The long and winding road

Notice that $\nabla(x^2 \sin y)=\vec{F}$ , so $\vec{F}$ is conservative, i.e., the integral $\displaystyle \int\limits_{C} \vec{F} \cdot \mathrm{d}\vec{r}$ is independent of the path, we only need the first and the last point of the path $C$ . So, let's take the line segment that goes from $(0,0)$ to $\left(\pi,\frac{\pi}{2}\right)$ . Then the integral is: $\begin{aligned} \int\limits_C \vec{F} \cdot \mathrm{d} \vec{r} = x^2\sin y \Bigr|_{(0,0)}^{\left(\pi,\frac{\pi}{2}\right)} = \pi^2 \sin \dfrac{\pi}{2} - 0^2\sin 0=\boxed{\pi^2} \end{aligned}$