Find the parametric equation of the surface $S$, where $S$ is the part of the paraboloid $z=x^2 + y^2 + 1$ bounded by the plane $z=2x+3$
My attempt
The OXY projection of $S$ is $x^2 + y^2 + 1 = 2x + 3$. Then, it is the disk
$$(x-1)^2 + y^2 \leq 3$$ Then, the first two components of a parametrization of $S$ would be
\begin{align*}
x(r,\theta) &= r\cos\theta + 1\\
y(r, \theta)&= r\sin\theta
\end{align*}
The answer gives $z(r, \theta) = r^2 + 1$. But why?
Thanks in advance!
Best Answer
If $x(r,\theta) = r \cos \theta$ and $y(r, \theta) = r \sin \theta$, then $$z(r, \theta) = x^2 + y^2 + 1 = (r \cos \theta)^2 + (r \sin \theta)^2 + 1 = r^2 + 1.$$ However, the limits of $r$ under this parametrization will depend on $\theta$ in a nontrivial way. To rectify this, note that the projection of the boundary in the $xy$-plane is the circle $(x-1)^2 + y^2 = 3$, hence we may write $$(r \cos \theta - 1)^2 + (r \sin \theta)^2 = 3,$$ and solving for $r$ yields its dependence on $\theta$: $$0 \le r \le \sqrt{\frac{5+\cos 2\theta}{2}} + \cos \theta.$$ Consequently, we can formulate a parametrization as follows: $$\begin{align*} x &= r \Bigl(\sqrt{\tfrac{1}{2}(5+\cos 2\theta)} + \cos \theta\Bigr)\cos \theta, \\ y &= r \Bigl(\sqrt{\tfrac{1}{2}(5+\cos 2\theta)} + \cos \theta\Bigr)\sin \theta, \\ z &= \biggl(r \Bigl(\sqrt{\tfrac{1}{2}(5+\cos 2\theta)} + \cos \theta\Bigr)\biggr)^{\!2} + 1,\end{align*} \qquad r \in [0,1], \; \theta \in [0,2\pi).$$
This removes the dependency of the radius parameter $r$ on the angle $\theta$ by rescaling it according to how "far out" you need to go from the origin at a particular angle. You can see this effect in the picture: the curves of constant $r$-values are more widely spaced in the region of the surface that is farther from the origin. It isn't necessary, but it does make the parametrization "nice" in the sense that the range of the parameters do not depend on each other.