Radial differential form Hi have to determinate a primitive of this differential form:
$$\omega = \frac{xy}{\sqrt{(x^2+y^2)}}dx + \frac{x^2 + 2y^2}{\sqrt{(x^2+y^2)}}dy$$
As far as I know this should be a radial form which I could simplify as follow:
$$ \omega = xh(\rho)dx + yh(\rho)dy$$
So it's a radial form => closed => exact form in $R^2 \backslash \{(0,0)\}$ and then I can determinate the primitive by solving this integral: $\int h(\rho)\rho d\rho $
I managed to do that and I have arrived at this point:
$$\omega = xh(\rho)dx + yh(\rho)dy + \rho dy$$
Assuming $h(\rho)=y/\rho$ . I am quite sure I'm on the right road but I really don't know how to manage the $\rho dy$ part of the form...
is there something I am doing wrong?
Thank you in advance!
EDIT: The $h(\rho)$ function should be correct, in fact solving $\int y/\rho * \rho d\rho = y\sqrt{x^2+y^2}$ that is a primitive of the given form.
 A: In polar coordinates you have
$$
\omega=r\cos\phi\sin\phi \,d(r\cos\phi)+r(1+\sin^2\phi)\,d(r\sin\phi)=\\
r\cos^2\phi\sin\phi\, dr- r^2\cos\phi\sin^2\phi\, d\phi+r(\sin\phi+\sin^3\phi)\,dr+r^2(\cos\phi+\cos\phi\sin^2\phi)\,d\phi=\\
2r\sin\phi\, dr +r^2\cos\phi\, d\phi=d(r^2\sin\phi).
$$ Therefore,
$$
\omega=d\left(y\sqrt{x^2+y^2}\right).
$$
A: I see two problems:
First, $h(\rho) = y / \rho$  seems very suspicious, since $y$ is not a function of $\rho$.
Secondly, your premise is surely wrong. If the form was of the form $x h(\rho) \mathrm{d}x + yh(\rho) \mathrm{d}y$, you could divide the coefficients and get $ \frac{y h(\rho)}{x h(\rho)} = \frac{y}{x} $, but the given form clearly does not satisfy this.

Anyways, it is fairly easy to antidifferentiate the coefficient on $\mathrm{d}x$ with respect to $x$:
$$ \int \frac{xy}{\sqrt{x^2 + y^2}} \mathrm{d}x = y \sqrt{x^2 + y^2} + g(y)$$
and then to solve for $g$ we compute the total derivative
$$ \frac{xy}{\sqrt{x^2 + y^2}} \mathrm{d}x
+ \frac{y^2}{\sqrt{x^2 + y^2}} \mathrm{d}y
+ \sqrt{x^2 + y^2} \mathrm{d}y
+ g'(y) \mathrm{d}y $$
Combining the middle two terms gives
$$ \frac{xy}{\sqrt{x^2 + y^2}} \mathrm{d}x
+ \frac{x^2 + 2 y^2}{\sqrt{x^2 + y^2}} \mathrm{d}y
+ g'(y) \mathrm{d}y $$
so we get a solution by picking $g$ to be a constant function.
