Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.5 Curl and Divergence - 16.5 Exercises - Page 1149: 5


a) $-\dfrac{\sqrt z}{(1+y)^2}i-\dfrac{\sqrt x}{(1+z)^2}-\dfrac{\sqrt y}{(1+x)^2}$ b) $\dfrac{1}{2 \sqrt x(1+z)}+\dfrac{1}{2 \sqrt y(1+x)}+\dfrac{1}{2 \sqrt z(1+y)}$

Work Step by Step

a) When $F=ai+bj+ck$, then we have $curl F=[c_y-b_z]i+[a_z-c_z]j+[b_x-a_y]k$ $curl F=(\dfrac{-\sqrt z}{(1+y)^2}-0)i+(\dfrac{-\sqrt x}{(1+z)^2}-0)j+(\dfrac{-\sqrt y}{(1+x)^2}-0)k$ or, $=-\dfrac{\sqrt z}{(1+y)^2}i-\dfrac{\sqrt x}{(1+z)^2}-\dfrac{\sqrt y}{(1+x)^2}$ b) $div F=\dfrac{\partial (\dfrac{\sqrt x}{1+z})}{\partial x}+\dfrac{\partial (\dfrac{\sqrt y}{1+x})}{\partial y}+\dfrac{\partial (\dfrac{\sqrt z}{1+y})}{\partial z}=\dfrac{1}{2 \sqrt x(1+z)}+\dfrac{1}{2 \sqrt y(1+x)}+\dfrac{1}{2 \sqrt z(1+y)}$
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