## Calculus 8th Edition

$\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$ $\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$ This implies that $\iint_S fc \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot c dV$ and $\implies \iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot c dV$ $\implies \iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$ and $\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$ Hence the result has been verified.