Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.5 Surface Integrals of Vector Fields - Preliminary Questions - Page 968: 4

Answer

(b)

Work Step by Step

We have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S$. If ${\bf{F}}\left( P \right) = {\bf{n}}\left( P \right)$ at each point on $S$, then $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{n}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S$ Since $Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$, so the answer is (b).
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