Chapter 14: Q. 24 (page 1150)

In Exercises $21 - 24$ , compute the divergence of the given vector field.
$F (x, y, z) = \cos x i - y \sin z y j + \cos z k$

Short Answer

Expert verified

The divergence of the given vector field is $- \sin x - \sin z y - y z \cos y z - \sin z$

Step by step solution

Step 1

Think about the vector field,

$F (x, y, z) = \cos x i - y \sin z y j + \cos z k$

Our goal is to determine the vector field's divergence.

Make use of the formula, $div F = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}$

Compare and contrast the vectors $F (x, y, z) = y z \cos z i + (z - x) j + e^{x} y k$ and $F (x, y, z) = F_{1} (x, y, z) + F_{2} (x, y, z) + F_{3} (x, y, z)$

So, $F_{1} (x, y, z) = \cos x$

$F_{2} (x, y, z) = - y \sin z y$

$F_{3} (x, y, z) = \cos z$

Calculation

Estimate $div F$

$div F = \frac{\partial F_{1}}{\partial x} + \frac{\partial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z}$

$= \frac{\partial}{\partial x} (\cos x) + \frac{\partial}{\partial y} (- y \sin z y) + \frac{\partial}{\partial z} (\cos z)$

$= (- \sin x) + (- \sin z y - y z \cos y z) + (- \sin z)$

$= - \sin x - \sin z y - y z \cos y z - \sin z$

Therefore, $div F = - \sin x - \sin z y - y z \cos y z - \sin z$

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In Exercises $21 - 24$ , compute the divergence of the given vector field.
$F (x, y, z) = \cos x i - y \sin z y j + \cos z k$

Short Answer

Step by step solution

Step 1

Calculation

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In Exercises 21-24, compute the divergence of the given vector field.F(x,y,z)=cosxi-ysinzyj+coszk

Short Answer

Step by step solution

Step 1

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

In Exercises $21 - 24$ , compute the divergence of the given vector field.
$F (x, y, z) = \cos x i - y \sin z y j + \cos z k$