Integrate over a disk In a proove that I have to do in DoCarmo Differential Forms and aplications, There is a step that I can't understand. 
If $D$ is a disk, $u(x,y);v(x,y)$ then:
$$\int_{\partial D} \frac{u du+vdv}{u^2+v^2} = 0 $$
Another way to see (That the book give as a hint) is to see that is the same as:
$$ \int_{\partial D} d(\log(u^2+v^2)) = 0 $$
Is clear the reason that is the same but I don't know why is 0.
If I have that, I can prove what I need, but I don't know how to argue why that integral is 0. 
Please help with this.
 A: If $D$ is a disk where $u^2+v^2$ does not vanish, then this is a consequence of Stokes' theorem.
Stokes' thoerem says for any domain $D$ with boundary $\partial D$, the exact differential $d\omega$ has integral given by
$$
\int_Dd\omega=\int_{\partial D}\omega.
$$
In your case, we have that 
$$d(\log(u^2+v^2))=\frac{\partial (\log z)}{\partial z}\,dz=\frac{\partial \log z}{\partial z}\left(\frac{\partial z}{\partial u}du+\frac{\partial z}{\partial v}dv\right)\Big|_{z=u^2+v^2}\\
=\frac{1}{u^2+v^2}\left(2u\,du+2v\,dv\right)=2\frac{u\,du+v\,dv}{u^2+v^2}.
$$
Which up to a factor of $2$ is the integrand you started with.
So that becomes
$$
\int_D\frac{u\,du+v\,dv}{u^2+v^2}=\frac{1}{2}\int_{\partial D}d(\log(u^2+v^2))=\frac{1}{2}\int_Dd^2(\log(u^2+v^2)).
$$
The exterior derivative satisfies $d^2=0$, which is ultimately a consequence of the equality of mixed partial derivatives on $C^2$ functions, so $d^2(\log(u^2+v^2))=0,$ and so we have $\int_{\partial D}d(\log(u^2+v^2))=0.$ 
Note that this fails on any domain $D$ which includes $u=v=0.$
