Let C be the square oriented counterclockwise with corners (0,0), (6,0), (6,6), and (0,6). If we label the edges of the square as $C_1, C_2, C_3, C_4$ starting from the bottom edge going counterclockwise, then the edges may be linearly parameterized, with $0 \le t \le 6$, by:
edge $C_1$
$x_1(t) = t$
$y_1(t) = 0$
edge $C_2$
$x_2(t) = 6$
$y_2(t) = t$
edge $C_3$
$x_3(t) = 6$
$y_3(t) = 6$
edge $C_4$
$x_4(t) = 0$
$y_4(t) = t$
$\int_C y^2 dx + x^2 dy = \int_{C_1} y^2 dx + x^2 dy + \int_{C_2} y^2 dx + x^2 dy + \int_{C_3} y^2 dx + x^2 dy + \int_{C_4} y^2 dx + x^2 dy$
$= \Box + \Box + \Box + \Box$
Applying Green's theorem,
$\int_C y^2 dx + x^2 dy = \iint_D \Box \Box dx dy$
$= \Box \Box dy = \Box$
The vector field $F = y^2i + x^2j$ is: not conservative
