Green's first identity : why $\iint \left | \triangledown f \right |^2dA= 0$?

Question

The question is: Use Green's first identity to show that if $f$ is harmonic on $D$, and if $f(x,y) = 0$ on the boundary curve $C$, then $$\iint \left | \triangledown f \right |^2dA = 0.$$

Green's first identity looks like this

$$\iint_D (f\triangledown ^2 g )dA = \oint _C f(\triangledown g)\cdot ds - \iint_D(\triangledown f \cdot \triangledown g)dA.$$

If I place $f$ instead of $g$, it would look like this

$$ \iint_D (f\triangledown ^2 f )dA = \oint _C f(\triangledown f)\cdot ds - \iint_D(\triangledown f \cdot \triangledown f)dA. $$

Am I doing right? How can I go further?

Answer 1

$f$ is harmonic, so $\nabla^2f=0$, and the integral on the left hand side of the last identity is $=0$. $f$ vanishes on $C$, so that the first integral on the right hand side of the last identity is also $=0$.

Answer 2

Yes, your are doing right. Note that on the equality

$$\iint_D (f\triangledown ^2 f )dA - \oint _C f(\triangledown f)\cdot ds =- \iint_D(\triangledown f \cdot \triangledown f)dA$$

the first integral vanishes since $f$ is harmonic (that is, $\nabla^2 f=0$ on $D$) and the second integral vanishes because $f=0$ on $C.$ Thus, it is

$$0=\iint_D(\triangledown f \cdot \triangledown f)dA=\iint_D |\triangledown f|^2dA.$$