Answer
(a) $\int \textbf F \cdot \textbf r_1= 1$
(a) $\int \textbf F \cdot \textbf r_2= 1$
Work Step by Step
$ \textbf{F} (x,y) = 2xy \textbf i +x^2 \textbf j \\
\nabla \times \textbf F =\begin{vmatrix}
\textbf i & \textbf j & \textbf k\\
\frac{\delta}{\delta x} & \frac{\delta}{\delta y}& \frac{\delta}{\delta z} \\
2xy & x^2 & 0\\
\end{vmatrix}
= 0 \\
$
So, $\textbf F(x,y)$ is a conservative field and let $f(x,y)$ be the corresponding potential function.
$\therefore f(x,y) = \int F_x dx= \int 2xy dx = x^2y +g(y)\\
\therefore f(x,y) = \int F_y dy= \int x^2 dy = x^2y +h(x)$
from these two versions of $f(x,y)$ we get , $g(y) = h(x) = A (\text{constant})\\
\therefore f(x,y) = x^2y+A$
(a) $\textbf r_1 = x\textbf i + y\textbf j =t\textbf i + t^2\textbf j $, for $0\le t \le 1$
$\therefore x(t) = t, y(t) = t^2$ for (0,0) to (1,1) point
$\therefore \int \textbf F \cdot \textbf r_1 = f(1,1)-f(0,0) = 1+A-A = 1$
(b) $\textbf r_2 = x\textbf i + y\textbf j =t\textbf i + t^3\textbf j $, for $0\le t \le 1$
$\therefore x(t) = t, y(t) = t^3$ for (0,0) to (1,1) point
$\therefore \int \textbf F \cdot \textbf r_2 = f(1,1)-f(0,0) = 1+A-A = 1$
