Divergence change of variables (to polar) I would wish to simplify this integral $$\oint_{\partial D}(f\nabla f)\cdot\hat{n}\,ds$$ in terms of a line integral of $g$ on $[0,2\pi]$ where $g(\theta)=f(e^{i\theta})$.
Background info: $D=\{(x,y)\in\mathbb{R}^2: x^2+y^2<1\}$, $f$ is a harmonic function on $D$ and twice continuously differentiable on $\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq 1\}$. (This background info may or may not be needed. This is the second part of a question. I already know how to do the first part, where I used the fact that $f$ is harmonic. )

My working (not entirely sure if it is correct):
Parametrize the circle $\partial D$ by $x=\cos\theta$, $y=\sin\theta$. We may take $\hat n\,ds=(dy,-dx)$.
$\begin{align*}
&\oint_{\partial D}(f\nabla f)\cdot(dy,-dx)\\
&=\oint_{\partial D}(ff_x,ff_y)\cdot(dy,-dx)\\
&=\oint_{\partial D} (ff_x\,dy-ff_y\,dx)
\end{align*}$
Making the change of variable $g(\theta)=f(e^{i\theta})$, then $f_x=g'\cdot\frac{1}{-\sin\theta}$, $f_y=g'\frac{1}{\cos\theta}$, $dx=-\sin\theta\, d\theta$, $dy=\cos\theta\,d\theta$.
Substituting all of these in, the eventual integral becomes $$\int_0^{2\pi} gg'(\tan\theta-\cot\theta)\,d\theta.$$

Is the above correct? Anything wrong with it? Thanks for any help.
 A: Okay, so we have
$$
\int_{\partial D} f\nabla f\cdot \hat{n} ds
$$
Define this change of variables:
$$
x = r\cos(\theta)\\
y = r\sin(\theta)
$$
Now let $g(r,\theta) = f(r\cos(\theta),r\sin(\theta))$. Note that $\hat{n}$ is normal to the circle, i.e., it points radially, which is $\hat{n} = \hat{r}$. The line element transforms as $ds = r\;d\theta$ under this transformation. So:
$$
\int_{\partial D} g\nabla g\cdot \hat{r} \;r\;d\theta,
$$
where now we understand the gradient to be the one in polar coordinates. The gradient of $g$ dotted into a unit vector is just the derivative in the direction of the unit vector, i.e. 
$$
\nabla g\cdot \hat{r} = \frac{\partial g}{\partial r},
$$
leaving
$$
\int_{\partial D} g\frac{\partial g}{\partial r} \;r\;d\theta,
$$
On the circle, $r=1$, and $\theta\in[0,2\pi)$, so the integral becomes
$$
\int_{0}^{2\pi} g(1,\theta)\left.\frac{\partial g(r,\theta)}{\partial r}\right|_{r=1}\;d\theta
$$
This is an integral over the circle strictly in terms of the value of the original function on the circle, and the radial derivative of the function on the circle. It's not really clear that this is in fact a function of a single variable $z = x + iy$, but the context of the question (harmonic functions of two variables) lends itself to the complex plane interpretation, so this is the final answer you want.
A: Generically there is no relationship between the value of a function $f(p)$, on the boundary of the unit disk, and the value of its normal derivative $\nabla f \cdot \hat{n}$ on the boundary. Consider that you can start by prescribing $f$ on the boundary, and then perturb $f$ away from the boundary to change $\nabla f \cdot \hat{n}$ without changing $f$; or alternatively, you can solve the biharmonic PDE
\begin{align*}
\Delta^2 f(x) = 0 &\quad x\in B\\
f(x) = f_0(x) &\quad x \in \partial B\\
\nabla f(x)\cdot \hat{n} = f_1(x) &\quad x \in \partial B
\end{align*}
to find such an $f$.
In your case, you have extra information: you know that $f$ is harmonic, and hence its value and normal derivative at the boundary are coupled by the Laplacian Poincaré–Steklov operator mapping Dirichlet to Neumann boundary conditions. More specifically, give $d:\partial B \to \mathbb{R}$, let $[Pd](x) = \nabla u(x)\cdot \hat{n}$, where $u$ is the unique solution to the PDE
\begin{align*}
\Delta u(x) = 0, \quad &x\in B\\
u(x) = d(x), \quad &x \in \partial B.
\end{align*}
Now you can write down your boundary integral as
$$\int_{\partial B} fPf\,ds.$$
It should be possible to compute an explicit formula for $P$ given the Green's function of the Dirichlet Laplace equation on the unit disk; it's unlikely to be pleasant.
