Answer
$\iint_D (f \nabla^2 g-g \nabla^2 f) dA=\oint_C (f\nabla g-g\nabla f) \cdot n ds$
Work Step by Step
$\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$ ..(Equation 1)
The above equation (1) can be re-written as:
$\iint_D g \nabla^2 g dA=\oint_C g(\nabla f) \cdot n ds-\iint_D \nabla g \cdot \nabla f dA$ ..(Equation 2)
After subtracting equation-1 from the equation 1, we have
$\iint_D (f \nabla^2 g-g \nabla^2 f) dA=\oint_C (f\nabla g-g\nabla f) \cdot n ds$
Hence, the result is proved.
