Answer
$\phi(x,y)=\dfrac{x^2}{2}+\dfrac{y^2}{2}$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$
We have: $f(x,y)=x$ and $g(x,y)=y$
Thus, $\dfrac{\partial f}{\partial y}=0$ and $\dfrac{\partial g}{\partial x}=0$
Therefore, a vector field $F$ is Conservative.
Now, potential function $F=\nabla \phi$
So, $\dfrac{\partial \phi}{\partial x}=x $ and $\phi(x,y)=\dfrac{x^2}{2}+h(y)$
and $\dfrac{\partial \phi}{\partial y}=y =h^{\prime}(y)$ and $h(y)=\dfrac{y^2}{2}$
Thus, $\phi(x,y)=\dfrac{x^2}{2}+\dfrac{y^2}{2}$
