semi ellipsoid and cylinder parametrize the curve the curve of intersection of the semi ellipsoid $x^2 + y^2 +4z^2 = 4, y &gt;=0 $ (i.e. the half of the ellipsoid with positive y coordinates), and the cylinder $ x^2 + z^2 =1 $.
Consider the polar angle in the xz-plane with the positive z axis (see figure). Use this coordinate as the parameter to parametrize the curve with the counterclockwise direction looking into the y axis. Sketch the curve (hint: simplify the expression for y(t) so that it is given by only one
term).

figure :http://i.imgur.com/m1uENAg.jpg

How do I solve this? Solution if possible.
Plug $z=\cos t, x=\sin t$ into the first equation:
$$\sin^2t+y^2+4\cos^2t=4$$
We get $y^2=3-3\cos^2t=3\sin^2t$, since $y&gt;0$, $y=\sqrt{3}|\sin t|$.
The projection of the curve C of intersection onto the $xz$ -plane is the circle
$x^2+z^2 = 1, y =0$, so we can write $x=\cos t, z=\sin t, 0\le t \le 2\pi$.

$C$ also lies on the surface $x^2+y^2+4z^2 =4$ and since $y\ge 0$ we can write

$$y=\sqrt{4-x^2-4z^2}=\sqrt{4-\cos^2t-4\sin^2t}=\sqrt(4-\cos^2t-4(1-\cos^2t)=\sqrt(3\cos^2t)$$

Hence the parametric equation
$x=cost y=\sqrt(3)|\cos t|,z=\sin t,0\le t\le 2\pi$ and the corresponding vector
$$r(t)=\cos t i +\sqrt(3)|\cos t|j +\sin t k,0\le t\le 2\pi$$