Answer
$r(t)=\cos \theta i+\sqrt 3|\cos \theta| j+\sin \theta k$ ; $0 \leq \theta \leq 2 \pi$
Work Step by Step
Here, we have $x^2+y^2+4z^2=4; y \geq 0$ and $x^2+z^2=1$
$(x^2+z^2)+y+3z^2=4$
or, $y^2+3z^2=3$
or, $y^2=3-3z^2$
The parametric equations of a circle having radius $r$ are:
$x=r \cos t ; y =r \sin t$
Thus, we have $y=\sqrt {3-3z^2}$
Consider $z= \sin \theta$ and $y=\sqrt {3-3(\sin \theta)^2}=\sqrt 3|\cos \theta|$
and $x^2+z^2=1 \implies x= \cos \theta$
Hence, the parametric equations in vector form are:
$r(t)=\cos \theta i+\sqrt 3|\cos \theta| j+\sin \theta k$ ; $0 \leq \theta \leq 2 \pi$
