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Find the curl of each vector field \(\mathbf{F}\). $$\mathbf{F}=y e^{z} \mathbf{i}+z e^{x} \mathbf{j}-x e^{y} \mathbf{k}$$

Short Answer

Expert verified
\( \nabla \times \mathbf{F} = - (x e^{y} + e^{x}) \mathbf{i} + (y e^{z} + e^{y}) \mathbf{j} + (z e^{x} - e^{z}) \mathbf{k} \)

Step by step solution

01

Understanding the Curl Operator

The curl of a vector field \( \mathbf{F} \) is represented as \( abla \times \mathbf{F} \). It measures the rotation or the tangling of the vector field in space. For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), the curl is given by \( abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ P & Q & R \end{vmatrix} \).
02

Identify Components of \( \mathbf{F} \)

In the given vector field \( \mathbf{F} = y e^{z} \mathbf{i} + z e^{x} \mathbf{j} - x e^{y} \mathbf{k} \), identify the components as follows:- \( P = y e^{z} \)- \( Q = z e^{x} \)- \( R = -x e^{y} \).
03

Compute \( \frac{\partial R}{\partial y} \) and \( \frac{\partial Q}{\partial z} \)

Determine \( \frac{\partial R}{\partial y} \) and \( \frac{\partial Q}{\partial z} \):- \( \frac{\partial R}{\partial y} = \frac{\partial (-x e^{y})}{\partial y} = -x e^{y} \).- \( \frac{\partial Q}{\partial z} = \frac{\partial (z e^{x})}{\partial z} = e^{x} \).
04

Compute \( \frac{\partial P}{\partial z} \) and \( \frac{\partial R}{\partial x} \)

Determine \( \frac{\partial P}{\partial z} \) and \( \frac{\partial R}{\partial x} \):- \( \frac{\partial P}{\partial z} = \frac{\partial (y e^{z})}{\partial z} = y e^{z} \).- \( \frac{\partial R}{\partial x} = \frac{\partial (-x e^{y})}{\partial x} = -e^{y} \).
05

Compute \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \)

Determine \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \):- \( \frac{\partial Q}{\partial x} = \frac{\partial (z e^{x})}{\partial x} = z e^{x} \).- \( \frac{\partial P}{\partial y} = \frac{\partial (y e^{z})}{\partial y} = e^{z} \).
06

Substitute into the Curl Formula

Substitute the computed derivatives into the curl formula:\[abla \times \mathbf{F} = \left( -x e^{y} - e^{x} \right) \mathbf{i} + \left( y e^{z} + e^{y} \right) \mathbf{j} + \left( z e^{x} - e^{z} \right) \mathbf{k} \].
07

Simplify the Result

After substituting, simplify the expression for the curl:\[abla \times \mathbf{F} = - (x e^{y} + e^{x}) \mathbf{i} + (y e^{z} + e^{y}) \mathbf{j} + (z e^{x} - e^{z}) \mathbf{k}\].

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Key Concepts

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

Curl of a Vector Field
The curl of a vector field is a fundamental concept in vector calculus, often represented as \( abla \times \mathbf{F} \). It helps us understand the rotation or circulatory nature of a field in three-dimensional space. Imagine water swirling around—this swirling action is akin to the curl of the vector field. For a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the curl is computed using:
  • \( abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ P & Q & R \end{vmatrix} \)
The curl is essentially a cross-product involving partial derivatives and the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), which are standard basis vectors in 3D coordinates. To physically interpret the curl, remember that it reflects how much and in what direction a tiny paddle wheel would rotate if placed in the field.
Vector Field Components
A vector field assigns a vector to every point in a region of space, making it a powerful tool in physics and engineering. In mathematical terms, if you have a vector field like \( \mathbf{F} = y e^{z} \mathbf{i} + z e^{x} \mathbf{j} - x e^{y} \mathbf{k} \), you can break it down into:
  • \( P = y e^{z} \)
  • \( Q = z e^{x} \)
  • \( R = -x e^{y} \)
These components represent the field's influence in the \( x \), \( y \), and \( z \) directions, respectively. Understanding vector components is crucial because they allow us to apply operations like divergence and curl systematically by working step by step on each element. The components help describe complicated vector fields in more manageable terms.
Differential Operators
Differential operators like \( \frac{\partial}{\partial x} \), \( \frac{\partial}{\partial y} \), and \( \frac{\partial}{\partial z} \) form the backbone of calculus operations on functions. In the context of calculating the curl of a vector field, they allow us to understand how parts of a vector field change over their respective directions. This information is necessary to compute the derivatives that make up the components of the curl:
  • \( \frac{\partial R}{\partial y} = \frac{\partial (-x e^{y})}{\partial y} = -x e^{y} \)
  • \( \frac{\partial Q}{\partial z} = \frac{\partial (z e^{x})}{\partial z} = e^{x} \)
  • \( \frac{\partial P}{\partial z} = \frac{\partial (y e^{z})}{\partial z} = y e^{z} \)
  • \( \frac{\partial R}{\partial x} = \frac{\partial (-x e^{y})}{\partial x} = -e^{y} \)
  • \( \frac{\partial Q}{\partial x} = \frac{\partial (z e^{x})}{\partial x} = z e^{x} \)
  • \( \frac{\partial P}{\partial y} = \frac{\partial (y e^{z})}{\partial y} = e^{z} \)
These operators enable us to measure how a function or its components respond to change along each axis, making them indispensable in vector calculus.
Determinants in Vector Calculus
In vector calculus, we often use determinants to simplify and solve complex problems involving multiple variables. A determinant provides a scalar representation of a square matrix, which can be a helpful tool in computing operations like the curl. The curl of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is calculated using the determinant:
  • \( abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ P & Q & R \end{vmatrix} \)
This arrangement is akin to a cross product in matrix form, where the top row consists of unit vectors, the middle row contains partial derivatives, and the bottom row lists the vector field components \( P, Q, R \). Solving the determinant involves calculating the sum of products across diagonals and reflects how multiple factors interact within the field.

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Most popular questions from this chapter

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \( \mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through the portion of the cylinder \(x^{2}+y^{2}=1\) cut by the planes \(z=0\) and \(z=a\) in the direction away from the z-axis

a. Show that the outward flux of the position vector field \(\mathbf{F}=\) \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through a smooth closed surface \(S\) is three times the volume of the region enclosed by the surface. b. Let \(n\) be the outward unit normal vector field on \(S\). Show that it is not possible for \(\mathbf{F}\) to be orthogonal to \(\mathbf{n}\) at every point of \(S .\)

Find the area of the surface \(x^{2}-2 y-2 z=0\) that lies above the triangle bounded by the lines \(x=2, y=0,\) and \(y=3 x\) in the \(x y\) -plane.

Bendixson's criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) of the flow's velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region \(R\) (no holes or missing points \()\) and that if \(M_{x}+N_{y} \neq 0\) throughout \(R,\) then none of the streamlines in \(R\) is closed. In other words, no particle of fluid ever has a closed trajectory in \(R\). The criterion \(M_{x}+N_{y} \neq 0\) is called Bendixson's criterion for the nonexistence of closed trajectories.

A revealing experiment By experiment, you find that a force field \(\mathbf{F}\) performs only half as much work in moving an object along path \(C_{1}\) from \(A\) to \(B\) as it does in moving the object along path \(C_{2}\) from \(A\) to \(B .\) What can you conclude about F? Give reasons for your answer.

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