University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 15 - Section 15.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises - Page 850: 19

Answer

$ 9 \ln (2)$

Work Step by Step

Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k$ $\dfrac{\partial f}{ \partial x}= 3x^2$ and $\dfrac{\partial f}{ \partial y}=\dfrac{z^2}{y} $ and $ \dfrac{\partial f}{ \partial z}=2z \ln y$ Now, $f=x^3+g(y,z) $ or, $\dfrac{\partial g(y,z)}{ \partial x}=z^2 \ln y$ $f=x^3 +z^2 \ln y +h(z) $ or, $h(z)=C$ Now, we will substitute all the above values in the given integral. $ \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)$ or, $= 9 \ln (2)$
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