Answer
The work required to move an object against ${\bf{F}}$ from $\left( {1,1} \right)$ to $\left( {3,4} \right)$ along any path in the first quadrant is
$W = \ln \frac{4}{3}$
Work Step by Step
1. Write ${\bf{F}} = \left( {\frac{1}{x}, - \frac{1}{y},0} \right)$.
2. Next, we check if ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\frac{1}{x}, - \frac{1}{y},0} \right)$ 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 ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right) = \left( {\frac{1}{x}, - \frac{1}{y},0} \right)$ 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 ${\bf{F}} = - \nabla V = - \left( {\frac{{\partial V}}{{\partial x}},\frac{{\partial V}}{{\partial y}},\frac{{\partial V}}{{\partial z}}} \right)$. So,
a. taking the integral of $\dfrac{{\partial V}}{{\partial x}}$ with respect to $x$ gives
$V\left( {x,y,z} \right) = - \smallint \frac{1}{x}{\rm{d}}x = - \ln x + f\left( {y,z} \right)$
b. taking the integral of $\dfrac{{\partial V}}{{\partial y}}$ with respect to $y$ gives
$V\left( {x,y,z} \right) = - \smallint \left( { - \frac{1}{y}} \right){\rm{d}}y = \ln y + g\left( {x,z} \right)$
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
$ - \ln x + f\left( {y,z} \right) = \ln y + g\left( {x,z} \right) = h\left( {x,y} \right)$
From here we conclude that the potential function is $V\left( {x,y,z} \right) = - \ln x + \ln y + C$, where $C$ is a constant.
4. The work required to move an object against ${\bf{F}}$ from $\left( {1,1} \right)$ to $\left( {3,4} \right)$ along any path in the first quadrant is
$W = - \mathop \smallint \limits_{\left( {1,1} \right)}^{\left( {3,4} \right)} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{\left( {1,1} \right)}^{\left( {3,4} \right)} \nabla V\cdot{\rm{d}}r = V\left( {3,4,0} \right) - V\left( {1,1,0} \right)$
$ = - \ln 3 + \ln 4 + \ln 1 - \ln 1$
$ = \ln \frac{4}{3}$
