Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.5 - Curl and Divergence - 16.5 Exercise - Page 1109: 5

Answer

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) Consider $F=A i+B j+C k$ Then $curl F=\begin{vmatrix}i&j&k\\\dfrac{\partial}{\partial x}&\dfrac{\partial }{\partial y}&\dfrac{\partial }{\partial z}\\A&B&C\end{vmatrix}$ $curl F=[C_y-B_z]i+[A_z-C_z]j+[B_x-A_y]k$ $curl F=(\dfrac{-\sqrt z}{(1+y)^2}-0)i+(\dfrac{-\sqrt x}{(1+z)^2}-0)j+(\dfrac{-\sqrt y}{(1+x)^2}-0)k=-\dfrac{\sqrt z}{(1+y)^2}i-\dfrac{\sqrt x}{(1+z)^2}-\dfrac{\sqrt y}{(1+x)^2}$ b) $div F=\dfrac{\partial A}{\partial x}+\dfrac{\partial B}{\partial y}+\dfrac{\partial C}{\partial z}$ $div F=\dfrac{\partial (\dfrac{\sqrt x}{1+z})}{\partial x}+\dfrac{\partial (\dfrac{\sqrt y}{1+x})}{\partial y}+\dfrac{\partial (\dfrac{\sqrt z}{1+y})}{\partial z}=\dfrac{1}{2 \sqrt x(1+z)}+\dfrac{1}{2 \sqrt y(1+x)}+\dfrac{1}{2 \sqrt z(1+y)}$
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