How to show this differential form is exact using definition?

Question

I was given this vector field: $$F(x, y, z)=(x^2, xyz, yz^2)$$ I calculated its line integral on $$\gamma(t)=(\cos(t), \sin(t), 0)$$ and I found it to be equal to $0$. Therefore I know for sure the field is conservative. I now want to find a function $f$ so that, given the differential form associated to the field $$dw=x^2dx+xyzdy+yz^2dz$$ we have $df=dw$, where $$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$$ I set $\frac{\partial f}{\partial x}=x^2$ and so on. I found three different forms for $f(x, y, z)$: $$f(x, y, z)=\frac{x^3}{3}+h_1(y, z)$$ $$f(x, y, z)=\frac{xy^2z}{2}+h_2(x, z)$$ $$f(x, y, z)=\frac{yz^3}{3}+h_3(x, y)$$ I set the first one equal to the third and I solved it for $h_1(y, z)$: $$h_1(y, z)=-\frac{x^3}{3}+\frac{yz^3}{3}+h_3(x, y)$$ Since $h_1$ only depends on y and z, we have that: $$h_3(x, y)=\frac{x^3}{3}+h_4(y)$$ Therefore: $$h_1(y, z)=\frac{yz^3}{3}+h_4(y)$$ And: $$f(x, y, z)=\frac{x^3}{3}+\frac{yz^3}{3}+h_4(y)$$ Then I set this equal to the second form of f: $$\frac{x^3}{3}+\frac{yz^3}{3}+h_4(y)=\frac{xy^2z}{2}+h_2(x, z)$$ I did the same thing as before and solved it for $h_4(y)$: $$h_4(y)=\frac{xy^2z}{2}-\frac{x^3}{3}-\frac{yz^3}{3}+h_2(x, z)$$ At this point though I can't find an expression for $h_2(x, z)$ so that $h_4$ only depends on y. I don't understand if it exists and I am unable to find it, or if it doesn't exist at all, but this would mean the form is not exact, which is absurd given that its integral on $\gamma(t)$ is $0$. The other option is I solved the integral in the wrong way, nut I checked it many times, so if I got it wrong it must mean I didn't understand how to solve the integral of a differential form at all: in that case, could you explain to me how you would do it?

Answer 1

“I computed the line integral on $\gamma(t)=(\cos t,\sin t,0)$ and found it to be equal to $0$. Therefore I know for sure the field is conservative.”

I stopped reading after that. So what? Having the line integral equal to $0$ on one curve says NOTHING about exactness of the $1$-form. If this was the case for EVERY closed loop, then you can say the $1$-form is exact. This is a very serious mistake, make sure you understand the difference.

In your case, the 1-form is obviously not exact, and you can do this in a 2-line calculation; let \begin{align} \alpha=x^2\,dx+xyz\,dy+yz^2\,dz, \end{align} and you’ll easily find that $d\alpha\neq 0$ (what is it?). If $\alpha$ was exact, then there would be some $f$ such that $\alpha=df$, and so $d\alpha=d(df)=0$, which is absurd. In short, exactness implies closedness, so by contraposition, not-closed forms cannot be exact (in more 3-dimensional vector calculus language, a vector field with non-zero curl (essentially what $d\alpha$ is computing) cannot be conservative).