Chapter 14: Q. 22 (page 1150)

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

Short Answer

Expert verified

The divergence of the given vector field is $0$

Step by step solution

Step 1

Think about the vector field,

$F (x, y, z) = y z \cos z i + (z - x) j + e^{x} y k$

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

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

In comparison to the supplied vector, $F (x, y, z) = y z \cos z i + (z - x) j + e^{x} y k$ using the vector, $F (x, y, z) = F_{1} (x, y, z) + F_{2} (x, y, z) + F_{3} (x, y, z)$

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

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

$F_{3} (x, y, z) = e^{x} y$

Calculation

Estimate $div F$

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

$\frac{\partial}{\partial x} (y z \cos z) + \frac{\partial}{\partial y} (z - x) + \frac{\partial}{\partial z} (e^{x} y)$

$= 0 + 0 + 0$

$= 0$

Hence, $div F = 0$

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

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)=yzcoszi+(z-x)j+exyk

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

Discrete Mathematics

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

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