Cylinders From Constant-Radius Profiles

From a Parallel Profile to a Cylinder

A cylinder can be generated by rotating a straight line segment around a parallel axis. The segment stays at a fixed distance from the axis as it turns, so every point on it traces a circle of the same radius. As the segment rotates, those circles form the cylinder’s curved lateral surface.

Two measurements determine that lateral area: the radius of the circles and the length of the segment along the axis. It is important to distinguish the curved surface generated by the segment from the flat disks that may close the ends of a solid cylinder. A segment by itself generates the lateral surface, not the disks.

Definition: A constant-distance profile for an axis is a curve whose points remain at one fixed perpendicular distance \(r\) from that axis. In this tutorial, the profile is a straight segment parallel to the axis, with length \(L\). If \(r>0\) and \(L>0\), revolving the segment generates the lateral surface of a cylinder of radius \(r\) and axial length \(L\).

Distance and the Two Sides of an Axis

In a plane containing the axis, the points at a fixed positive distance \(r\) from the axis lie on two lines parallel to it, one on each side. For example, if the axis is \(y=0\), the points at distance \(r\) satisfy \(y=r\) or \(y=-r\). A connected profile segment at that distance lies on one of those lines.

Rotating either of those parallel segments around the axis produces the same cylinder. The two possible sides do not represent two separate lateral surfaces to add together: each segment, considered on its own, sweeps out the same set of points as the other.

The radius is always the perpendicular distance from the profile to the axis. For a horizontal axis \(y=c\) and a parallel profile \(y=d\), that distance is \(|d-c|\). For a vertical axis \(x=c\) and a parallel profile \(x=d\), it is \(|d-c|\) as well. When the lines are oblique, use their perpendicular separation, not simply the difference between their vertical coordinates.

Deriving the Lateral-Area Formula

To see exactly what surface the profile generates and why its area has the usual cylinder formula, choose coordinates with the axis as the \(z\)-axis. A profile segment at distance \(r\), extending a length \(L\) along the axis, can be written as the points \((r,0,t)\), where \(0\leq t\leq L\). Rotating each point through an angle \(\theta\), where \(0\leq\theta\leq2\pi\), gives the parameterization

\mathbf{X}(\theta,t) = (r\cos\theta,r\sin\theta,t).

The partial derivatives point in the directions of increasing angle and increasing axial position:

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

The magnitude of their cross product gives the area factor for this parameterization:

\left|\mathbf{X}_{\theta}\times\mathbf{X}_{t}\right| = \left|(r\cos\theta,r\sin\theta,0)\right| = r.

Integrating this area factor over one full rotation and the segment’s full axial length gives the surface area:

A = \int_0^L\int_0^{2\pi} r\,d\theta\,dt = \int_0^L 2\pi r\,dt = 2\pi rL.

Theorem: Revolving a straight segment of length \(L\), parallel to and at a positive perpendicular distance \(r\) from an axis, generates a cylindrical lateral surface of area \(A=2\pi rL\). The formula requires \(r\) to be the perpendicular radius and \(L\) to be the segment’s length parallel to the axis.

The derivation also explains the geometry: each small axial strip contributes the circumference \(2\pi r\) times its small length. Adding those strips over length \(L\) gives circumference times axial length. The parameterization covers the curved surface; it does not sweep out the end disks.

Worked Examples

Worked Example: A Horizontal Profile Revolved About the \(x\)-Axis

Revolve the segment \(y=3\), for \(1\leq x\leq6\), about the \(x\)-axis. Find the generated lateral area.

The axis is \(y=0\), so the perpendicular distance from the profile to the axis is

$$ r=|3-0|=3. $$

The segment is horizontal and parallel to the axis. Its length is

$$ L=6-1=5. $$

Therefore, its lateral area is

A=2\pi rL =2\pi(3)(5) =30\pi.

Check: the circumference of each circular cross-section is \(2\pi(3)=6\pi\), and the segment extends for \(5\) units along the axis. Thus circumference times length is \((6\pi)(5)=30\pi\), as calculated. This is only the lateral area. If flat disks are added to close the cylinder, each has area \(\pi(3)^2=9\pi\); together the two caps add \(18\pi\), making the closed cylinder’s total surface area \(48\pi\).

Worked Example: A Vertical Profile Revolved About a Shifted Axis

Revolve the segment \(x=5\), for \(-1\leq y\leq4\), about the vertical line \(x=2\). Find the lateral area.

The profile and axis are both vertical. Their horizontal separation is the perpendicular distance:

$$ r=|5-2|=3. $$

The profile’s vertical length, and hence its length along the axis, is

$$ L=4-(-1)=5. $$

Substitution into the cylinder formula gives

A=2\pi rL =2\pi(3)(5) =30\pi.

