the plane, whose boundary is the curve \(C\), may be computed using line integrals with the formula $$\text { area of } R=\int_{C} x d y=-\int_{C} y d x$$ Let \(R\) be the rectangle with vertices \((0,0),(a, 0),(0, b),\) and \((a, b),\) and let \(C\) be the boundary of \(R\) oriented counterclockwise. Use the formula \(A=\int_{C} x d y\) to verify that the area of the rectangle is \(a b\).

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T(\text {the Laplacian of } T) .\) Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-x^{2}+y^{2}+z^{2}}$$

Maximum surface integral Let \(S\) be the paraboloid \(z=a\left(1-x^{2}-y^{2}\right),\) for \(z \geq 0,\) where \(a>0\) is a real number. Let \(\mathbf{F}=\langle x-y, y+z, z-x\rangle .\) For what value(s) of \(a\) (if any) does \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) have its maximum value?

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) \(T(x, y, z)=100+x^{2}+y^{2}+z^{2} ; D\) is the unit sphere centered at the origin.

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.