I am attempting the following problem in Robert Adams' Calculus textbook:
Here is the solution:
I am just wondering about the part in red: why is it not $2y(x-z)$? I am just learning this concept of conservative fields.
Is the potential function $\phi=\frac{1}{2}x^{2}+yx+xy-zy+\frac{1}{2}z^{2}-yz=\frac{x^{2}+z^{2}}{2}+2xy-2yz$?
If the part in red were $2y(x-z)$ then this would result in the first component of the gradient being $\frac{\partial}{\partial x} [\frac{x^2+z^2}{2}+2y(x-z)] = x+2y$. This would imply that the $i$th component of $F$ would be $x+2y$ instead of $x+y$. Note that $y(x-z) = yx - yz$, not the $yx+xy-zy$ as you interpreted.