TITLE: Derivative with respect to the normal?
QUESTION [5 upvotes]: I am trying to use greens theorem to show the following:
$$\int \int (f_{xx}+f_{yy}) \, dx \, dy=\int\frac{\partial f}{\partial n} \, ds$$
I am not completely sure how to treat the $d/dn$.  I have simplified to a point where I have
$$\int f_x \, dy-f_y \, dx=\int \int (f_{xx}+f_{yy}) \, dx \, dy$$
So I need to go from the left side of the last equation to the right side of the first equation.  Any intuition on what the derivative implies?

REPLY [6 votes]: The derivative with respect to the normal is the rate of change in the direction of the normal:
$$
\partial_n f=n\cdot\nabla f
$$
where $n$ is the unit normal.
Along a curve piece $(\mathrm{d}x,\mathrm{d}y)$, the unit outward normal to a counterclockwise curve times the length of the curve piece is $(\mathrm{d}y,-\mathrm{d}x)=n\,\mathrm{d}s$. Thus, we get
$$
\begin{align}
\int n\cdot\nabla f\,\mathrm{d}s
&=\int \nabla f\cdot(\mathrm{d}y,-\mathrm{d}x)\\
&=\int f_x\,\mathrm{d}y-f_y\,\mathrm{d}x
\end{align}
$$