Answer
Conservative.
$f(x,y)=xy^2-x^2+k$
Work Step by Step
Given: $F(x,y)=(y^2-2x)i+2xy)$
If $F(x,y)=Ai+Bj$ is a conservative field, then throughout the domain $D$, we get
$\dfrac{\partial A}{\partial y}=\dfrac{\partial B}{\partial x}$
Here, $A$ and $B$ are first-order partial derivatives on the domain $D$.
Then, we have $A_x=2y; B_y=2y$
Here, $\dfrac{\partial A}{\partial y} = \dfrac{\partial Q}{\partial x}$
Thus, the vector field $F(x,y)$ is conservative.
Next, we will find $f$ such that $F=\nabla f$
Here, we have $f_x=y^2-2x..(1) \\ f_y=2xy ...(2)$
Now, integrate $f_x$ with respect to $x$, so we get
$f(x,y)=xy^2-x^2+g(y)$ ...(3)
Here, $g(y)$ represents a constant of integration with respect to $x$ with a function of $y$.
Now, we will differentiate $f(x,y)=xy^2-x^2+g(y)$ with respect to $y$.
$f_y(x,y)=2xy+g'(y) ...(4)$
On comparing the equations (2) and (4) , we get
$f_y(x,y)=2xy$ and $g'(y)=0$
Now, $g(y)=K$ [ Integrate with respect to $y$]
Here, $K$ is a constant.
Hence, the function $f(x,y)$ becomes:
$f(x,y)=xy^2-x^2+k$
