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Chapter 14: Problem 24

Consider the following vector fields, where \(\mathbf{r}=\langle x, y, z\rangle\). a. Compute the curl of the field and verify that it has the same direction as the axis of rotation. b. Compute the magnitude of the curl of the field. $$\mathbf{F}=\langle 1,-1,0\rangle \times \mathbf{r}$$

Short Answer

Expert verified
Answer: The curl of the vector field \(\mathbf{F}\) is \(\langle 0, 0, 0\rangle\), and its magnitude is \(0\). Since the curl is a zero vector, it does not really have a direction, and the curl does not show any rotation around any axis in this case.

Step by step solution

01

Find the explicit form of the vector field

We are given that the vector field is formed by the cross product of \(\langle 1, -1, 0\rangle\) and \(\mathbf{r} = \langle x, y, z\rangle\). Let's find their cross product: $$ \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -1 & 0\\ x & y & z \end{vmatrix} = \mathbf{i}(0 - (-z)) - \mathbf{j}(0 - z) + \mathbf{k}(1y - (-1x)) = \langle z, z, x+y \rangle $$
02

Compute the curl of the vector field

To compute the curl of the vector field, we need to evaluate the following expression: $$ \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ z & z & x + y \end{vmatrix} $$ Calculating the determinants, we get: $$ \nabla \times \mathbf{F} = \mathbf{i}\left(\frac{\partial (x+y)}{\partial y} - \frac{\partial z}{\partial z}\right) - \mathbf{j}\left(\frac{\partial (x+y)}{\partial x} - \frac{\partial z}{\partial z}\right) + \mathbf{k}\left(\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y}\right) = \langle 0, 0, 0 \rangle $$
03

Verify the direction of the curl

The curl of the vector field \(\nabla \times \mathbf{F}\) is found to be \(\langle 0, 0, 0\rangle\). Since the curl is a zero vector, it does not really have a direction. In other words, the curl does not show any rotation around any axis in this case.
04

Compute the magnitude of the curl

The magnitude of the curl is given by the formula: $$ ||\nabla \times \mathbf{F}|| = \sqrt{(0)^2 + (0)^2 + (0)^2} = 0 $$ So, the magnitude of the curl of the field is \(0\).

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Most popular questions from this chapter

Find a vector field \(\mathbf{F}\) with the given curl. In each case, is the vector field you found unique? $$\operatorname{curl} \mathbf{F}=\langle 0,1,0\rangle$$

Consider the vector field \(\mathbf{F}=\frac{-y}{x^{2}+y^{2}} \mathbf{i}+\frac{x}{x^{2}+y^{2}} \mathbf{j}+z \mathbf{k}\) a. Show that \(\nabla \times \mathbf{F}=\mathbf{0}\) b. Show that \(\oint_{C} \mathbf{F} \cdot d \mathbf{r}\) is not zero on a circle \(C\) in the \(x y\) -plane enclosing the origin. c. Explain why Stokes' Theorem does not apply in this case.

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