Calculus 8th Edition

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

Chapter 16 - Vector Calculus - 16.3 The Fundamental Theorem for Line Integrals - 16.3 Exercises - Page 1135: 35

Answer

a) $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$ b) $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of path.

Work Step by Step

The vector field $F(x,y)=Pi+Qj$ is known as conservative field throughout the domain $D$, when we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ $P$ and $Q$ represents the first-order partial derivatives on the domain $D$. From the given problem, we get $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$ Thus, we have the given the vector field $F$ a conservative field. Hence, we get $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$ b) The line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is independent of path when $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}=0$ for every closed curve $C$. Here, we have $\int_{C_1} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{\pi} (-\sin \theta i+\cos \theta j) (-\sin \theta i+\cos \theta j) d\theta$ and $ \int_0^{\pi} 1 d\theta =\pi$ Also, $\int_{C_2} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{-\pi} (-\sin \theta i+\cos \theta j) (-\sin \theta i+\cos \theta j) d\theta$ and $ \int_0^{-\pi} 1 d\theta =-\pi$ It has been seen that the two integrals are differ, thus the line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of path.
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