Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.3 Conservative Vector Fields - Exercises - Page 946: 30

Answer

(a)${\cal D}$ is not simply connected. (b) No, this does not guarantee that ${\bf{F}}$ is conservative. (c) We show that there exists a potential function for ${\bf{F}}$. So, ${\bf{F}}$ is path-independent. By Theorem 2, ${\bf{F}}$ is conservative. (d) No

Work Step by Step

(a) ${\cal D}$ is not simply connected because there is a "hole" at the origin $\left( {0,0} \right)$. (b) Write ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\dfrac{x}{{{x^2} + {y^2}}},\dfrac{y}{{{x^2} + {y^2}}},0} \right)$. In the following, we check if ${\bf{F}}$ satisfies the cross-partial condition: $\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ }$ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ }$ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$ We get $ - \dfrac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = - \dfrac{{2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$, ${\ \ \ }$ $0=0$, ${\ \ \ }$ $0=0$ These results imply that ${\bf{F}}$ satisfies the cross-partials condition. However, this does not guarantee that ${\bf{F}}$ is conservative since from part (a) we know that the domain ${\cal D}$ is not simply connected. (c) We find a potential function for ${\bf{F}}$. Let the potential function for ${\bf{F}} = \left( {\dfrac{x}{{{x^2} + {y^2}}},\dfrac{y}{{{x^2} + {y^2}}},0} \right)$ be $V\left( {x,y,z} \right)$ such that ${\bf{F}} = - \nabla V = - \left( {\dfrac{{\partial V}}{{\partial x}},\dfrac{{\partial V}}{{\partial y}},\dfrac{{\partial V}}{{\partial z}}} \right)$. So, 1. taking the integral of $\dfrac{{\partial V}}{{\partial x}}$ with respect to $x$ gives $V\left( {x,y,z} \right) = - \smallint \dfrac{x}{{{x^2} + {y^2}}}{\rm{d}}x = - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) + f\left( {y,z} \right)$ 2. taking the integral of $\dfrac{{\partial V}}{{\partial y}}$ with respect to $y$ gives $V\left( {x,y,z} \right) = - \smallint \dfrac{y}{{{x^2} + {y^2}}}{\rm{d}}y = - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) + g\left( {x,z} \right)$ 3. taking the integral of $\dfrac{{\partial V}}{{\partial z}}$ with respect to $z$ gives $V\left( {x,y,z} \right) = \smallint 0{\rm{d}}z = 0 + h\left( {x,y} \right)$ Since the three ways of expressing $V\left( {x,y,z} \right)$ must be equal, we get $ - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) + f\left( {y,z} \right) = - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) + g\left( {x,z} \right) = h\left( {x,y} \right)$ From these results we conclude that the potential function is $V\left( {x,y,z} \right) = - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) + C$, where $C$ is a constant. Note that $V$ is not defined at the origin $\left( {0,0} \right)$. Since there exists a potential function for ${\bf{F}}$, it is path-independent. By Theorem 2, ${\bf{F}}$ is conservative. (d) The requirement of Theorem 4 is ${\bf{F}}$ be defined on a simply-connected domain. If ${\bf{F}}$ is not defined on a simply-connected domain, we cannot use Theorem 4. However, in this case, although the domain ${\cal D}$ is not simply-connected, there exists a potential function for ${\bf{F}}$. Hence ${\bf{F}}$ is conservative. So, it does not contradict Theorem 4.
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