Calculus: Early Transcendentals 8th Edition

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

Chapter 16 - Section 16.3 - The Fundamental Theorem for Line Integrals - 16.3 Exercise - Page 1095: 19

Answer

$\dfrac{4}{e}$

Work Step by Step

When the vector field $F(x,y)=Pi+Qj$ is a conservative field, then throughout the domain $D$, we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ $P$ and $Q$ are the first-order partial derivatives on the domain $D$. Here, we have $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}=-2xe^{-y}$ This implies that the vector field $F$ is conservative. Now, $f(x,y)=x^2e^{-y}+g(y)$ $f_y(x,y)=-x^2e^{-y}+g'(y)$ Here, $g(y)=C$ where $C$ is a constant. Thus, $f(x,y)=x^2e^{-y}+C$ Now, $\int_C F \cdot dr =f(2,1)-f(1,0)=(1+4e^{-1}+C)-(1+C)=\dfrac{4}{e}$
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