Cylindrical Surface Patches From Axes And Curves

Introduction

A cylinder is determined not only by its cross-section but also by the line along which that cross-section is carried. A parametrization makes both features explicit: one parameter moves around a circle or ellipse, and another moves along the axis.

With that description, the tangent vectors and a normal vector can be computed directly. The same parameters also make it possible to identify which curves bound a particular patch and which apparent edges are merely artifacts of the chosen parameter range.

Describing an Axis and a Cross-Section

Let $\mathbf{a}$ be a unit vector in the direction of the cylinder’s axis, and let $\mathbf{p}$ be a point on that axis. The axis line is then $\mathbf{p}+s\mathbf{a}$, where $s$ is a real number. Choose unit vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ that are perpendicular to $\mathbf{a}$ and to each other. We choose their order so that $\mathbf{e}_1\times\mathbf{e}_2=\mathbf{a}$.

The vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ give perpendicular directions across the cylinder. For a circular cylinder of radius $R>0$, the cross-section at a fixed axial value is a circle in those directions. For an elliptic cylinder, the cross-section is an ellipse with positive semiaxis lengths $A$ and $B$ along $\mathbf{e}_1$ and $\mathbf{e}_2$, respectively.

Definition: A lateral patch of a circular cylinder about the axis $\mathbf{p}+s\mathbf{a}$ has parametrization $$ \mathbf{r}(\theta,z)=\mathbf{p}+z\mathbf{a} +R\cos\theta\,\mathbf{e}_1+R\sin\theta\,\mathbf{e}_2. $$ A lateral patch of an elliptic cylinder with the same axis and transverse directions has parametrization $$ \mathbf{r}(\theta,z)=\mathbf{p}+z\mathbf{a} +A\cos\theta\,\mathbf{e}_1+B\sin\theta\,\mathbf{e}_2. $$ Here $\theta$ is the angular parameter and $z$ is the axial parameter. The parameter domain specifies which portion of the cylinder is included.

For a full turn, $\theta$ ranges over an interval of length $2\pi$, such as $[0,2\pi]$. A shorter interval selects only part of the lateral surface. The axial parameter may range over a fixed interval, or its lower and upper limits may depend on $\theta$.

Tangent Vectors and Normals

The tangent vectors are the partial derivatives with respect to the two parameters. For the elliptic parametrization, differentiating with respect to $\theta$ moves around the cross-section, while differentiating with respect to $z$ moves parallel to the axis:

\mathbf{r}_{\theta} =-A\sin\theta\,\mathbf{e}_1+B\cos\theta\,\mathbf{e}_2, \qquad \mathbf{r}_{z}=\mathbf{a}.

The cross product depends on the order of the tangent vectors. With the chosen right-handed basis, $\mathbf{r}_{\theta}\times\mathbf{r}_{z}$ points away from the axis. Indeed, $\mathbf{e}_1\times\mathbf{a}=-\mathbf{e}_2$ and $\mathbf{e}_2\times\mathbf{a}=\mathbf{e}_1$, so

\mathbf{r}_{\theta}\times\mathbf{r}_{z} =B\cos\theta\,\mathbf{e}_1+A\sin\theta\,\mathbf{e}_2.

This is a normal vector, not necessarily a unit vector. For the circular case, take $A=B=R$; the same formula becomes $R\cos\theta\,\mathbf{e}_1+R\sin\theta\,\mathbf{e}_2$, which points radially outward.

Regularity condition: If $A>0$, $B>0$, and $\mathbf{a},\mathbf{e}_1,\mathbf{e}_2$ are mutually perpendicular unit vectors, then $\mathbf{r}_{\theta}\times\mathbf{r}_{z}\ne\mathbf{0}$ for every $\theta$. Thus the parametrization has two independent tangent directions throughout the patch. Reversing the cross-product order reverses the normal: $\mathbf{r}_{z}\times\mathbf{r}_{\theta}=-(\mathbf{r}_{\theta}\times\mathbf{r}_{z})$.

