Answer
$r(t)=\cos \theta i+\sqrt 3|\cos \theta| j+\sin \theta k$ ; $0 \leq \theta \leq 2 \pi$
Work Step by Step
Given: $x^2+y^2+4z^2=4; y \geq 0$ and $x^2+z^2=1$
Re-write as: $(x^2+z^2)+y+3z^2=4$
This implies that $y^2+3z^2=3$
This yields: $y^2=3-3z^2$
Now, we get $y=\sqrt {3-3z^2}$
We need to plug $z= \sin \theta$ and $y=\sqrt {3-3(\sin \theta)^2}=\sqrt 3|\cos \theta|$ in order to get the vector form.
Thus, we have $x^2+z^2=1 $ and $x= \cos \theta$
Thus, $r(t)=\cos (\theta) i+\sqrt 3|\cos \theta| j+\sin (\theta) k$ ; $0 \leq \theta \leq 2 \pi$
