Cross section of a sphere cut by cylinder This is not a calculus question regarding the area of cross section, but it is more of just a geometry question. This must have an easy answer, but I'm sort of confused now.
Suppose you cut out a sphere(say, $x^2+y^2+z^2=16$, a sphere centered at the origin of radius $4$) by a cylinder which has an axis parallel to the z-axis, but not centered at the origin(say, $(x-2)^2+y^2=1$ in the coordinate space). Then, think of the figure which has a boundary made by the cylinder and the sphere.(My English fails me,sorry.) Then, the figure could be explained as a cut of a cylinder which is not parallel to the $xy$-plane, so it must be an ellipse, yet the cut is on the sphere, so it must be a circle. Where does this contradiction come from?
Can anyone explain it to me? Thanks in advance.
 A: If you try to solve the system $x^2+y^2+z^2=16; (x-2)^2+y^2=1$ to find the line of intersection, you will get, subtracting, $4(x-1) = 15 - z^2.$ Note that dependence between $x$ and $z$ is not linear. Rather, taking $z$ as a parameter, we get that the $x$-coordinate draws a parabola while you are moving up or down along your line. In particular, it has a constant "bend". Note that $y$ also does not exhibit a linear dependence on $z$.
A: A cylinder of radius $\dfrac{R}{2}$ intersecting a sphere of radius $R$ looks like this.  The thick boundary is the intersection of the two.

A: Here are more visual options:
Front-left side , top and perspective views:

A: The question asked about removing $(x-2)^2+y^2\le1$ from $x^2+y^2+z^2\le16$. This gives the solid:

It is hard to be sure that the edge of the cut is not contained in a plane. Here is the surface of the piece removed and a red line added between two extreme points:

This illustrates that that edge of the cut does not lie in a plane.

Removing $(x-2)^2+y^2\le4$ from $x^2+y^2+z^2\le16$ gives the solid mentioned by Lexi Belle Fan:

It is a bit easier to see that the edge of the cut does not lie in a plane. However, just to be sure that the edge of the cut is not contained in a plane, here is the surface of the piece removed and a red line added between two extreme points:

This illustrates that that edge of the cut does not lie in a plane.

More simply, when a plane cuts a sphere, the intersection is a circle, so if the edge of the cut lies in a plane then it is a circle. The cut is made by a circular cylinder, so if the edge is also in a plane, the cut will be the intersection of the cylinder and that plane. This is a circle only when that plane is perpendicular to the axis of the cylinder. This is not the case when the axis of the cylinder does not pass through the center of the sphere.
Here is a parametrization of the edge of the upper cut:
$$
\left(2+4\cos(u),4\sin(u),\sqrt{12-r^2-4r\cos(u)}\right)
$$
The lower cut is
$$
\left(2+4\cos(u),4\sin(u),-\sqrt{12-r^2-4r\cos(u)}\right)
$$
