/*! 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 12 Find the divergence of the follo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 12

Find the divergence of the following vector fields. $$\mathbf{F}=\left\langle x^{2} y z,-x y^{2} z,-x y z^{2}\right\rangle$$

Short Answer

Expert verified
Question: Calculate the divergence of the given vector field $$\mathbf{F}=\left\langle x^{2} y z,-x y^{2} z,-x y z^{2}\right\rangle$$. Answer: The divergence of the vector field is $$\nabla \cdot \mathbf{F} = -2 x y z$$.

Step by step solution

01

Identify the components of the vector field

The given vector field is represented as: $$\mathbf{F}=\left\langle x^{2} y z,-x y^{2} z,-x y z^{2}\right\rangle$$ The components of the vector field are: $$ F_x = x^{2} y z \\ F_y = -x y^{2} z \\ F_z = -x y z^{2} $$
02

Calculate the partial derivatives of the components

We will now calculate the partial derivatives of the components to their corresponding variables: $$ \frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x}(x^{2} y z) \\ \frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y}(-x y^{2} z) \\ \frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z}(-x y z^{2}) $$ Calculating the partial derivatives, we get: $$ \frac{\partial F_x}{\partial x} = 2 x y z \\ \frac{\partial F_y}{\partial y} = -2 x y z \\ \frac{\partial F_z}{\partial z} = -2 x y z $$
03

Find the divergence of the vector field

Finally, we find the divergence by adding the partial derivatives we computed in Step 2: $$ \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} $$ Plugging in the values from Step 2: $$ \nabla \cdot \mathbf{F} = 2 x y z - 2 x y z - 2 x y z $$ Simplifying, we get the divergence of the vector field: $$ \nabla \cdot \mathbf{F} = -2 x y z $$

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!

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

A square plate \(R=\\{(x, y): 0 \leq x \leq 1,\) \(0 \leq y \leq 1\\}\) has a temperature distribution \(T(x, y)=100-50 x-25 y\) a. Sketch two level curves of the temperature in the plate. b. Find the gradient of the temperature \(\nabla T(x, y)\) c. Assume that the flow of heat is given by the vector field \(\mathbf{F}=-\nabla T(x, y) .\) Compute \(\mathbf{F}\) d. Find the outward heat flux across the boundary \(\\{(x, y): x=1,0 \leq y \leq 1\\}\) e. Find the outward heat flux across the boundary \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\)

Consider the rotational velocity field \(\mathbf{v}=\mathbf{a} \times \mathbf{r},\) where \(\mathbf{a}\) is a nonzero constant vector and \(\mathbf{r}=\langle x, y, z\rangle .\) Use the fact that an object moving in a circular path of radius \(R\) with speed \(|\mathbf{v}|\) has an angular speed of \(\omega=|\mathbf{v}| / R\). a. Sketch a position vector a, which is the axis of rotation for the vector field, and a position vector \(\mathbf{r}\) of a point \(P\) in \(\mathbb{R}^{3}\). Let \(\theta\) be the angle between the two vectors. Show that the perpendicular distance from \(P\) to the axis of rotation is \(R=|\mathbf{r}| \sin \theta\). b. Show that the speed of a particle in the velocity field is \(|\mathbf{a} \times \mathbf{r}|\) and that the angular speed of the object is \(|\mathbf{a}|\). c. Conclude that \(\omega=\frac{1}{2}|\nabla \times \mathbf{v}|\).

Use the procedure in Exercise 57 to construct potential functions for the following fields. $$\mathbf{F}=\langle-y,-x\rangle$$

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T\) (the Laplacian of \(T\)). Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100(1+\sqrt{x^{2}+y^{2}+z^{2}})$$

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.)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

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.