Answer
$\iint_D |\nabla f|^2 dA=0$
Work Step by Step
When $f$ is harmonic then $\nabla^2 f=0$ on $D$, then we have
$\iint_Df \nabla^2 f dA=0$
Since, we have $\iint_Df \nabla^2 g dA=\oint_C f(\nabla g) \cdot n ds-\iint_D \nabla f \cdot \nabla g dA$
$\iint_D f \nabla^2 f dA=\oint_C f(\nabla f) \cdot n ds-\iint_D \nabla f \cdot \nabla f dA$ ...(Equation-1)
Also, when $f(x,y)=0$ on the curve $C$, then we have
$\iint_Cf( \nabla f) \cdot n ds=0$
From the equation- 1, we have
$\iint_D \nabla f \cdot \nabla f dA=0$
Hence, we get $\iint_D |\nabla f|^2 dA=0$ (proved)
