Answer
$\phi (x, y) =\sqrt {x^2+y^2}$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial F_1}{\partial y}=\dfrac{\partial F_2}{\partial x}$
We are given that $F(x,y)=\dfrac{(x, y)}{\sqrt {x^2+y^2}}$
Also, the curve is $r_1(t)= (t, 0) \implies r_1'(t) = (1, 0)$ and $r_2(t)= (x, t) \implies r_2'(t) = (0, 1)$
Therefore, the integral is:
$\oint F \ dr=\int_0^{x} [F[r_1(t)] r_1'(t) \ dt +\int_0^{y} F[r_2(t)] r_2'(t)] \ dt\\=\int_0^{x} \dfrac{(t, 0)}{t} \cdot (1, 0) +\int_0^y \dfrac{(x, y)}{\sqrt {x^2+y^2}} \cdot (0, 1) \ dt \\=\sqrt {x^2+y^2}$
Thus, $\phi (x, y) =\sqrt {x^2+y^2}$
