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: 18


$\ln (\dfrac{\pi}{2})$

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}= 2 \cos y ; \dfrac{\partial f}{ \partial y}=\dfrac{1}{y}-2 x \sin y \\ \dfrac{\partial f}{ \partial z}=\dfrac{1}{z}$ Now, $f=2x \cos y +g(y,z) ...(2)$ $\implies \dfrac{\partial g(y,z)}{ \partial x}=\dfrac{1}{y}$ and $g(y,z)=\ln y + h(z)$ Equation (2) becomes: $f=2x \cos y +\ln y+h(z) .....(3)$ and $h(z)=\ln z+C$ Substitute all the above values in the given integral as follows: $ \int_{0,2,1}^{1,0.5 \pi,2} 2 \cos y dx+(\dfrac{1}{y}-2 x \sin t ) dy +\dfrac{dz}{z}=\ln (\pi/2)+\ln 2 -\ln 2=\ln (\dfrac{\pi}{2})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.