Why the Cross Product Gives an Outward Normal

A normal vector must be perpendicular to both tangent directions. The cross product has this property by definition. Its direction can also be checked geometrically in the transverse plane: at angle $\theta$, the outward direction from the axis is proportional to $\cos\theta\,\mathbf{e}_1+\sin\theta\,\mathbf{e}_2$. The elliptic cross product is not generally a multiple of that radial direction, but it points to the outward side of the ellipse.

To verify the direction without relying on a sketch, consider the implicit equation of the cross-sectional ellipse in transverse coordinates $x,y$: $x^2/A^2+y^2/B^2=1$. Its gradient in that plane is $2x\mathbf{e}_1/A^2+2y\mathbf{e}_2/B^2$. At $x=A\cos\theta$, $y=B\sin\theta$, this is $2\cos\theta\,\mathbf{e}_1/A+2\sin\theta\,\mathbf{e}_2/B$. Multiplying by the positive number $AB/2$ gives $B\cos\theta\,\mathbf{e}_1+A\sin\theta\,\mathbf{e}_2$, exactly the cross product above. The gradient points toward increasing values of the ellipse’s defining expression, hence outward.

For completeness, the tangent and normal relationships can also be checked directly. Since the basis vectors are perpendicular,

\begin{aligned} \mathbf{r}_{\theta}\cdot\mathbf{r}_{z}&=0,\\ (\mathbf{r}_{\theta}\times\mathbf{r}_{z})\cdot\mathbf{r}_{z} &=0,\\ (\mathbf{r}_{\theta}\times\mathbf{r}_{z})\cdot\mathbf{r}_{\theta} &=(B\cos\theta)(-A\sin\theta)+(A\sin\theta)(B\cos\theta)=0. \end{aligned}

The cross product is nonzero because its squared length is $B^2\cos^2\theta+A^2\sin^2\theta$, which is positive when both semiaxis lengths are positive. These facts establish that the two parameter directions are independent and that the stated cross product is a valid normal everywhere on the patch.

Boundary Curves from the Parameter Domain

The boundary of a patch is traced by the boundary of its parameter domain. If $\theta$ ranges from $\alpha$ to $\beta$ and $z$ ranges from $z_0$ to $z_1$, the four parameter edges give four surface curves. Fixing $z$ gives an elliptical arc around the axis; fixing $\theta$ gives a straight generator parallel to the axis.

\begin{aligned} z=z_0 &: \quad \mathbf{r}(\theta,z_0),\\ z=z_1 &: \quad \mathbf{r}(\theta,z_1),\\ \theta=\alpha &: \quad \mathbf{r}(\alpha,z),\\ \theta=\beta &: \quad \mathbf{r}(\beta,z). \end{aligned}

If $\theta$ covers a full turn, the edges at $\theta=0$ and $\theta=2\pi$ describe the same generator. That matching curve is a seam in the parametrization, not a boundary of the complete lateral cylinder. In contrast, for a restricted angular interval, its two endpoint generators are genuine boundary curves of the patch.

Axial limits can also vary with angle. If $g_0(\theta)\leq g_1(\theta)$, the parameter domain may be described by $\alpha\leq\theta\leq\beta$ and $g_0(\theta)\leq z\leq g_1(\theta)$. Its two curved boundary traces are $\mathbf{r}(\theta,g_0(\theta))$ and $\mathbf{r}(\theta,g_1(\theta))$, together with the endpoint curves at $\theta=\alpha$ and $\theta=\beta$. The curves need not be planar merely because the cylinder is circular or elliptic.

Worked Examples

Worked Example: A Quarter of a Circular Cylinder

Consider the circular cylinder of radius $3$ about the $z$-axis, restricted to $0\leq\theta\leq\pi/2$ and $1\leq z\leq4$. Its parametrization is

\mathbf{r}(\theta,z)=(3\cos\theta,3\sin\theta,z).

