Curves of Intersection 2 curves? Draw the surfaces defined by $x^2 + z^2 = 4$ and $y^2 + z^2 = 1$ in $\mathbb{R}^3$. Draw their curve(s) of intersection.
That is the problem I am trying to do right now. I have it so far that $x^2 + z^2 = 4$ is a circle in the $xz$ plane with a radius $2$. The next one I graphed is a circle in the $yz$ plane with a radius $1$. The 2nd circle is not big enough to have a radius that is big enough to intersect the first circle with radius two? Am I missing something?.
 A: Your error in interpreting the equations is that you forgot about the third coordinate.
The choice of value of $y$ is irrelevant to the question whether the equation $x^2+z^2=4$ holds or not. Meaning that you get a circle for each choice of $y$. Together those circles form a cylinder – a tube of radius two around the $y$-axis.
Similarly, the equation $y^2+z^2=1$ describes a cylindrical tube of radius one around the $x$-axis. This latter surface has a parametrization $y=\cos t, z=\sin t$ (from the unit circle) with $t\in[0,2\pi]$ leaving $x$ to have a value of our choice. Plugging these into the former equation gives
$$
x^2+\sin^2t=4
$$
from which we can solve $x=\pm\sqrt{4-\sin^2t}$.
Below you will see a picture of the two cylinders as well as thick black curves showing the intersection (the choice of sign of $x$ gives two non-intersecting curves).

A: The given curves are two cylinder and their intersection can be described in parametric form as follows

*

*$z=\cos t$

*$y=\sin t$

*$x=\pm \sqrt{4-\cos^2 t}$
with $t\in[0,2\pi)$.
Refer also to the related

*

*Is $z=x^2$ a cylinder?
