Determine $f: \mathbb{R}^3 \to \mathbb{R}$ in $C^1$ such that the differential form $\omega = fdx +zdy+ydz$ is exact in $\mathbb{R}^3$. Then calculate a potential of $\omega$.
Since $\mathbb{R}^3$ is a star domain, $\omega$ is exact iff the curl of the vector field associated is zero, that is iff $\partial_3 f=\partial_2f=0$. This happens iff f depends only on x; let $g(x)=f(x,y,z)$, then a potential is $\int_{0}^{x}g(t)dt+yz$.
Does it seems correct?
Best Answer
Your arguments are fine ! Everything is O.K.