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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

01

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

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

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

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

02

Calculation

Estimate $div F$

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

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

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

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

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

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