Answer
The vector field $\overrightarrow{F}$ is not conservative.
Work Step by Step
The vector field $F(x,y)=ai+bj$ is known as conservative field throughout the domain $D$, when we have
$\dfrac{\partial a}{\partial y}=\dfrac{\partial b}{\partial x}$
$a$ and $b$ represents the first-order partial derivatives on the domain $D$.
The work integral (the work done is line integral of force) $W=\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not dependent of path when the line integral $W=\int_C \overrightarrow{F} \cdot \overrightarrow{dr}=0$ for every closed curve $C$. When we draw a closed loop around the center of the vector field, then we will have to show non-zero amount of work in order to get the initial point.
However, for a conservative field, there must have an equal amount of positive and negative work or zero work done in order to get the initial point.
We can see from the above discussion that the line integral of $\overrightarrow{F}$ is not path independent and thus, the vector field $\overrightarrow{F}$ is not conservative.
