Use Green's Theorem to evaluate $\oint_{C} \vec{F} \cdot d \vec{r}$, where $\vec{F}(x, y)=\langle\sin (x) \cos (y), x y+\cos (x) \sin (y)\rangle$ and is the boundary of the region between the graphs of $y=x$ and $y=\sqrt{x}$ oriented counterclockwise.
The integral obtained from from Green's Theorem is
\[
\iint_{D} \square d A
\]
where $D$ is the interior of the region.
This evaluates to

Step 1 ：First, we need to find the curl of the vector field \(\vec{F}(x, y)=\langle\sin (x) \cos (y), x y+\cos (x) \sin (y)\rangle\). The curl of a vector field in two dimensions is given by the formula \(\nabla \times \vec{F} = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\), where \(F_1\) and \(F_2\) are the components of the vector field. After finding the curl, we find that the curl of \(\vec{F}\) is \(y\).

Step 2 ：Next, we need to set up the double integral over the region D bounded by the curves \(y=x\) and \(y=\sqrt{x}\). The bounds for \(x\) are from 0 to 1, and the bounds for \(y\) are from \(\sqrt{x}\) to \(x\).

Step 3 ：Finally, we evaluate the double integral, which gives us the result -1/12. This is the value of the line integral \(\oint_{C} \vec{F} \cdot d \vec{r}\) over the given vector field and path, according to Green's Theorem.

Step 4 ：So, the final answer is \(\boxed{-\frac{1}{12}}\).