Answer
$\phi(x,y)=\sqrt {x^2+y^2}$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$
We have: $f(x,y)=\dfrac{x}{\sqrt {x^2+y^2}}$ and $g(x,y)=\dfrac{y}{\sqrt {x^2+y^2}}$
Thus, $\dfrac{\partial f}{\partial y}=\dfrac{-xy}{(x^2+y^2)^{3/2}}$ and $\dfrac{\partial g}{\partial x}=\dfrac{-xy}{(x^2+y^2)^{3/2}}$
Therefore, a vector field $F$ is Conservative.
Now, potential function $F=\nabla \phi$
So, $\dfrac{\partial \phi}{\partial x}=\dfrac{x}{(x^2+y^2)^{1/2}} $ and $\phi(x,y)=\sqrt{x^2+y^2}+h(y)$
and $\dfrac{\partial \phi}{\partial y}=\dfrac{y}{(x^2+y^2)^{1/2}}+h^{\prime}(y)$ and $h^{\prime \prime}(y)=0$
Thus, $\phi(x,y)=\sqrt {x^2+y^2}$