Check: each cross-section has circumference \(2\pi(3)=6\pi\), and the cylinder extends \(5\) units along the vertical axis, so \(6\pi\cdot5=30\pi\). The other parallel line at the same distance is \(x=-1\). A segment of the same length on that line would generate the same cylinder, not an additional lateral surface. The two end caps, if included to form a closed solid, would contribute \(2\pi(3)^2=18\pi\), which is not part of the area swept out by the given segment.

Worked Example: Parallel Oblique Lines

Revolve the segment of \(y=2x+7\) with \(0\leq x\leq3\) about the parallel line \(y=2x+1\). Find the lateral area. Because the lines are oblique, neither the vertical separation \(6\) nor the change in \(x\) alone is the required distance or length.

A direction vector for both lines is \((1,2)\), whose magnitude is

\sqrt{1^2+2^2}=\sqrt{5}.

The segment’s \(x\)-coordinate changes by \(3\), so its length along the axis direction is

L=3\sqrt{5}.

A unit normal to the lines is \(\frac{1}{\sqrt{5}}(-2,1)\). The difference between points on the profile and axis at the same \(x\)-coordinate is \((0,6)\). Its component in the unit-normal direction gives the perpendicular separation:

r = \left|(0,6)\mathbin{\cdot}\frac{1}{\sqrt{5}}(-2,1)\right| = \frac{6}{\sqrt{5}}.

Now calculate the lateral area:

A = 2\pi rL = 2\pi\left(\frac{6}{\sqrt{5}}\right)(3\sqrt{5}) = 36\pi.

Check using circumference times axial length: the circumference is \(2\pi r=\frac{12\pi}{\sqrt{5}}\). Multiplying by \(3\sqrt{5}\) gives \(\frac{12\pi}{\sqrt{5}}(3\sqrt{5})=36\pi\), agreeing with the result. Each cap, if included, would have area \(\pi r^2=\frac{36\pi}{5}\); the two caps together would add \(\frac{72\pi}{5}\) to the closed cylinder’s total surface area.

Lateral Area Is Not Total Surface Area

The surface generated by revolving the profile segment consists of the cylinder’s curved side. The segment’s endpoints trace circles, but tracing a circle is not the same as sweeping out the disk bounded by that circle. Consequently, the two disks are not automatically part of the generated surface.

For a closed cylinder with radius \(r\) and axial length \(L\), the curved lateral area is \(2\pi rL\). Each of its two end caps has area \(\pi r^2\), so the total surface area of that closed cylinder is

2\pi rL+2\pi r^2.

This total applies when the end caps are included. They may be generated by revolving additional profile segments that extend across the circular ends, but they are not generated by revolving the parallel segment alone.

A Common Mistake

Suppose the profile from the first example is revolved around the \(x\)-axis. A common incorrect calculation is to treat the endpoint circles as disks and write

2\pi(3)(5)+2\pi(3)^2 = 30\pi+18\pi = 48\pi.

The arithmetic is correct, but the quantity is wrong for the stated profile: \(48\pi\) includes two end caps that the segment does not generate. For the segment’s lateral surface alone, the correct calculation is

A=2\pi(3)(5)=30\pi.

Use \(48\pi\) only if the question asks for the total surface area of the closed cylinder, including both caps. A separate geometric pitfall is to use vertical separation as the radius for oblique lines. The radius is perpendicular distance; in the oblique example it is \(6/\sqrt{5}\), not \(6\).

Why the Constant-Radius Case Matters

In a general surface-of-revolution problem, the radius can vary along the profile, so both the radius and the length of each small curve segment must be accounted for. Here the profile is parallel to the axis, so its distance from the axis stays constant. The entire surface therefore has one fixed radius, and the calculation reduces to a cylinder’s circumference multiplied by its axial length.

Recognizing this special geometry helps prevent unnecessary integration and makes it easier to check a result. The radius must be a perpendicular distance, the length must be measured parallel to the axis, and the requested area must specify whether it includes end caps.

Check Your Understanding

Use the radius and axial length of each profile to find its generated lateral area. Unless a question explicitly asks for a closed cylinder, do not include end caps.

A segment \(y=-4\), \(0\leq x\leq7\), is revolved about the \(x\)-axis. What are the radius, axial length, and lateral area?
A segment \(x=1\), \(2\leq y\leq8\), is revolved about \(x=-2\). What is the lateral area?
Explain why a profile at \(y=r\) and a profile at \(y=-r\) generate the same cylinder when revolved about \(y=0\).
For the oblique lines \(y=3x+5\) and \(y=3x-1\), what is their perpendicular separation? Why is it not \(6\)?
What additional area must be added to the lateral area to obtain the total surface area of a closed cylinder?