/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Find the divergence of the vecto... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 10

Find the divergence of the vector field. (Note: \(\vec{r}=x \vec{i}+y \vec{j}+z \vec{k} .)\) $$\vec{F}(\vec{r})=\vec{a} \times \vec{r}$$

Short Answer

Expert verified
The divergence of \(\vec{F}(\vec{r}) = \vec{a} \times \vec{r}\) is zero.

Step by step solution

01

Understand the Vector Field

The vector field \(\vec{F}(\vec{r}) = \vec{a} \times \vec{r}\) represents the cross product of a constant vector \(\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}\) with the position vector \(\vec{r} = x \vec{i} + y \vec{j} + z\vec{k}\). Our goal is to find the divergence of this vector field.
02

Express the Cross Product

Calculate the cross product \(\vec{a} \times \vec{r}\): \\[ \vec{a} \times \vec{r} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \ a_1 & a_2 & a_3 \ x & y & z \end{vmatrix} = (a_2z - a_3y)\vec{i} + (a_3x - a_1z)\vec{j} + (a_1y - a_2x)\vec{k}. \] \Thus, \(\vec{F}(\vec{r}) = (a_2z - a_3y)\vec{i} + (a_3x - a_1z)\vec{j} + (a_1y - a_2x)\vec{k}.\)
03

Calculate the Divergence

The divergence of a vector field \(\vec{F} = \vec{F}_x \vec{i} + \vec{F}_y \vec{j} + \vec{F}_z \vec{k}\) is given by: \ \[ abla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}. \] \For our vector field, with \(F_x = a_2z - a_3y\), \(F_y = a_3x - a_1z\), and \(F_z = a_1y - a_2x\): - \(\frac{\partial}{\partial x}(a_2z - a_3y) = 0\)- \(\frac{\partial}{\partial y}(a_3x - a_1z) = 0\)- \(\frac{\partial}{\partial z}(a_1y - a_2x) = 0\).
04

State the Result

Adding these results gives \(abla \cdot \vec{F} = 0 + 0 + 0 = 0\). Thus, the divergence of the vector field \(\vec{F}(\vec{r}) = \vec{a} \times \vec{r}\) is zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vector Field
In multivariable calculus, a vector field assigns a vector to every point in a given space. Think of it like a map where each point has a little arrow attached to it, showing direction and magnitude.
  • Each vector is usually described in terms of its components, typically in the x, y, and z dimensions, and written as \( \vec{F}(x, y, z) = F_x \vec{i} + F_y \vec{j} + F_z \vec{k} \).
  • Vector fields are useful for visualizing many physical phenomena, such as magnetic or gravitational fields.
For our problem, the vector field is defined as the cross product of a constant vector \( \vec{a} \) with the position vector \( \vec{r} \). The focus here is to analyze how this particular combination behaves, specifically by exploring its divergence.
Cross Product
The cross product is a mathematical operation applied to two vectors in three-dimensional space, resulting in a third vector that is perpendicular to both. It is denoted as \( \vec{a} \times \vec{b} \).
  • This operation is crucial when dealing with rotational motion and fields like electromagnetism where perpendicular vectors arise naturally.
  • The magnitude of the cross product vector corresponds to the area of the parallelogram defined by the original two vectors.
In this exercise, \( \vec{F}(\vec{r}) = \vec{a} \times \vec{r} \) represents the cross product between a constant vector and the position vector. Understanding this result helps us take the next step towards computing the divergence.
Position Vector
The position vector is essentially a vector that describes the position of a point in space relative to an origin. It has components along the x, y, and z axes and is usually written as \( \vec{r} = x \vec{i} + y \vec{j} + z \vec{k} \).
  • This tool helps us not only locate a point in three-dimensional space but also is pivotal in defining the vector field in our problem.
  • When interacting with other vectors, say for operations like the cross product, this becomes a cornerstone of calculating resultant vectors.
The position vector in our example helps define the vector field when combined with the constant vector, providing a basis for investigating properties such as divergence.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. It involves the study of calculus over three-dimensional spaces.
  • Key operations include partial differentiation, multiple integration, and the use of vector calculus tools like divergence and curl.
  • It's broadly used in fields requiring advanced spatial analysis, such as physics, engineering, and computer graphics.
This calculus allows us to break down complex three-dimensional problems into manageable calculations, such as finding the divergence of our vector field \( \vec{F} \). By using multivariable calculus techniques鈥攍ike calculating partial derivatives鈥攚e efficiently discover that the divergence of this particular field is zero, revealing important characteristics about its behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write an iterated integral for the flux of \(\vec{F}\) through the spherical surface \(S\) centered at the origin, oriented away from the origin. Do not evaluate the integral. \(\vec{F}(x, y, z)=x \vec{i}+2 y \vec{j}+3 z \vec{k}\) \(S:\) radius \(5,\) entire sphere

Calculate the flux of the vector field through the surface. \(\vec{F}=6 \vec{i}+x^{2} \vec{j}-\vec{k},\) through the square of side 4 in the plane \(y=3,\) centered on the \(y\) -axis, with sides parallel to the \(x\) and \(z\) axes, and oriented in the positive \(y\) -direction.

Electric charge is distributed in space with density (in coulomb/m \(^{3}\) ) given in spherical coordinates by $$\delta(\rho, \phi, \theta)=\left\\{\begin{array}{ll}\delta_{0}(\text { a constant }) & \rho \leq a \\\0 & \rho>a\end{array}\right.$$ (a) Describe the charge distribution in words. (b) Find the electric field \(\vec{E}\) due to \(\delta .\) Assume that \(\vec{E}\) can be written in spherical coordinates as \(\vec{E}=\) \(E(\rho) \vec{e}_{\rho},\) where \(\vec{e}_{\rho}\) is the unit outward normal to the sphere of radius \(\rho .\) In addition, \(\vec{E}\) satisfies Gauss's Law for any simple closed surface \(S\) enclosing a volume \(W:\) $$\int_{S} \vec{E} \cdot d \vec{A}=k \int_{W} \delta d V, \quad k \text { a constant }$$

Explain what is wrong with the statement. For a certain vector field \(\vec{F}\) and oriented surface \(S,\) we have \(\int_{S} \vec{F} \cdot d \vec{A}=2 \vec{i}-3 \vec{j}+\vec{k}.\)

Compute the flux of the vector field \(\vec{F}\) through the surface \(S\). \(\vec{F}=y z^{4} \vec{i}-x z^{4} \vec{j}+e^{z^{2}} \vec{k}\) and \(S\) is the cone \(z=\) \(\sqrt{x^{2}+y^{2}}\) for \(1 \leq z \leq 2,\) oriented upward.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.