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Chapter 17: Problem 45

Line integrals of vector fields on closed curves Evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) for the following vector fields and closed oriented curves \(C\) by parameterizing C. If the integral is not zero, give an explanation. \(\quad \mathbf{F}=\langle x, y\rangle ; C\) is the circle of radius 4 centered at the origin oriented counterclockwise.

Short Answer

Expert verified
Question: Compute the line integral of the vector field F(x, y) = on the closed curve C, a circle of radius 4 centered at the origin and oriented counterclockwise. Solution: The line integral of the given vector field on the closed curve C is zero. The integral is zero because the dot product of the vector field and the derivative of the parameterization is zero for all points on the curve, indicating that the vector field is orthogonal to the curve at every point.

Step by step solution

01

Parameterize the curve C

First, we need to parameterize the circle of radius 4 centered at the origin. We can use the following parameterization: $$\mathbf{r}(t) = \left \langle 4\cos(t), 4\sin(t) \right \rangle \hspace{1cm} 0 \leq t \leq 2\pi$$
02

Find dr/dt

Next, we need to compute the derivative of the parameterization with respect to the parameter \(t\). Differentiate \(\mathbf{r}(t)\) component-wise: $$\frac{d \mathbf{r}}{dt} = \left \langle -4\sin(t), 4\cos(t) \right \rangle$$
03

Evaluate F(r(t))

Now, we need to evaluate the vector field \(\mathbf{F}\) at the points of the curve given by the parameterization \(\mathbf{r}(t)\). Substitute the parameterization into the vector field: $$\mathbf{F}(\mathbf{r}(t)) = \langle 4\cos(t), 4\sin(t)\rangle$$
04

Calculate the dot product of F(r(t)) and dr/dt

Now, we will calculate the dot product of \(\mathbf{F}(\mathbf{r}(t))\) and \(\frac{d\mathbf{r}}{dt}\): $$\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} = \langle 4\cos(t), 4\sin(t) \rangle \cdot \langle -4\sin(t), 4\cos(t) \rangle = -16\cos(t)\sin(t) + 16\sin(t)\cos(t) = 0$$
05

Evaluate the line integral

Finally, we will evaluate the line integral using the result from the previous step: $$\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt = \int_0^{2\pi} 0 \, dt = 0$$ The line integral of the given vector field on the closed curve C is zero. The integral is zero because the dot product of the vector field and the derivative of the parameterization is zero for all points on the curve, indicating that the vector field is orthogonal to the curve at every point.

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Key Concepts

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

Parameterization of Curves
In vector calculus, parameterization of a curve means describing the curve using a parameter, typically denoted as 't'. This approach allows us to express the coordinates of any point on the curve as functions of this parameter. Consider a circle centered at the origin with radius 4. Its parameterization is achieved using the sine and cosine functions due to their periodic and circular nature.

For our circle, we can write: \[\mathbf{r}(t) = \left \langle 4\cos(t), 4\sin(t) \right \rangle, \quad 0 \leq t \leq 2\pi\]

The parameter 't' represents the angle in radians from the positive x-axis, thereby sweeping the circle as it ranges from 0 to 2蟺. We can visualize 't' as the tool that allows us to traverse along the circle, describing its outline in a way that can be used for further calculations, such as line integrals.

Vector Field Evaluation
Vector field evaluation is the process of finding the value of a vector field at specific points. A vector field \( \mathbf{F} \) assigns a vector to each point in space, which we can express using coordinate functions. When we have a curve parameterized, as in the previous section, we substitute the parameterization into the vector field to get the vector field's values along the curve.

Using the parameterization \( \mathbf{r}(t) \) for our circle, the evaluated vector field \( \mathbf{F}(\mathbf{r}(t)) \) becomes: \[\mathbf{F}(\mathbf{r}(t)) = \langle 4\cos(t), 4\sin(t)\rangle\]

This represents how the vector field 'behaves' along the path of the curve. During a line integral calculation, which we're exploring in this exercise, these evaluated values are crucial as they'll be combined with the rate of change of the curve's parameterization to determine the work done by the vector field.

Dot Product in Vector Calculus
The dot product is a crucial operation in vector calculus, especially when computing line integrals. It is a scalar representation of two vectors' magnitudes and the cosine of the angle between them.

For two vectors \(\mathbf{A} = \langle a_1, a_2 \rangle \) and \(\mathbf{B} = \langle b_1, b_2 \rangle\), the dot product is defined as: \[\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2\]

In the context of line integrals of vector fields, the dot product combines the vector field value at a point on the curve, \( \mathbf{F}(\mathbf{r}(t)) \) and the derivative of the position vector \( \frac{d\mathbf{r}}{dt} \) with respect to 't'. If the result of this operation is zero at all points along the curve, as in our exercise, it indicates that the vector field is orthogonal to the direction of the curve at every point, resulting in a line integral value of zero. This scenario reflects a balance where there is no net 'work' done by the field on a particle moving along the curve.

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Most popular questions from this chapter

Prove that for a real number \(p,\) with \(\mathbf{r}=\langle x, y, z\rangle, \nabla \cdot \frac{\langle x, y, z\rangle}{|\mathbf{r}|^{p}}=\frac{3-p}{|\mathbf{r}|^{p}}\)

Miscellaneous surface integrals Evaluate the following integrals using the method of your choice. Assume normal vectors point either outward or upward. $$\begin{aligned} &\iint_{S} \nabla \ln |\mathbf{r}| \cdot \mathbf{n} d S, \text { where } S \text { is the hemisphere } x^{2}+y^{2}+z^{2}=a^{2}\\\ &\text { for } z \geq 0, \text { and where } \mathbf{r}=\langle x, y, z\rangle \end{aligned}$$

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2},\) for \(-L \leq z \leq L\) a. Find the outward flux of the field \(\mathbf{F}=\langle x, y, 0\rangle\) across \(S\) b. Find the outward flux of the field \(\mathbf{F}=\frac{\langle x, y, 0\rangle}{\left(x^{2}+y^{2}\right)^{p / 2}}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}}\) across \(S\), where \(|\mathbf{r}|\) is the distance from the \(z\) -axis and \(p\) is a real number. c. In part (b), for what values of \(p\) is the outward flux finite as \(a \rightarrow \infty(\text { with } L\) fixed)? d. In part (b), for what values of \(p\) is the outward flux finite as \(L \rightarrow \infty\) (with \(a\) fixed)?

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