In order for F to be conservative, there must be a scalar function f such that the gradient of f is equal to F. This means
[tex]\dfrac{\partial f}{\partial x}=3x^2-2y^2[/tex]
[tex]\dfrac{\partial f}{\partial y}=4xy+4[/tex]
Integrate both sides of the first equation with respect to x :
[tex]f(x,y)=x^3-2xy^2+g(y)[/tex]
Differentiate both sides with respect to y :
[tex]\dfrac{\partial f}{\partial y}=-4xy+\dfrac{\mathrm dg}{\mathrm dy}=4xy+4\implies\dfrac{\mathrm dg}{\mathrm dy}=8xy+4[/tex]
But we assume g is a function of y, which means its derivative can't possibly contain x, so there is no scalar function f whose gradient is F. Therefore F is not conservative.
In this problem, since the condition of equal derivatives does not apply, the vector field is not conservative.
A vector field can be described as:
[tex]F = <P,Q>[/tex]
It is conservative if:
[tex]\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}[/tex]
In this problem, the field is:
[tex]F = <3x^2 - 2y^2, 4xy + 4>[/tex]
Then:
[tex]P(x,y) = 3x^2 - 2y^2[/tex]
[tex]\frac{\partial P}{\partial y} = -4y[/tex]
[tex]Q(x,y) = 4xy + 4[/tex]
[tex]\frac{\partial Q}{\partial x} = 4y[/tex]
Since [tex]\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}[/tex], the field is not conservative.
A similar problem is given at https://brainly.com/question/15236009
