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Chapter 20: Problem 3

for the circle \(C\) and \(\operatorname{curl} \vec{F}\) as described, is the circulation of \(\vec{F}\) around \(C\) positive, negative, or zero? \(C\) is in the \(y z\) -plane oriented clockwise when viewed from the positive \(x\) -axis, and curl \(\vec{F}\) points parallel to and in the direction of \(-\vec{i}\)

Short Answer

Expert verified
The circulation of \(\vec{F}\) around \(C\) is positive.

Step by step solution

01

Understanding the Problem

We need to determine if the circulation of the vector field \(\vec{F}\) around a circle \(C\) is positive, negative, or zero. The circle \(C\) is oriented clockwise when viewed from the positive \(x\)-axis and lies in the \(yz\)-plane. The curl of the vector field \(\operatorname{curl} \vec{F}\) is directed in the negative \(x\)-direction, parallel to \(-\vec{i}\).
02

Analyzing the Orientation of the Circle

The circulation is affected by the orientation of the curve and the direction of the curl. When viewed from the positive \(x\)-axis, the circle \(C\) is oriented clockwise. Thus, the normal vector to the plane \(yz\), as per the right-hand rule, is going into the \(x\)-axis (\(-\vec{i}\) direction).
03

Relating Curl and Circulation

The circulation is calculated using Stokes' Theorem, which states that the circulation of \(\vec{F}\) around \(C\) equals the surface integral of its curl, \(\operatorname{curl} \vec{F}\), over the surface bounded by \(C\). Mathematically, it's \(\int_C \vec{F} \cdot d\vec{r} = \int_S (\operatorname{curl} \vec{F}) \cdot d\vec{S}\). Here, \(d\vec{S}\) points towards \(-\vec{i}\) due to the clockwise orientation.
04

Calculating the Dot Product

Since \(\operatorname{curl} \vec{F}\) points in the \(-\vec{i}\) direction and \(d\vec{S}\) also points in the \(-\vec{i}\) direction, their dot product is positive (as both are in the same direction). This results in a positive circulation value.
05

Conclusion

Because the dot product \((\operatorname{curl} \vec{F}) \cdot d\vec{S}\) is positive, we conclude that the circulation of \(\vec{F}\) around the circle \(C\) is positive.

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Key Concepts

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

Stokes' Theorem
Stokes' Theorem is a fundamental tool in vector calculus that relates a surface integral over a surface \( S \) to a line integral around its boundary curve \( C \). It bridges the gap between the concepts of curl and circulation. The theorem is expressed as:\[ \int_C \vec{F} \cdot d\vec{r} = \int_S (\operatorname{curl} \vec{F}) \cdot d\vec{S} \]
  • \( C \) is the closed curve that bounds the surface \( S \).
  • \( \vec{F} \) is a vector field.
  • \( d\vec{r} \) is the differential vector along curve \( C \).
  • \( d\vec{S} \) is the vector normal to the surface \( S \).
This theorem allows us to convert a difficult line integral into a simpler surface integral that is usually easier to evaluate, especially when understanding the flow or rotation of a field through a surface. By using the theorem, the calculation of circulation, a complex task at times, becomes more manageable.
Curl of a Vector Field
The curl of a vector field is a vector that describes the rotation or rotational tendency at a point within the field. It tells us how much and in what direction the field "curls" around that point. For a vector field \( \vec{F} = \langle P, Q, R \rangle \), the curl is defined as: \[ \operatorname{curl} \vec{F} = abla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \]
  • The curl provides a measure of the "twisting" or rotational influence of the vector field.
  • A zero curl implies a field is irrotational, lacking any local spinning.
  • In the context of the problem, the direction of \( \operatorname{curl} \vec{F} \) affects circulation based on how it aligns with the normal vector of the surface.
In simple terms, the curl can be visualized as the way water swirls as it goes down a drain, indicating the rotational component of the field.
Orientation of a Curve
The orientation of a curve is crucial when calculating line integrals as it affects the evaluation of the integral. Orientation can be thought of as the "direction" in which a curve \( C \) is traversed. In the practice of vector calculus, particularly with Stokes' Theorem, the orientation of curve \( C \) needs to be consistent with a right-hand rule applied to the surface normal.
  • Curve orientation is clockwise or counterclockwise in the context of the viewpoint.
  • The right-hand rule assists by curling the fingers in the curve's traversal direction; the thumb then points along the normal vector.
  • The orientation determines the direction of \( d\vec{r} \), which impacts the line integral's sign and value.
In our example, the curve is oriented clockwise as viewed from the positive \( x \)-axis. This means that our surface normal, and hence \( d\vec{S} \), aligns with \(-\vec{i}\).
Circulation of a Vector Field
The circulation of a vector field describes the cumulative "flow" of the field around a closed curve. It is a fundamental concept that measures how much the vector field pushes along the curve.
  • It's represented as \( \oint_C \vec{F} \cdot d\vec{r} \), the integral of the tangent component of \( \vec{F} \) along the curve \( C \).
  • Circulation can indicate whether a field is consistent with a certain rotational tendency within a region.
  • A positive, negative, or zero circulation often reveals how forces or flow in the field behave.
In the given exercise, the positive circulation is indicated by the alignment of \( \operatorname{curl} \vec{F} \) with \( d\vec{S} \), showing that the vector field \( \vec{F} \) effectively "rotates" in agreement with the curve's orientation around \( C \).

