Given a vector field F, the divergence theorem states that

\oint_{s} \to F . d ¯ S = \oint_{V} \nabla . \to F d v

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\oint_{s} \to F . d \to S = \oint_{V} \nabla \times \to F d v

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\oint_{s} \to F . d \to S = \oint_{V} (\nabla \to F) d v

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\oint_{s} \to F . d ¯ S = \oint_{V} (\nabla \times \to F) d v

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Solution

The correct option is A $\oint_{s} \to F . d ¯ S = \oint_{V} \nabla . \to F d v$
The divergence theorem states that the total outward flux of a vector field $\to F$ through the closed surface S is same as the volume integral of the divergence of $\to F$
$\Rightarrow$ $\oint_{s} \to F . d ¯ S = \oint_{V} \nabla . \to F d v$

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