Let $C$ be the intersection curve between cylinder $x^{2}+y^{2}=4$ and the plane $z =y+3$ in $R^3$
Find a parameterization of $C$ which goes in a positive direction of rotation (counter-clockwise) relative to the $xy$ plane.
what I did: we know that $x^2+y^2=4 \rightarrow (x-0)^2+(y-0)^2=2^2$ which gives us that $r=2$.
So the parameterization would be $x=2\cos(t), y=2\sin(t)$ for $t\in[0, 2\pi]$
is this correct?
Your calculations for the $x$ and $y$ coordinates are correct since they represent the cylinder $x^2+y^2=2^2=4$. Since the cutting curve is the intersection of this cylinder and the plane $z=y+3$, the remaining step is to write $z = y + 3 = 2\sin(t)+3$ for $t \in [0,2\pi]$.