## Calculus: Early Transcendentals 8th Edition

$\iint_D div (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$ and $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
$\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA+\iint_D \nabla f \cdot \nabla g dA$ This implies that $\iint_Df \nabla^2 g dA+\iint_D \nabla f \cdot \nabla g dA=\oint_C f(\nabla g) \cdot n ds$ Re-arrange as: $\iint_D \nabla (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$ This implies that $f\nabla^2 g+\nabla f \cdot \nabla g=f\nabla \cdot (\nabla g)+\nabla f \cdot \nabla g=\nabla \cdot (f \nabla g)$ $\iint_D div (f \nabla g) dA=\oint_C f(\nabla g) \cdot n ds$ Hence, the result has been proved. $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$