Answer
$\int_{0}^{\pi} \mathbf{F} \cdot d\mathbf{r} = 0$
Work Step by Step
\[ \begin{aligned} \int_{0}^{\pi} \mathbf{F} \cdot d\mathbf{r} &= \int_{0}^{\pi} (x^2\mathbf{i} + xy\mathbf{j}) \cdot (-2\sin{t}\mathbf{i} + 2\cos{t}\mathbf{j}) \, dt \\ &= \int_{0}^{\pi} (-2x^2\sin{t} + 2xy\cos{t}) \, dt \end{aligned} \] Here, \(\mathbf{r} = 2\cos{t}\mathbf{i} + 2\sin{t}\mathbf{j}\), \(x = 2\cos{t}\), and \(y = 2\sin{t}\). Continuing the calculation: \[ \begin{aligned} &= \int_{0}^{\pi} (-24\cos^2{t}\sin{t} + 2 \cdot 4\cos{t}\sin{t}\cos{t}) \, dt \\ &= \int_{0}^{\pi} (-8\cos^2{t}\sin{t} + 8\cos^2{t}\sin{t}) \, dt \\ &= \int_{0}^{\pi} 0 \, dt \\ &= 0 \end{aligned} \] Result: $\int_{0}^{\pi} \mathbf{F} \cdot d\mathbf{r} = 0$
