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Chapter 17: Problem 48

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. \(\quad \mathbf{F}=\langle y,-x\rangle ; C\) is the circle of radius 3 centered at the origin oriented counterclockwise.

Short Answer

Expert verified
Based on the solution provided, the line integral of the given vector field around the closed curve is 9蟺. This nonzero value indicates that the vector field is not conservative, and the work done by the field to move a particle around the closed path is not zero.

Step by step solution

01

Parameterize the curve C

First, we need to find a parameterization of the circle with radius 3. On a circle with radius 3 centered at the origin, all points can be represented as \((x,y) = (3\cos t, 3\sin t),\) for \(t\in[0,2\pi]\). Thus, we can describe the closed curve as \(\mathbf{r}(t) = \langle 3\cos t, 3\sin t \rangle.\)
02

Compute dr/dt

Now that we have a parameterized curve, we need to compute its derivative, \(\mathbf{r}'(t)\). Differentiating \(\mathbf{r}(t)\) componentwise, we get: \(\mathbf{r}'(t) = \langle -3\sin t, 3\cos t \rangle.\)
03

Compute F(r(t))

We are given the vector field \(\mathbf{F}=\langle y, -x \rangle\). We now compute the value of the vector field along the curve C by substituting \(\mathbf{r}(t)\): \(\mathbf{F}(\mathbf{r}(t)) = \langle 3\sin t, -3\cos t \rangle.\)
04

Compute the dot product F(r(t)) 鈥� r'(t)

Next, we need to compute the dot product of the vector field and the derivative of the curve: $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \langle 3\sin t, -3\cos t \rangle \cdot \langle -3\sin t, 3\cos t \rangle = -9 \sin^2{t} + 9\cos^2{t}.$
05

Evaluate the line integral

Now we need to compute the line integral: \(\oint_{C}\mathbf{F} \cdot d \mathbf{r} = \int_{0}^{2\pi}\left(-9\sin^2{t}+9\cos^2{t}\right)dt.\) We can simplify the integral as follows: \(\oint_{C}\mathbf{F} \cdot d \mathbf{r} = 9\int_{0}^{2\pi}\left(\cos^2{t}-\sin^2{t}\right)dt.\) Now, by using the sine and cosine double angle formula, we can rewrite the integrand as: \(\oint_{C}\mathbf{F} \cdot d \mathbf{r} = \frac{9}{2}\int_{0}^{2\pi}\left(1 + \cos(2t)\right)dt.\) Integrating and evaluating the integral, we get: \(\oint_{C}\mathbf{F} \cdot d \mathbf{r} = \frac{9}{2} \left[t + \frac{1}{2}\sin(2t) \right]_{0}^{2\pi} = \frac{9}{2} \times 2\pi = 9\pi.\) The line integral is not zero, as we found \(\oint_{C}\mathbf{F} \cdot d \mathbf{r} = 9\pi\), this indicates that the vector field has nonzero circulation around the closed curve C. In other words, the vector field is not conservative, and the work done by the field to move a particle around the closed path is not zero.

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Key Concepts

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

Parameterization of Curves
Understanding the parameterization of curves is crucial when dealing with vector calculus. It allows us to describe complex paths using simpler representations, often in terms of a single variable. Specifically, it involves finding a function that maps a range of real numbers to points on the curve.

For instance, given a circle centered at the origin with radius 3, all points on the circle can be mapped through a function of an angle, usually denoted by t. The function \( \mathbf{r}(t) = \langle 3\cos t, 3\sin t \rangle \) expresses all points (x, y) on the said circle for \( t \) in the interval \( [0,2\pi] \). This concept is essential because it allows us to compute line integrals over curves by transitioning from a problem in space to a problem in time, so to speak, where 'time' is the parametric variable t.
Dot Product in Vector Fields
The dot product is a fundamental operation in vector calculus used to project one vector onto another and measure the extent to which vectors align. In vector fields, we use the dot product to calculate work, flux, and when computing line integrals, among other applications.

For example, given two vectors \( \mathbf{F} \) and \( \mathbf{r}\' \), their dot product \( \mathbf{F} \cdot \mathbf{r}\' \) would represent the instantaneous rate at which \( \mathbf{F} \) does work on a particle moving along path \( \mathbf{r} \) with respect to 'time' t. This computation is vital for evaluating line integrals in a vector field.
Closed Curve Circulation
The notion of closed curve circulation pertains to the total work done by a vector field along a closed loop. When we discuss the circulation, it's akin to assessing how much the field 'circulates' around the path. A non-zero circulation signifies that there is a net work done by the field in moving a particle around the curve.

For the given vector field \( \mathbf{F}=\langle y,-x \rangle \) and the closed curve C, which is a circle, we found that \( \oint_C \mathbf{F} \cdot d\mathbf{r} = 9\pi \). This non-zero result tells us there's a rotational effect of the field around C, illustrating the presence of a kind of 'vortex' within the field that is causing the circulation.
Conservative Vector Field
A conservative vector field is one where the work done to move an object along a path between two points is independent of the path taken. In other words, the line integral around any closed curve in a conservative field is zero. This concept is equivalent to saying the field has no 'whirlpools' or circulation.

In our exercise, as the line integral over the closed curve C yielded a value of \( 9\pi \) instead of zero, this strongly indicates that the vector field \( \mathbf{F}=\langle y,-x \rangle \) is not conservative. There is notable work done by the vector field when moving along the closed curve, contrasting with the behavior of conservative fields where such work would always net to zero.

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Most popular questions from this chapter

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