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If  
\[\mathbf{F}=F_1 \mathbf{i} + F_2 \mathbf{j}\]
  is a vector field then  
\[div \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}\]

Gauss's Theorem says  
\[\oint_C \mathbf{F} \cdot \mathbf{n} \: ds = \int \int_B div \mathbf{F} \: dx \: dy\]

If  
\[f\]
  is a function such that  
\[grad f = \mathbf{\nabla} f = \mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} \]
  then  
\[div \mathbf{F} = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}= \nabla^2 f\]

Hence  
\[\oint_C (\mathbf{\nabla} f) \cdot \mathbf{n} \: ds = \int \int_B \nabla^2 f \: dx \: dy\]
.
If  
\[f\]
  is harmonic so that 
\[\nabla^2 f=0\]
  then  
\[\oint_C (\mathbf{\nabla} f) \cdot \mathbf{n} \: ds =0\]