The potential energy function associated with the force \[\vec F = 4xy\hat i + 2{x^2}\hat j\] is:
A. \[U = - 4{x^2}y + \] constant
B. \[U = {x^2}y + \] constant
C. \[U = {x^3}y + \] constant
D. Not defined

Hint: In this question, we are given a conservative force \[\vec F\] in a two-dimensional plane and we need to find out the potential energy function. Now the change in potential energy due to internal conservative force is given by \[U = - \int {F \cdot dr} \] . Using this, we can find out the potential energy function.

Formula used:
Potential energy due to internal conservative force is, \[U = \int {dU = } - \int\limits_{{r_i}}^{{r_f}} {F \cdot dr} = - W\]
Where \[F\] is the conservative force
\[dr\] is the small displacement

Complete step by step answer:
We have given,
\[\vec F = 4xy\hat i + 2{x^2}\hat j\]
Now as we know potential energy for a conservative internal force is given by,
\[U = \int {dU = } - \int\limits_{{r_i}}^{{r_f}} {F \cdot dr} = - W\]
We can also say that the potential energy of a body is negative of the work done by the conservative forces in bringing the body from infinity to the present position
\[U = - \int {F \cdot dr} \]
As the motion is in two-dimensional plane
Thus, we can write \[dr = dx\hat i + dy\hat j\]
Substituting the value of \[\vec F\] in the equation we get,
\[U = - \int {\left( {4xy\hat i + 2{x^2}\hat j} \right)} \cdot \left( {dx\hat i + dy\hat j} \right)\]
Taking dot product we get,
\[ \Rightarrow U = - \int {4xydx + 2{x^2}dy} \]
\[ \Rightarrow U = - \int {4xydx - \int {2{x^2}dy} } \]
On integrating we get,
\[ \Rightarrow U = - 2{x^2}y - 2{x^2}y + c\]
Where \[c\] is a constant
\[ \Rightarrow U = - 4{x^2}y + c\]
The potential energy function associated with the force \[\vec F = 4xy\hat i + 2{x^2}\hat j\] is \[ - 4{x^2}y + c\]
Hence, option A. is the correct option.

Additional information:
We know \[F = - \dfrac{{dU}}{{dr}}\] . If the force is conservative and is in equilibrium, then \[F = - \dfrac{{dU}}{{dr}} = 0\]. Thus, at these points the body is said to be in an equilibrium.
An object is said to be in stable equilibrium if the body tends to return to the same position after a slight displacement. If \[\dfrac{{{d^2}U}}{{d{r^2}}}\] is positive body is in stable equilibrium
If \[\dfrac{{{d^2}U}}{{d{r^2}}}\] is negative, the body is said to be in unstable equilibrium
If \[\dfrac{{{d^2}U}}{{d{r^2}}}\] is zero, the body is said to be in neutral equilibrium

Note: Potential energy is defined for a conservative field only. For non-conservative forces, it has no meaning. Potential energy should be considered as the property of the entire system rather than assigning it to some specific particle. Potential energy depends on the frame of reference.