Let C be the square oriented counterclockwise with corners (0,0),(3,0),(3,3), and (0,3). 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 $0\leq t\leq 3$, by:

edge $C_1$
$x_1(t)=$
$y_1(t)=$

edge $C_2$
$x_2(t)=$
$y_2(t)=$

edge $C_3$
$x_3(t)=$
$y_3(t)=$

edge $C_4$
$x_4(t)=$
$y_4(t)=$

$\displaystyle\oint\limits_C y^2\,dx+x^2\,dy$$\displaystyle{}= \int\limits_{C_1} y^2\,dx+x^2\,dy$ $\displaystyle{}+\int\limits_{C_2} y^2\,dx+x^2\,dy$ $\displaystyle{}+\int\limits_{C_3} y^2\,dx+x^2\,dy$$\displaystyle{}+ \int\limits_{C_4} y^2\,dx+x^2\,dy$
$\hspace{15pt} =$ + + +

Applying Green's theorem,

$\displaystyle\oint\limits_C y^2\,dx+x^2\,dy$

=

$\displaystyle\int$

$\displaystyle\int$

$dx\,dy$

=

$\displaystyle\int$

$dy$

=

The vector field $F=y^2 i+ x^2 j$ is: conservative not conservative