Answer
$\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
Work Step by Step
$\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA+\iint_D \nabla f \cdot \nabla g dA$
This implies that $\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds$
That is, $\iint_D \nabla (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
Now, we have $\iint_D div (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$
Hence, we have $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
