Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises 16.3 - Page 967: 20


$-\ln 2$

Work Step by Step

Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k ...(1)$ We have: $ \dfrac{\partial f}{ \partial x}= 2x \ln y -yz ; \dfrac{\partial f}{ \partial y}=\dfrac{x^2}{y} -xz \\ \dfrac{\partial f}{ \partial z}=-xy$ Now, $f=x^2\ln y -xyz +g(y,z) ...(2)$ $\implies \dfrac{\partial g(y,z)}{ \partial x}=0 $ Equation (2) becomes: $f=x^2\ln y -xyz+h(z) .....(3)$ and $h(z)=C=0$ Substitute all the above values in the given integral as follows: $ \int_{1,2,1}^{2,1,1} (2 x \ln y ) dx +(\dfrac{x^2}{y} -xz) dy -xy dz=f(2,1,1)-f(1,2,1)=-\ln 2$
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