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Most popular questions from this chapter

is the statement true or false? Give a reason for your answer. If \(S\) is the upper unit hemisphere \(x^{2}+y^{2}+z^{2}=1, z \geq\) 0, oriented upward, then the boundary of \(S\) used in Stokes" Theorem is the circle \(x^{2}+y^{2}=1, z=0\) with orientation counterclockwise when viewed from the positive \(z\) -axis.

An electric charge \(q\) at the origin produces an electric field \(\vec{E}=q \vec{r} /\|\vec{r}\|^{3}\) (a) Does curl \(\vec{E}=\overrightarrow{0} ?\) (b) Does \(\overrightarrow{\boldsymbol{E}}\) satisfy the curl test? (c) Is \(\overrightarrow{\boldsymbol{E}}\) a gradient field?

Let \(\vec{F}=y \vec{i}-x \vec{j}\) and let \(C\) be the unit circle in the \(x y\) -plane centered at the origin and oriented counterclockwise when viewed from above. (a) Calculate \(\int_{C} \vec{F} \cdot d \vec{r}\) by parameterizing the circle. (b) Calculate curl \(\overrightarrow{\boldsymbol{F}}\). (c) Calculate \(\int_{C} \vec{F} \cdot d \vec{r}\) using your result from part (b). (d) What theorem did you use in part (c)?

Water in a bathtub has velocity vector field near the drain given, for \(x, y, z\) in \(\mathrm{cm},\) by $$\vec{F}=-\frac{y+x z}{\left(z^{2}+1\right)^{2}} \vec{i}-\frac{y z-x}{\left(z^{2}+1\right)^{2}} \vec{j}-\frac{1}{z^{2}+1} \vec{k} \mathrm{cm} / \mathrm{sec}$$ (a) Rewriting \(\vec{F}\) as follows, describe in words how the water is moving: $$\vec{F}=\frac{-y \vec{i}+x \vec{j}}{\left(z^{2}+1\right)^{2}}+\frac{-z(x \vec{i}+y \vec{j})}{\left(z^{2}+1\right)^{2}}-\frac{\vec{k}}{z^{2}+1}$$ (b) The drain in the bathtub is a disk in the \(x y\) -plane with center at the origin and radius \(1 \mathrm{cm} .\) Find the rate at which the water is leaving the bathtub. (That is, find the rate at which water is flowing through the disk.) Give units for your answer.

Is the statement true or false? Give a reason for your answer. If \(\vec{F}\) is a gradient field, then \(\int_{S} \operatorname{curl} \vec{F} \cdot d \vec{A}=0,\) for any smooth oricnted surface, \(S,\) in 3 -space.
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