Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - Chapter Review Exercises - Page 970: 11

Answer

$div\left( {\bf{F}} \right) = - 1$ $curl\left( {\bf{F}} \right) = - {\bf{k}}$

Work Step by Step

We have ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {y,0, - z} \right)$. 1. $div\left( {\bf{F}} \right) = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}$ $div\left( {\bf{F}} \right) = - 1$ 2. $curl\left( {\bf{F}} \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ y&0&{ - z} \end{array}} \right|$ $curl\left( {\bf{F}} \right) = - {\bf{k}}$
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