Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.9 - The Divergence Theorem - 16.9 Exercise - Page 1146: 29


$\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$

Work Step by Step

The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $ Here, $S$ is a closed surface and $E$ is the region inside that surface. $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ Since, we have $D_nf=(\nabla f) \cdot n$ This implies that $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV$ Since, $F=\nabla g$ Therefore, $\iint_S (f \nabla g) dS=\iiint_E div (f \nabla g) dV=\iint_E \nabla (F \nabla g) dV=\iiint_E f(\nabla \cdot ( \nabla g) +\nabla f \cdot (\nabla g) dV$ Hence, it has been verified that $\iint_S (f \nabla g) \cdot n dS=\iiint_E (f \nabla^2g+\nabla f \cdot \nabla g)dV$
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.