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Chapter 16: Problem 5

Which fields are conservative, and which are not? $$\mathbf{F}=(z+y) \mathbf{i}+z \mathbf{j}+(y+x) \mathbf{k}$$

Short Answer

Expert verified
The field \( \mathbf{F} = (z+y) \mathbf{i} + z \mathbf{j} + (y+x) \mathbf{k} \) is not conservative.

Step by step solution

01

Understanding Conservative Vector Fields

A vector field \( \mathbf{F} \) is conservative if it can be expressed as the gradient of some scalar potential function \( \phi \). Mathematically, this means that \( \mathbf{F} = abla \phi \). A necessary condition for \( \mathbf{F} \) to be conservative is that its curl must be zero everywhere in the domain it is defined.
02

Compute the Curl of \( \mathbf{F} \)

To determine if \( \mathbf{F} = (z+y) \mathbf{i} + z \mathbf{j} + (y+x) \mathbf{k} \) is conservative, calculate its curl: \( abla \times \mathbf{F} \). This is given by the determinant of the matrix: \[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ z + y & z & y + x \end{vmatrix} \]We expand this determinant to find each component.
03

Calculate Each Component of the Curl

Evaluate the components:1. \( \mathbf{i} \)-component: \( \frac{\partial}{\partial y}(y+x) - \frac{\partial}{\partial z}(z) = 1 - 1 = 0 \)2. \( \mathbf{j} \)-component: \( \frac{\partial}{\partial z}(z+y) - \frac{\partial}{\partial x}(y+x) = 1 - 1 = 0 \)3. \( \mathbf{k} \)-component: \( \frac{\partial}{\partial x}(z) - \frac{\partial}{\partial y}(z+y) = 0 - 1 = -1 \)
04

Determine if \( \mathbf{F} \) is Conservative

The curl of \( \mathbf{F} \) is \( abla \times \mathbf{F} = 0\mathbf{i} + 0\mathbf{j} - 1\mathbf{k} = -\mathbf{k} \). Since the curl is not zero, \( \mathbf{F} \) is not a conservative vector field.

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Key Concepts

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

Gradient
In vector calculus, the gradient is a vector operation that acts on scalar fields. Think of a scalar field as a landscape and the gradient as the slope of that landscape. The gradient of a scalar function \( \phi(x, y, z) \) is a vector field, represented as \( abla \phi \). This vector points in the direction of the greatest rate of increase of the function.
  • The gradient consists of the partial derivatives of \( \phi \) with respect to each variable: \( \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) \).
  • It provides valuable information about how the function changes at any point in space.
This operation is crucial because it can help identify the direction of the steepest ascent and is often used to find the necessary condition for a conservative vector field.
Scalar Potential Function
A scalar potential function is a smooth function \( \phi \) such that the vector field \( \mathbf{F} \) can be expressed as its gradient, \( \mathbf{F} = abla \phi \). This means that the field lines of \( \mathbf{F} \) originate from the changes or differences in the scalar potential values.
  • If you can find such a function \( \phi \), your vector field is conservative.
  • In practical terms, energy conservation and gravitational force can often be represented using scalar potential functions.
However, if the curl of \( \mathbf{F} \) is not zero, finding a scalar potential function becomes impossible, indicating that the vector field is non-conservative.
Curl
The curl of a vector field is a vector that measures the rotation or "twirling" of the field at any point in space.
  • The mathematical expression for curl is \( abla \times \mathbf{F} \), a determinant that evaluates how the field circles around a given point.
  • For a conservative vector field, the curl must be zero throughout its defined domain.
In contexts like fluid dynamics or electromagnetism, understanding the curl helps in analyzing rotations or vortices. If the curl is non-zero, like in the exercise above, the vector field is not conservative, as it exhibits some kind of rotational movement.
Vector Calculus
Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields, primarily in three-dimensional Euclidean space. It extends the concepts of single-variable calculus into multiple dimensions using vectors.
  • Its fundamental building blocks include the gradient, divergence, and curl, as well as line and surface integrals.
  • Vector calculus is crucial for fields like physics, engineering, and computer graphics, as it provides methods to model and analyze physical phenomena.
The connection between these operations and theorems like Green's, Stokes', and Gauss's is essential in providing relationships between various forms of integrals and derivatives in vector fields.

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Most popular questions from this chapter

Evaluate $$\iint_{S} \nabla \times(y \mathbf{i}) \cdot \mathbf{n} d \sigma $$where \(S\) is the hemisphere \(x^{2}+y^{2}+z^{2}=1, z \geq 0\)

Find the outward flux of the field \(\mathbf{F}=x z \mathbf{i}+y z \mathbf{j}+\mathbf{k}\) across the surface of the upper cap cut from the solid sphere \(x^{2}+y^{2}+z^{2} \leq 25\) by the plane \(z=3\).

If a simple closed curve \(C\) in the plane and the region \(R\) it encloses satisfy the hypotheses of Green's Theorem, the area of \(R\) is given by $$\text { Area of } R=\frac{1}{2} \oint_{C} x d y-y d x$$ The reason is that by Equation (4), run backward, $$\begin{aligned} \text { Area of } R &=\iint_{R} d y d x=\iint_{R}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x \\ &=\oint_{C} \frac{1}{2} x d y-\frac{1}{2} y d x \end{aligned}.$$ Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves. The ellipse \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(b \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Find the counterclockwise circulation and the outward flux of the field \(\mathbf{F}=(-\sin y) \mathbf{i}+(x \cos y) \mathbf{j}\) around and over the square cut from the first quadrant by the lines \(x=\pi / 2\) and \(y=\pi / 2.\)

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 \(S\)
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