Answer
$\oint_C D_n g ds=0$
Work Step by Step
Here, we have
$\oint_C F \cdot n ds=\iint_D div F(x,y) dA$
when $\nabla^2 g=0$, then we have $\oint_C \nabla g \cdot n ds=0$
when $F=\nabla g$
$\oint_C \nabla g \cdot n ds=\iint_D div (\nabla g) dA=\iint_D \nabla \cdot (\nabla g) dA=\iint_D \nabla^2 g dA=\iint_D (0) dA=0$
Since, $D_ng$ can be defined as $\nabla g \cdot n$
This implies that $\oint_C D_n g ds=0$
Hence, the result has been proved.
