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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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Most popular questions from this chapter

Let \(S\) be a surface that represents a thin shell with density \(\rho .\) The moments about the coordinate planes (see Section 13.6 ) are \(M_{y z}=\iint_{S} x \rho(x, y, z) d S, M_{x z}=\iint_{S} y \rho(x, y, z) d S\) and \(M_{x y}=\iint_{S} z \rho(x, y, z) d S .\) The coordinates of the center of mass of the shell are \(\bar{x}=\frac{M_{y z}}{m}, \bar{y}=\frac{M_{x z}}{m}, \bar{z}=\frac{M_{x y}}{m},\) where \(m\) is the mass of the shell. Find the mass and center of mass of the following shells. Use symmetry whenever possible. The cylinder \(x^{2}+y^{2}=a^{2}, 0 \leq z \leq 2,\) with density \(\rho(x, y, z)=1+z\)

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

Find the exact points on the circle \(x^{2}+y^{2}=2\) at which the field \(\mathbf{F}=\langle f, g\rangle=\left\langle x^{2}, y\right\rangle\) switches from pointing inward to outward on the circle, or vice versa.

Prove Green's First Identity for twice differentiable scalar-valued functions \(u\) and \(v\) defined on a region \(D\) : $$\iiint_{D}\left(u \nabla^{2} v+\nabla u \cdot \nabla v\right) d V=\iint_{S} u \nabla v \cdot \mathbf{n} d S$$ where \(\nabla^{2} v=\nabla \cdot \nabla v .\) You may apply Gauss' Formula in Exercise 48 to \(\mathbf{F}=\nabla v\) or apply the Divergence Theorem to \(\mathbf{F}=u \nabla v\)

Prove the following identities. a. \(\iiint_{D} \nabla \times \mathbf{F} d V=\iint_{S}(\mathbf{n} \times \mathbf{F}) d S\) (Hint: Apply the Divergence Theorem to each component of the identity.) b. \(\iint_{S}(\mathbf{n} \times \nabla \varphi) d S=\oint_{C} \varphi d \mathbf{r}\) (Hint: Apply Stokes' Theorem to each component of the identity.)
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