Let f : $R^2$ → R be a twice continuously differentiable scalar field which is harmonic, that is, $∇^2$ f = 0. Use Green’s Theorem to prove that the line integral $$\int_c {df\over dy}dx-{df\over dx}dy$$
is independent of curve C in the (xy)-plane.
$$so far $$ I have worked out that ∇ x f=0 for the graph being irrotational I'm not sure this is relevant to be honest I know f is conservative if it is the gradient of a scalar field as well and I think conservative and independent are the same in this sense I'm not sure how to take the gradient I've tried to apply greens theorem making ${df\over dy}$ = and $-{df\over dx}$=Q but the answers just look plain wrong to me can anyone help
my application gave me $2 \int \int_R {d^2f\over dxdy}$dxdy if that is right I don't know how it proves independence.
To apply Green's Theorem, we have to have a closed curve that encloses a region. $C$ itself might not be closed. The phrase "independent of curve $C$" means that as long as two curves $C_1$ and $C_2$ start and end at the same point, then they will give the same line integral. On a piece of paper, draw a random curve $C_1$ and then draw another curve $C_2$ starting and ending at the same point. Notice that this now encloses a region. Call it $R$.
Then Green's Theorem says that along the curve $C=C_1-C_2$, which is the path along $C_1$ and then back to the beginning along $C_2$, we have $$\int_C Pdx+Qdy=\iint_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$$
So we actually get $$\iint_R\left(-\frac{\partial^2 f}{\partial x^2}-\frac{\partial^2 f}{\partial y^2}\right)dA=-\iint_R\left(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\right)dA$$
Now we need to use the fact that $f$ is harmonic. What does this tell us?