Answer
$\phi (x, y) =\dfrac{x^4+x^2y^2+y^4}{2}$
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 $F(x,y)=(2x^3 +xy^2, 2y^3+x^2 y)$
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:
$\oint F \ dr=\int_0^{x} [F[r_1(t)] r_1'(t) \ dt +\int_0^{y} F[r_2(t)] r_2'(t)] \ dt\\=\int_0^{x} (2t^3, 0) \cdot (1, 0) +\int_0^y (2x^3 +xt^2, 2t^3+x^2 t) \cdot (0, 1) \ dt \\=\int_0^ x 2t^3 \ dt+\int_0^y (2t^3 +x^2 t) \ dt\\=\dfrac{2 x^4}{4}+\dfrac{x^2y^2+y^4}{2} \\=\dfrac{x^4+x^2y^2+y^4}{2}$
Thus, $\phi (x, y) =\dfrac{x^4+x^2y^2+y^4}{2}$
