Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.9 The Divergence Theorem - 16.9 Exercises - Page 1186: 28


$\iint_S D_n f dS=\iiint_E \nabla^2 f dV $

Work Step by Step

We know that $D_nf=(\nabla f) \cdot n$ ...(1) Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$ From equation (1), we have $\iint_S D_n f dS=\iiint_S (\nabla f) \cdot dS=\iint_S (\nabla F) \cdot dS $ This gives: $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV $ and $\iint_S (\nabla f) \cdot dS=\iiint_E div (\nabla F) dV$ or, $\iiint_E \nabla \cdot (\nabla F) dV=\iiint_E \nabla^2 f dV $ Thus, $\iint_S D_n f dS=\iiint_E \nabla^2 f dV $
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.