Answer
The work expended when a particle is moved from $O$ to $Q$ along segments $\overline {OP} $ and $\overline {PQ} $ is
${W_{OQ}} = - \frac{2}{3}$
The work expended moving in a complete circuit around the square is zero.
Work Step by Step
1. Write ${\bf{F}} = \left( {{x^2},{y^2},0} \right)$.
2. Next, we check if ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {{x^2},{y^2},0} \right)$ satisfies the cross-partials condition:
$\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ }$ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ }$ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$
We get
$0=0$, ${\ \ \ }$ $0=0$, ${\ \ \ }$ $0=0$
These results imply that ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {{x^2},{y^2},0} \right)$ satisfies the cross-partials condition. Therefore, by Theorem 4, there is a potential function for ${\bf{F}}$.
3. Now, we find a potential function for ${\bf{F}}$.
Let the potential function for ${\bf{F}}$ be $V\left( {x,y,z} \right)$ such that ${\bf{F}} = - \nabla V = - \left( {\dfrac{{\partial V}}{{\partial x}},\dfrac{{\partial V}}{{\partial y}},\dfrac{{\partial V}}{{\partial z}}} \right)$. So,
a. taking the integral of $\dfrac{{\partial V}}{{\partial x}}$ with respect to $x$ gives
$V\left( {x,y,z} \right) = - \smallint {x^2}{\rm{d}}x = - \frac{{{x^3}}}{3} + f\left( {y,z} \right)$
b. taking the integral of $\dfrac{{\partial V}}{{\partial y}}$ with respect to $y$ gives
$V\left( {x,y,z} \right) = - \smallint {y^2}{\rm{d}}y = - \frac{{{y^3}}}{3} + g\left( {x,z} \right)$
c. taking the integral of $\dfrac{{\partial V}}{{\partial z}}$ with respect to $z$ gives
$V\left( {x,y,z} \right) = \smallint 0{\rm{d}}z =0+ h\left( {x,y} \right)$
Since the three ways of expressing $V\left( {x,y,z} \right)$ must be equal, we get
$ - \frac{{{x^3}}}{3} + f\left( {y,z} \right) = - \frac{{{y^3}}}{3} + g\left( {x,z} \right) = h\left( {x,y} \right)$
From here we conclude that the potential function is $V\left( {x,y,z} \right) = - \frac{{{x^3}}}{3} - \frac{{{y^3}}}{3} + C$, where $C$ is a constant.
4. The work expended when a particle is moved from $O$ to $Q$ along segments $\overline {OP} $ and $\overline {PQ} $ is
${W_{OQ}} = - \mathop \smallint \limits_{OQ} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{OQ} \nabla V\cdot{\rm{d}}r = V\left( Q \right) - V\left( O \right) = - \frac{1}{3} - \frac{1}{3} + 0 = - \frac{2}{3}$
5. Since ${\bf{F}} = - \nabla V$, so ${\bf{F}}$ is conservative. By Theorem 1, the circulation of ${\bf{F}}$ around a closed curve is zero, so the work expended moving in a complete circuit around the square is zero.
