Answer
a) The vector field $F$ is conservative. For all three paths in the given graph, the end points are same, so the line integral is the same for all three paths.
b) $16$
Work Step by Step
a) When $F(x,y)=Ai+Bj$ is a conservative field, then throughout the domain $D$, we get
$\dfrac{\partial A}{\partial y}=\dfrac{\partial B}{\partial x}$
$a$ and $b$ are the first-order partial derivatives on the domain $D$.
Here, we have $\dfrac{\partial A}{\partial y} = \dfrac{\partial B}{\partial x}=2x$
Thus, the vector field $F$ is conservative.
For all three paths in the given graph, the end points are same, so the line integral is the same for all the three paths.
b) Now, $f(x,y)=x^2y+g(y)$ [g(y) : A function of y]
$f_y(x,y)=x^2+g'(y)$
Here, $g(y)=k$
Thus, $f(x,y)=x^2y+k$
Now, $\int_C F \cdot dr =f(3,2)-f(1,2)=(18+k)-(2+k)=16$