Does an Implicit equation for an infinite Cylinder exist ? i.e f(x,y,z)=0

Question

I wanted to know if the surface for any arbitrary cylinder(infinte or restricted does not matter) can be expressed with an implicit equation, like that for a sphere: $$(X-x)^2 + (Y-y)^2 + (Z-z)^2 = R^2$$ where $(x,y,z)$ are the co-ordinates of the center of the sphere. I know the equation for general quadric surfaces are of the form $$ aX^2+bY^2+cZ^2 +dXY+eYZ+fXZ+gX+hY+iZ+j=0$$

What I want to know is how many of these terms are zero for any arbitrary cylinder or equivalently if the equation has a special form for a cylinder like it does for a plane and for the sphere?

If I assume that a set of points belong to a cylindrical surface, what is the minimum number of points that I need to determine its equation?

I would be happy to explain further if something in my question is unclear.

Answer 1

If the axis of the cylinder is the line through $\vec{x}_0 = (x_0,y_0,z_0)$ and in the direction of $\vec{v} = (a,b,c)$ with radius $R$, assuming $|\vec{v}| = 1$, we can write the equation of the cylinder as the set of all points $\vec{x} = (x,y,z)$ satisfying:

$$|\vec{x} - \vec{x}_0|^{2} = R^{2} + [(\vec{x} - \vec{x}_0)\bullet \vec{v}]^{2}$$

Answer 2

$$(x-h)^2 + (y-k)^2 - r^2 = 0$$ describes such a cylinder whose axis is the line $(x,y,z) = (h,k,z)$ and has radius $r$. So for example, $$x^2 + y^2 - 1 = 0$$ describes an infinite cylinder in $\mathbb R^3$ with axis coinciding with the $z$-axis, and with radius $1$. Now you might say, "this is just the equation of the unit circle in the plane," and you are right, but it is also the same equation for a cylinder in Euclidean 3-space. The fact that $z$ does not show up in the equation simply means that the value of $z$ is irrelevant. This is precisely analogous to the situation in the plane, where I can define a vertical line with an implicit equation such as $$x = 2.$$ This still describes a set of points in $\mathbb R^2$ if we regard this equation as representing a line in the plane.

Now, if you wish to have a less trivial example, it is not hard to think how this could work, much like we can write a general line $y = mx + b$ in the plane, we can generalize a cylinder in $\mathbb R^3$, and all three coordinates $(x,y,z)$ show up in the implicit equation. For example, $$(x-y)^2 + (y-z)^2 + (z-x)^2 = 3$$ describes a cylinder in $\mathbb R^3$ whose axis is the line satisfying $x = y = z$ and has radius $1$. The proof of this fact is left to the reader as an exercise.

In general, the equation of a cylinder with axis joining $(0,0,0)$ to some arbitrary point $(x_0, y_0, z_0)$ such that $x_0^2 + y_0^2 + z_0^2 = 1$ and has radius $r$, is given by $$(y_0 x - x_0 y)^2 + (z_0 y - y_0 z)^2 + (x_0 z - z_0 x)^2 = r^2.$$

Answer 3

From the equation in form $$ax^2+by^2+cz^2 +2dxy+2eyz+2fxz+2gx+2hy+2iz+j=0,$$ you can know if the quadric can be a cylinder.

For this consider the matrix of the quadratic part of the equation: $$Q=\begin{bmatrix} a&d&f\\ d&b&e \\f&e&c \end{bmatrix}$$ If $Q$ has rank $2$ and if its non-zero eigenvalues have the same sign, i.e. if the associated quadratic form has signature $(2,0)$ or $(0,2)$, then you know you have either an elliptic paraboloid or an elliptic cylinder.

The difference between both can be seen only once you've obtained a reduced form of the equation by a change of orthogonal basis, which will be:

• $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=Z\quad$ (elliptic paraboloid),
• $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1\quad$ (elliptic cylinder).