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Chapter 14: Problem 51

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

Short Answer

Expert verified
Answer: A possible vector field with the given curl is $\mathbf{F} = \langle \frac{z^2}{2}, -yx + \frac{z^3}{6}, R(x, y, z) \rangle$. The vector field is not unique, as any other vector field of the form $\mathbf{F} + \nabla \phi$ with $\phi(x,y,z)$ being a scalar function will have the same curl $\langle 0, z, -y \rangle$.

Step by step solution

01

Find the vector field components

To find the vector field \(\mathbf{F}\) given its curl, we need to find three functions \(P(x, y, z)\), \(Q(x, y, z)\), and \(R(x, y, z)\), such that the curl is \(\operatorname{curl} \mathbf{F} = \langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \rangle = \langle 0, z, -y \rangle\).
02

Solve the system of partial differential equations

To find the components of \(\mathbf{F}\), we establish a system of partial differential equations based on the curl components: \begin{align*} \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} &= 0, \\ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} &= z, \\ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} &= -y. \end{align*} Now we have to solve this system to obtain \(P(x, y, z)\), \(Q(x, y, z)\), and \(R(x, y, z)\).
03

Integrate the second and third equations

Integrate the second equation with respect to \(z\): $$P(x, y, z) - R_x(x, y, z) = z^2/2 + H(x, y),$$ where \(H(x, y)\) is an arbitrary function of \(x\) and \(y\). Integrate the third equation with respect to \(x\): $$Q(x, y, z) - P_y(x, y, z) = -yx + G(y, z),$$ where \(G(y, z)\) is an arbitrary function of \(y\) and \(z\).
04

Substitute \(P_y\) and \(R_x\)

Substitute \(P_y = Q+yx-G(y, z)\) and \(R_x = P-z^2/2-H(x, y)\) into the first equation: $$R_y - Q_z = \frac{\partial}{\partial y} (P-z^2/2-H(x, y)) - \frac{\partial}{\partial z} (Q+yx-G(y, z)) = 0.$$
05

Solve for \(H\) and \(G\)

Now we have the equation: $$\frac{\partial}{\partial y} (P-z^2/2-H(x, y)) - \frac{\partial}{\partial z} (Q+yx-G(y, z)) = 0.$$ To satisfy this equation, we can set \(H(x, y) = 0\) and \(G(y, z) = \frac{z^3}{6}\), which gives: $$P(x, y, z) = z^2/2,$$ $$Q(x, y, z) = -yx + \frac{z^3}{6},$$ and, therefore, $$\mathbf{F} = \langle z^2/2, -yx + z^3/6, R(x, y, z) \rangle.$$
06

Check for uniqueness

A vector field with a given curl is unique up to the gradient of a scalar potential. If a vector field \(\mathbf{F}\) has the curl \(\operatorname{curl} \mathbf{F} = \mathbf{G}\), then any other vector field \(\mathbf{F} + \nabla \phi\) with \(\phi(x,y,z)\) being a scalar function will have the same curl: $$\operatorname{curl} (\mathbf{F} + \nabla \phi) = \operatorname{curl} \mathbf{F} + \operatorname{curl} \nabla \phi = \mathbf{G}.$$ Since there is no additional information provided in the problem to determine the function \(\phi\), we can conclude that the vector field is not unique, and any other vector field of the form \(\mathbf{F} + \nabla \phi\) will have the same curl \(\langle 0, z, -y \rangle\).

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