Finding the anti Gradient? I have this question that says $f(x,y,z)$ such that $\nabla f(x,y,z) = <2xy,2yz+x^2,y^2>$ 
The way I see this is that they are asking for what can you take the partial derivative of with respect $x y$ and $z$ so that you get that $\nabla f(x,y,z)$
My answer would be $f(x,y,z) = (x^2y+y^2z+x^2y,y^2z) $
 A: $f$ is called the potential function. We have: $\nabla f = (f_x,f_y,f_z)$. Thus $f_x = 2xy\implies f = \int 2xydx + g(y,z)= x^2y+ g(y,z)\implies f_y = x^2 + \dfrac{\partial g(y,z)}{\partial y}=2yz + x^2\implies g(y,z) = \int 2yzdy + h(z)= y^2z + h(z)\implies f_z = y^2 + h'(z) = y^2\implies h'(z) = 0 \implies h(z) = C\implies f(x,y,z) = x^2y + y^2z + C$ ( constant ).
A: I've developed a sort of algorithm for myself in taking the "anti-gradient" and it personally expedites the process for me. 
For a vector valued function,
F = grad(f) = (H,G,I)
You start by sorting all of the linearly independent terms (LITs) in F into three categories (or subsets) p, q & r for each H,G and I.
That is,
{H}={p,q,r} ,{G}={p,q,r}, {I}={p,q,r}
p-terms:


*

*All LITs that themselves depend on the 2 variables that were not
apart of the partial differentiation (independent variables).


For the component H, p-terms would have to be a function of (y,z) as the partial derivative was with respect to x. 
q-terms:


*

*All LITs that depend on at least 2 variables in which one must be the dependent variable.


or 


*

*All LITs that depend on only one of the independent variables.


For H, q-terms would include functions of: (x,y), (x,z), (x,y,z), (z) or (y)
Basically q is anything that's not p or r.
r-terms:


*

*All LITs that depend only on the dependent variable. 


For H, r-terms are functions of x.
Once you have these sorted out you simply plug them into the following formula.
∫ (Hp+Hq+Hr)dx + (Gq+Gr)dy + (Ir)dz
in essence,
f = C1(x,y,z) + C2(y,z) + C3(z) + C4
As an aside, I found this particularly useful when calculating f for an irrotational vector field in a line integral. It saved me a ton of time on tests.
