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State the Divergence Theorem SHdA=WdivFdV\int _ { S } \vec { H } \cdot \vec { d A } = \int _ { W } { div } { \vec { F} } d V

Question 16

Multiple Choice

State the Divergence Theorem.


A) If W is a solid region whose boundary S is a piecewise smooth surface, then SHdA=WdivFdV\int _ { S } \vec { H } \cdot \vec { d A } = \int _ { W } { div } { \vec { F} } d V
B) If W is a solid region whose boundary S is a smooth surface oriented outward, then SHdA=WdivFdV\int _ { S } \vec { H } \cdot \vec { d A } = \int _ { W } { div } { \vec { F} } d V
C) If W is a solid region whose boundary S is a piecewise smooth surface oriented outward, and if F\vec { F } is a smooth vector field on an open region containing W and S, then
SFdA=WdivFdV\int_{S} \vec{F} \cdot \overrightarrow{d A}=\int_{W} \operatorname{div} \vec{F} d V
D) If W is a solid region whose boundary S is a piecewise smooth surface oriented outward, then SFdA=WdivFdVˉ\int _ { S } \vec { F } \cdot \vec { d A } = \int _ { W } { div } \vec { F } \cdot \overrightarrow { d \bar { V } }
E) If S is a solid region whose boundary W is a piecewise smooth surface oriented outward, then SFdA=WdivFdV\int _ { S } \vec { F } \cdot \vec { d A } = \int _ { W } { div } \vec { F } { d { V } }

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