Parameterize Intersection of Surfaces I need to parameterize the intersection of $$4x^2 + y^2 + z^2 = 9\tag{1}$$ and $$z=x^2+y^2\tag{2}$$. 
First, I'll solve (2) for $y^2$ and substitute the result into (1):
$$3x^2+z+z^2 = 9 \tag{3}$$
Next, I'll make the substitution $u=\sqrt{3}x$, such that we can complete the square in (3) by adding $1/4$ to each side and arrive at
$$\frac{u^2}{r^2} + \frac{(z+\frac{1}{2})^2}{r^2} = 1$$
where $r^2 = 9 + \frac{1}{4} = \frac{37}{4}$
Now I'll write a parameterization: 
$$u = r\cos \phi \implies x(\phi) = \frac{1}{\sqrt{3}}r\cos\phi$$
$$z(\phi) = r\sin \phi -\frac{1}{2}$$
$$y(\phi) = \pm \sqrt{z-x^2} = \pm\left(\sqrt{r\sin \phi - \frac{1}{2} - \left(\frac{1}{\sqrt{3}}r\cos\phi\right)^2}\right)$$
such that we have two branches:
$$\mathbf{r}(\phi)_1 = \big<x(\phi), y(\phi), z(\phi)\big>$$
$$\mathbf{r}(\phi)_2 = \big<x(\phi), -y(\phi), z(\phi)\big>$$
Is this correct?
 A: Yes your equation is correct. It was smart of you to do a substitution. 
If you [I] need a visual verification:

A: I think that taking cylindrical coordinates is rather natural. 
let us set $x=r \cos(t),y=r \sin(t)$, giving
$$\left\{\begin{array} .4 r^2 \cos^2 \theta +r^2 \sin^2 \theta+ z^2 & = & 9\\z & =& r^2 \end{array}\right. \ \iff \ \left\{\begin{array} . r^2 (1+ 3\cos^2 \theta) + r^4 & = & 9\\z & =& r^2 \end{array}\right.$$
Setting $R=r^2$, one has a quadratic equation:
$$R^2+R(1+ 3\cos^2 \theta)-9=0$$
whose solution is $R=\tfrac12(-1-3\cos^2 \theta + \sqrt{80 - 6 \cos^2\theta - 9 \cos^4 \theta})$
(we have not taken the other root, negative, which is impossible because $R=r^2 \geq 0$).
Thus the parametric representation of the intersection curve (of an ellipsoid and a paraboloid) in cylindrical coordinates is :
$$\left\{\begin{array}{rcl}r&=&\pm \sqrt{\tfrac12(-1-3\cos^2 \theta + \sqrt{80 - 6 \cos^2\theta - 9 \cos^4 \theta}})\\z & =& r^2 \end{array}\right.$$
The advantage of this representation is that it gives you a polar representation of the projection of the curve onto the horizontal plane.
A: It is not necessary to convert to polar coordinates, really. You have
$$z = x^2 + y^2 \tag{1}\label{1}$$
$$4 x^2 + y^2 + z^2 = 9 \tag{2}\label{2}$$
Equation $\eqref{1}$ can be written as
$$x^2 = z - y^2$$
and substituting it into equation $\eqref{2}$ yields
$$4 (z - y^2) + y^2 + z^2 = -3 y^2 + z^2 + 4 z = 9$$
which we can solve for $y$ by writing it as
$$3 y^2 = z^2 + 4 z - 9$$
nothing that
$$y^2 = \frac{1}{3} z^2 + \frac{4}{3} z - 3 \tag{3}\label{3}$$
Solving equation $\eqref{3}$ for $y$ we get
$$y = \pm \sqrt{\frac{1}{3} z^2 + \frac{4}{3} z - 3} \tag{4}\label{4}$$
The $\pm$ above means the intersection is symmetric with respect to the $y$ axis.
If we substitute equation $\eqref{3}$ into equation $\eqref{1}$, we get
$$z = x^2 + \frac{1}{3} z^2 + \frac{4}{3} z - 3$$
which we can trivially solve for $x^2$,
$$x^2 = - \frac{1}{3} z^2 - \frac{1}{3} z + 3$$
and therefore for $x$,
$$x = \pm \sqrt{-\frac{1}{3} z^2 - \frac{1}{3} z + 3} \tag{5}\label{5}$$
Here the $\pm$ means the intersection is also symmetric with respect to the $x$ axis.
By combining $\eqref{4}$ and $\eqref{5}$, we can write the intersection in Cartesian coordinates:
$$\begin{cases}
x = \pm \sqrt{-\frac{1}{3} z^2 - \frac{1}{3} z + 3} \\
y = \pm \sqrt{\frac{1}{3} z^2 + \frac{4}{3} z - 3}
\end{cases} \; \iff \; \begin{cases}
x = \pm \sqrt{\frac{1}{3}}\sqrt{ 9 - z (z + 1) }\\
y = \pm \sqrt{\frac{1}{3}}\sqrt{ z (z + 4) - 9 }
\end{cases}$$
It is a nice saddle-shaped curve. As JeanMarie already mentioned, it is much easier to visualize as a continuous curve in the cylindrical coordinates, though.
