How to show that (Change of variable) $ rF'(r)=\int_{\partial D(0, r)} \frac{\partial f}{\partial x}dy-\frac{\partial f}{\partial y}dx $

Question

Let $F(r)=\int_0^{2\pi} f(r\cos \theta, r\sin \theta) d\theta$ where $r>0$. Show that

$$ rF'(r)=\int_{\partial D(0, r)} \frac{\partial f}{\partial x}dy-\frac{\partial f}{\partial y}dx $$

where $\partial D(0, r)$ is the frontier (boundary) of the open ball of radius $r$.

I have no idea how to prove this. I know at some point there is a change of variables, but it is not a standard one with polar coordinates in a double integral, since $r$ is fixed.

Answer 1

Let $x = r \cos \theta$ and $y = r \sin \theta$. For fixed $r$ we have $dx = -r \sin \theta d\theta $ and $dy = r\cos\theta d\theta$ and so $$ rF'(r) = \int_B r\cos\theta f_x + r\sin\theta f_y d \theta = \int_B f_x dy - f_y dx. $$