Differentiate each coordinate:

\mathbf{r}_{\theta}=(-3\sin\theta,3\cos\theta,0), \qquad \mathbf{r}_{z}=(0,0,1).

The outward normal vector from this parameter order is

\mathbf{r}_{\theta}\times\mathbf{r}_{z} =(3\cos\theta,3\sin\theta,0).

For example, at $\theta=\pi/2$, the point on the cylinder is $(0,3,z)$, and the normal is $(0,3,0)$, directed away from the axis. The four boundary curves are the bottom arc $(3\cos\theta,3\sin\theta,1)$, the top arc $(3\cos\theta,3\sin\theta,4)$, and the two generators $(3,0,z)$ and $(0,3,z)$, with the stated parameter limits. Substitution into $\mathbf{r}$ recovers each curve, and the cross product is perpendicular to both displayed tangent vectors.

Worked Example: A Circular Cylinder About a Tilted Axis

Let the axis pass through $\mathbf{p}=(1,0,-1)$ in the direction $\mathbf{a}=(1,2,2)/3$. Choose $\mathbf{e}_1=(2,-1,0)/\sqrt{5}$. It has unit length and is perpendicular to $\mathbf{a}$, since $(2-2+0)/(3\sqrt{5})=0$. Set $\mathbf{e}_2=\mathbf{a}\times\mathbf{e}_1$, which gives

\mathbf{e}_2= \left(\frac{2}{3\sqrt{5}},\frac{4}{3\sqrt{5}},-\frac{\sqrt{5}}{3}\right).

Its squared length is $(4+16+25)/45=1$; it is perpendicular to both $\mathbf{a}$ and $\mathbf{e}_1$, and the definition gives $\mathbf{e}_1\times\mathbf{e}_2=\mathbf{a}$. For a cylinder of radius $2$, the parametrization is

\mathbf{r}(\theta,z)=\mathbf{p}+z\mathbf{a} +2\cos\theta\,\mathbf{e}_1+2\sin\theta\,\mathbf{e}_2, \qquad 0\leq\theta\leq\pi,\quad 0\leq z\leq2.

The tangent vectors and outward normal are

\mathbf{r}_{\theta}=-2\sin\theta\,\mathbf{e}_1+2\cos\theta\,\mathbf{e}_2, \qquad \mathbf{r}_{z}=\mathbf{a}, \qquad \mathbf{r}_{\theta}\times\mathbf{r}_{z} =2\cos\theta\,\mathbf{e}_1+2\sin\theta\,\mathbf{e}_2.

At $\theta=0,z=0$, the point is $\mathbf{p}+2\mathbf{e}_1=(1+4/\sqrt{5},-2/\sqrt{5},-1)$; the normal is $2\mathbf{e}_1=(4/\sqrt{5},-2/\sqrt{5},0)$. Their displacement from the axis point $\mathbf{p}$ is exactly $2\mathbf{e}_1$, confirming the radius and outward direction. The boundary consists of the two half-ellipse arcs $\mathbf{r}(\theta,0)$ and $\mathbf{r}(\theta,2)$, and the generators $\mathbf{r}(0,z)$ and $\mathbf{r}(\pi,z)$. Each is obtained by fixing the corresponding parameter in the displayed formula.

Worked Example: An Elliptic Cylinder with Variable Axial Limits

Take an elliptic cylinder about the $z$-axis with semiaxes $A=2$ and $B=1$. Restrict the patch to $0\leq\theta\leq\pi$, with $\cos\theta\leq z\leq2+\sin\theta$. These limits are in the correct order because $2+\sin\theta-\cos\theta\geq2-\sqrt{2}>0$. The parametrization and tangent data are

\begin{aligned} \mathbf{r}(\theta,z)&=(2\cos\theta,\sin\theta,z),\\ \mathbf{r}_{\theta}&=(-2\sin\theta,\cos\theta,0),\\ \mathbf{r}_{z}&=(0,0,1),\\ \mathbf{r}_{\theta}\times\mathbf{r}_{z}&=(\cos\theta,2\sin\theta,0). \end{aligned}

The lower and upper boundary curves are, respectively,

\mathbf{c}_0(\theta)=(2\cos\theta,\sin\theta,\cos\theta), \qquad \mathbf{c}_1(\theta)=(2\cos\theta,\sin\theta,2+\sin\theta).

