Find a parametrization of the intersection curve between two surfaces in $\mathbb{R^3}$ $x^2+y^2+z^2=1$ and $x^2+y^2=x$. Find a parametrization of the intersection curve between two surfaces in $\mathbb{R}^3$
$$x^2+y^2+z^2=1$$ and $$x^2+y^2=x.$$
I know that $x^2+y^2+z^2=1$ is a sphere and that $x^2+y^2=x$ is a circular cylinder. Any help is greatly appreciated, thank you. 

A point of the intersection belongs to the cylinder. So $(x-1/2)^2+y^2=1/4$. Take for parameter the angle $\theta$ such that $x-1/2=1/2 \cos \theta$ and $y=1/2 \sin \theta$. This is just the angle of the cylindrical coordinates.
Then you have $x+z^2=1$ which implies $z^2=1/2-1/2 \cos \theta=\sin^2(\theta/2)$. You then get $z=\sin(\theta/2)$.
