Given a 1-form, find the 0-form

So, I was doing some exercises of differential forms and exterior derivatives and I got stuck in one exercise that requested to find closed-forms, that is $d\omega =0$. All of them are 1-form and a need to find 0-form. Well, i know there's exist $d\alpha=\omega$. My question is how find $\alpha$. Do i have to use Stokes' Theorem?

Ps.: $\omega=2xydx+x^2dy+2zdz$

• Were you given a contour on which to integrate? – Larry B. Oct 26 '16 at 17:07
• No. So Stokes' Theorem is out. I didnt realize this. But now i have no ideas to proceed with this question.. – alvaro.louzi Oct 26 '16 at 17:11

A 0-form is just a function, and the exterior derivative is its differential:

$$df = \frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy+\frac{\partial f}{\partial z} dz$$

So, you need a function $f$, such that

$$\frac{\partial f}{\partial x}=2xy$$ $$\frac{\partial f}{\partial y}=x^2$$ $$\frac{\partial f}{\partial z}=2z$$

Integrating the first and second equation in variables $x$ and $y$, respectively, we obtain the same answer - $x^2y$. The last equation gives a term $z^2$. So, a function satisfying these equations is $$x^2y+z^2$$

In principal, you can integrate along a contour to solve such a problem (which contour does not matter, that is the point if a closed form). But for these small examples, guessing is actually a viable tactic:

A $0$-form is just a function $\alpha(x,y,z)$. You know that differentiating it w.r.t. $x$ should give $2xy$, so the first guess should be be $\alpha = x^2y$. Then the $y$-derivative is $x^2$ which is correct, but the $z$-derivative is $0$ which is not correct (should be $2z$ in your example). So you add another term: $\alpha=x^2y + z^2$. The additional term does not change the $x$- or $y$-derivative and gives the required $z$, so it is the solution. (You know that the solution is unique up to a constant).