Parametrization the curve of intersection of sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$. I've seen similar posts here but none of the answers helped me. I am trying to parametrize a curve of intersection of a (top half $z>0$) sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.
I tried $$x=\frac{1}{2}\cos t $$ $$y=\frac{1}{2}+\frac{1}{2}\sin t$$ $$z=\sqrt{5-\left(\frac{1}{2}\cos t\right)^2-\left(\frac{1}{2}+\frac{1}{2}\sin t\right)^2}=\sqrt{\frac{9}{2}-\frac{1}{2}\sin t}$$ for $t \in (0,2\pi)$,  but I don't think it is correct. Even if it was, is there a better (more simple) approach?
Note: I need to find the circulation of a field $F=(y+z,x-z,0)$ over this curve so I need a good parametrizatian so that I was able to integrate it.

A third alternative is using Stokes' theorem, parameterize the surface enclosed by the curve and integrate... And at first it seems like you are going to end up with a nicer integral
\begin{eqnarray}
&& \int_0^\pi d\phi \int_0^{\arcsin\left[\frac{1}{\sqrt{5}}\sin(\phi)\right]} d\theta \, \left[5 \sin(\theta)\right] (\nabla \times F)\cdot(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) =
\\
&& \int_0^\pi d\phi \int_0^{\arcsin\left[\frac{1}{\sqrt{5}}\sin\phi\right]} d\theta \, \, 5 \sqrt{2} \sin^2 \theta \sin \left( \phi + \frac{\pi}{4}\right) \\
&&
\int_0^\pi d\phi \, \frac{1}{2} \left(\cos \phi + \sin \phi \right) \left\{ 5 \arcsin \left[\frac{1}{\sqrt{5}}\sin\phi\right] - \sin\phi \sqrt{5 - \sin^2\phi}\right\}
\end{eqnarray}
but, at least at first sight, it does not seem specially simpler than the integral you get in the line integrals.

