Let $\omega = -\frac{4z}{(x^2+1)^2}dx+ \frac{2y}{y^2+1}dy+\frac{2x}{x^2+1}dz,$ and let $\eta = -\frac{4xz}{(x^2+1)^2}dx + \frac{2y}{y^2+1}dy+\frac{2}{x^2+1}dz$. Evaluate the line integrals over the straight line segment from $(0,0,0)$ to $(1,1,1)$ and determine which is exact.
Solution Attempt
For a straight line segment, we have
$$\gamma:[0,1] \to \Bbb{R}^3$$
via
$$t \mapsto (t,t,t).$$
And we know
$$\int_\gamma \omega = \int_{[0,1]} \gamma^* \omega.$$
So for
$$\int_\gamma \omega$$
I get
$$\int_0^1 -\frac{4t}{(t^2+1)^2}dt+2\int_0^1 \frac{2t}{t^2+1}dt=-1+2\ln(2).$$
For $\eta$ I get
$$\int_\gamma \eta = \int_0^1 - \frac{4t^2}{(t^2+1)^2}dt+\int_0^1 \frac{2t}{t^2+1}dt+2\int_0^1 \frac{1}{t^2+1}dt=1+\ln(2).$$
And for which is exact, I claim $\eta$ is exact with potential function
$$f(x,y,z)=\ln(y^2+1)+\frac{2z}{x^2+1}.$$
How do you generally prove something is not exact? Since showing exactness relies on existence of a potential function. So is $\omega$ not exact? I tried so hard but could not find a potential function whatsoever.
Added: I think I figured out to show $\omega$ isn’t exact. We’d need the partial of the first term wrt $z$ to be the same as the last term partial w respect to $x$ but they aren’t equal thus it’s not exact. I.e., If $\omega_1 = -\frac{4z}{(x^2+1)^2}$, and $\omega_3 = \frac{2x}{x^2+1}$ Then in order to have closed we need
$$\frac{\partial \omega_1}{\partial z}=\frac{\partial \omega_3}{\partial x}$$
But this does not hold.
For $$ \omega=a\,dx+b\,dy+c\,dz $$ to be closed we have to check if $d\omega\equiv 0\,.$ This means the three parentheses in $$ d\omega=(\partial_yc-\partial_zb)\,dy\wedge dz+(\partial_xc-\partial_za)\,dx\wedge dz+(\partial_xb-\partial_ya)\,dx\wedge dy $$ must vanish. In your case \begin{align} a &= -\frac{4z}{(x^2+1)^2}\,,&b&=\frac{2y}{y^2+1}&c&=\frac{2x}{x^2+1}\,. \end{align} So the first and last parenthesis vanish, but not the second: \begin{align} (\partial_yc-\partial_zb)&=(\partial_xb-\partial_ya)=0\,,\\[2mm] (\partial_xc-\partial_za)&=-\frac{2(-3+x^2)}{(x^2+1)^2}\,. \end{align} Since $\omega$ is not closed it cannot be exact.