Let the vector field $\vec{F}$ be defined by $\vec{F}=y\mathbf{i}+x\mathbf{j}$ evaluate the path integral: $$\int \vec{F}\cdot d\vec{r}$$ Along that part of the circular path $x^2+y^2=1,z=0$ from $(1,0,0)$ to $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)$ via the shortest route.
I know that the answer should be $\frac{1}{2}$ from my lecturers solutions but I can't seem to get it
My solution attempt : $$\vec{F}\cdot d\vec{r}=\int_{(1,0,0)}^{ \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\right)}ydx+xdy$$ Let $x=\cos(\theta)$ $y=\sin(\theta)$ then we have that $\theta=\frac{\pi}{4}$,$\theta=0$ so the integral becomes $$[yx+xy]^\frac{\pi}{4}_0=\int_0^\frac{\pi}{4}2\sin(\theta)\cos(\theta)\mathrm{d}\theta=1\ne \frac{1}{2}$$ Where is my mistake? I can't seem to see it, any help would be appreciated.