Compute $f$ so that $F=\nabla f$ for $$F(x,y,z)=(yz+x-y, xz-x+z, xy+y-z)$$
I need to see the method of finding this that is not using the theorem about the star shaped set. The method that is like integrating all three parts of $F$ and then comparing them. I cant get a correct answer.
On
Integrate the first w.r.t $x$, then differentiate w.r.t $y$, and compare with the second. Integrate the second w.r.t $y$, then differentiate w.r.t $z$ and compare with the third. And so on. $$f(x,y,z) = \int yz+x-y \,{\rm d}x = xyz + \frac{x^2}{2}-xy + c_1(y,z).$$ So, applying $\frac{\partial}{\partial y}$: $$\frac{\partial f}{\partial y}(x,y,z) = xz-x + \frac{\partial c_1}{\partial y}(y,z) = xz-x+z$$ gives us $\frac{\partial c_1}{\partial y}(y,z) = z$. Integrating: $$c_1(y,z) = \int z\,{\rm d}y = yz + c_2(z).$$ Plugging into $f$, we have: $$f(x,y,z) = xyz+\frac{x^2}{2}-xy+yz+c_2(z).$$ Now apply $\frac{\partial}{\partial z}$ and compare: $$\frac{\partial f}{\partial z}(x,y,z) = xy+y+c_2'(z) = xy+y-z,$$ hence $c_2'(z) = -z$, and $c_2(z) = -\frac{z^2}{2}+c$, with finally $c \in \Bbb R$.
Final answer: $$f(x,y,z) = xyz+\frac{x^2}{2}-xy+yz+-\frac{z^2}{2}+c.$$
Integrating you get: $$f(x,y,z) = xyz +x^2/2-yx +C_1(y,z)$$ $$f(x,y,z) = xyz +yz - yx +C_2(x,z)$$ $$f(x,y,z) = xyz +yz - z^2/2 +C_3(x,z)$$ So: $$f(x,y,z) = xyz +yz-xy+x^2/2-z^2/2 +C$$