For a direct check, their derivatives are $\mathbf{c}_0'(\theta)=(-2\sin\theta,\cos\theta,-\sin\theta)$ and $\mathbf{c}_1'(\theta)=(-2\sin\theta,\cos\theta,\cos\theta)$, which include the change in the axial limit as well as the angular motion. At the endpoints, the lower curve runs from $(2,0,1)$ to $(-2,0,-1)$; the upper curve runs from $(2,0,2)$ to $(-2,0,2)$. The other two boundary curves are the generators at $\theta=0$, where $1\leq z\leq2$, and at $\theta=\pi$, where $-1\leq z\leq2$. Substituting these parameter values into $\mathbf{r}$ gives the stated endpoints and traces. At $\theta=\pi/2$, the outward normal is $(0,2,0)$, pointing outward from the ellipse at $(0,1)$.

A Common Orientation Mistake

For the standard circular parametrization $\mathbf{r}(\theta,z)=(R\cos\theta,R\sin\theta,z)$, a common error is to reverse the order of the tangent vectors but keep calling the result outward. The correct calculation in the order $\mathbf{r}_{\theta}\times\mathbf{r}_{z}$ is

\begin{aligned} \mathbf{r}_{\theta}&=(-R\sin\theta,R\cos\theta,0),\\ \mathbf{r}_{z}&=(0,0,1),\\ \mathbf{r}_{\theta}\times\mathbf{r}_{z} &=(R\cos\theta,R\sin\theta,0). \end{aligned}

But reversing the order gives the negative:

\mathbf{r}_{z}\times\mathbf{r}_{\theta} =(-R\cos\theta,-R\sin\theta,0).

For instance, at $\theta=0$ this reversed vector is $(-R,0,0)$, which points toward the axis from the surface point $(R,0,z)$. The correction is to retain the chosen order when computing the normal, or explicitly negate the cross product if the opposite order is used. The two vectors are both valid normals; only one has the outward orientation.

A related boundary error is to count the seam at $\theta=0$ and $\theta=2\pi$ as two distinct physical edges of a full cylinder. Those parameter values trace the same generator. A full lateral cylinder has boundary at its axial ends, not along that seam.

Why the Parameter Description Helps

The axis and cross-section description separates the geometry into two simple motions. Changing $\theta$ moves around the cross-section; changing $z$ moves along the axis. This makes the tangent directions, cross-product normal, and boundary traces systematic to compute, even when the axis is tilted or the axial limits vary.

When setting up a patch, first identify its axis and choose a right-handed perpendicular basis. Then specify the cross-section dimensions and parameter domain. Finally, differentiate in both parameters, take the cross product in the desired order, and trace each actual edge of the domain. Keeping the parameter domain explicit prevents confusion between a true boundary curve and a seam introduced by the parametrization.

Check Your Understanding

Use the axis, cross-section, and parameter-domain descriptions to answer the following questions.

For an elliptic cylinder, what are $\mathbf{r}_{\theta}$, $\mathbf{r}_{z}$, and $\mathbf{r}_{\theta}\times\mathbf{r}_{z}$ in terms of $A$, $B$, $\mathbf{e}_1$, $\mathbf{e}_2$, and $\mathbf{a}$?
Why must both semiaxis lengths be positive for the stated cross product to be nonzero at every angle?
For a patch with $\alpha\leq\theta\leq\beta$ and fixed axial limits, which parameter edges give elliptical arcs and which give generators?
Why is the angular seam of a full-turn parametrization not a physical boundary of the lateral cylinder?
How does reversing the order of the tangent vectors affect the normal’s direction?