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: 19


$9 \ln (2)$

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}= 3x^2 ; \dfrac{\partial f}{ \partial y}=\dfrac{z^2}{y} \\ \dfrac{\partial f}{ \partial z}=2z \ln y$ Now, $f=x^3+g(y,z) ...(2)$ $\implies \dfrac{\partial g(y,z)}{ \partial x}=z^2 \ln y$ Equation (2) becomes: $f=x^3 +z^2 \ln y +h(z) .....(3)$ and $h(z)=C$ Substitute all the above values in the given integral as follows: $ \int_{1,1,1}^{1,2,3} 3x^2 dx +\dfrac{z^2 dy}{y}+2z \ln y dz=1+9 \ln 2 -(1+0)= 9 \ln (2)$
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