Parameterising the intersection of a plane and paraboloid Suppose we have the paraboloid $z=x^2+y^2$ and the plane $z=y$. Their intersection produces a curve $C$, and certain surfaces bounded by it, for example the disc $S$ which directly fills the area of $C$ and the paraboloid $S'$ given by $z=x^2+y^2$ which extends from $C$ downwards and is bounded by $C$.
My question is on how to parameterise these objects. My initial instinct is to substitute one equation into the other giving the circle $(y-\frac{1}{2})^2+x^2=\frac{1}{4}$. Now I could go ahead and parameterise this circle using polars or even do integrals directly using $x$ and $y$. But this gives the projection of the intersection of the plane and paraboloid in the xy-plane. But I have not directly parameterised $C$. Since the $z=y$ plane makes a $45$ degree angle with the xy-plane, I think that we can introduce a factor of $\sqrt2$ into our parameterisation, but I would like to know if there is a way to directly parameterise $C$ (and also the surfaces $S$ and $S'$)?
Note my question is similar to this one except instead of being level, my disc is slanted (and I would also like to find out how to parameterise the 'bowl shaped' surface).
Many thanks for any help.
 A: The surfaces intersect when $y = x^2 + y^2$ so the curve of intersection is 
$$ C = \left \{ (x, y, z) \, | \, \left( y - \frac{1}{2} \right)^2 + x^2 = \frac{1}{4}, z = y \right \}. $$
Assume for a minute that you would have the equation $x^2 - \left( y - \frac{1}{2} \right)^2 +  = \frac{1}{4}$ in $\mathbb{R}^2$. This is the equation of a circle with radius $\frac{1}{2}$ centered at $\left( 0, \frac{1}{2} \right)$. To parametrize this circle, you can use polar coordinates centered at $\left( 0, \frac{1}{2} \right)$. Namely, let 
$$x = \frac{1}{2} \cos \theta, \,\, y = \frac{1}{2} + \frac{1}{2} \sin \theta = \frac{1}{2}( 1 + \sin \theta)$$
and then the circle is described by 
$$\left \{ \frac{1}{2}(\cos \theta, 1 + \sin \theta) \, | \, \theta \in [0,2\pi] \right \}.$$
Returning to your original problem, you can do the same and then use the equation $z = y$ to get the "slanted circle"
$$ C = \left \{ \frac{1}{2} \left( \cos \theta, 1 + \sin \theta, 1 + \sin \theta \right) \, | \, \theta \in [0, 2\pi] \right \}. $$
Note that this curve is actually an ellipse on the $z = y$ plane and not a circle (precisely because you intersect the paraboloid with a slanted plane). Only the projection of this ellipse to the $xy$ plane results in a circle.
A: Let,
$$y-\frac{1}{2}=\frac{1}{2}\sin (\theta)$$
$$y=\frac{1}{2}(\sin (\theta)+1)$$
$$x=\frac{1}{2}\cos (\theta)$$
That way $(y-\frac{1}{2})^2+x^2=\frac{1}{4}$.
With $\theta \in [0,2\pi]$. Because we also intersect $z=x^2+y^2$. Then,
$$z=x^2+y^2$$
$$=(\frac{1}{2}(\sin (\theta)+1))^2+(\frac{1}{2}\cos (\theta))^2$$
