How a Plane Curve Extends into a Cylinder
A cylinder in three-dimensional space can be understood by taking a curve in a coordinate plane and extending it in a direction parallel to one of the coordinate axes. For circular and elliptic cylinders, the curve being extended is a circle or an ellipse.
The equation reveals this structure through a missing variable. If an equation involves \(x\) and \(y\) but not \(z\), then its solutions have the same allowed \(x\)- and \(y\)-coordinates at every value of \(z\). The resulting surface extends parallel to the \(z\)-axis. The earlier material on why missing variables produce cylinders explains this general pattern; here we identify the circular and elliptic cases and use their traces to sketch them.
Standard Equations and Their Axes
A circle centered at \((h,k)\) in the \(xy\)-plane has equation \((x-h)^2+(y-k)^2=r^2\), where \(r>0\) is its radius. If the same equation is used in three-dimensional space and does not involve \(z\), it describes a circular cylinder whose axis is parallel to the \(z\)-axis.
An ellipse centered at \((h,k)\) in the \(xy\)-plane has equation \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), where \(a>0\) and \(b>0\) are the semiaxis lengths. If this equation does not involve \(z\), it describes an elliptic cylinder with axis parallel to the \(z\)-axis. When \(a=b\), the ellipse is a circle, so the cylinder is circular.
The same forms apply when a different variable is absent. If \(x\) is absent, the cross-section curve lies in the \(yz\)-plane and the cylinder extends parallel to the \(x\)-axis. If \(y\) is absent, it extends parallel to the \(y\)-axis. In every case, the two variables present in the equation describe the cross-section perpendicular to the axis.
| Variable absent | Direction of extension | Cross-section lies in |
|---|---|---|
| \(z\) | Parallel to the \(z\)-axis | An \(xy\)-plane |
| \(x\) | Parallel to the \(x\)-axis | A \(yz\)-plane |
| \(y\) | Parallel to the \(y\)-axis | An \(xz\)-plane |
For an elliptic cylinder, \(a\) is the semiaxis length in the \(x\)-direction and \(b\) is the semiaxis length in the \(y\)-direction when the equation is written in the \(xy\)-variables. These are distances from the ellipse’s center to its endpoints, not the ellipse’s full widths. If \(a>b\), the ellipse is longer in the \(x\)-direction; if \(b>a\), it is longer in the \(y\)-direction.
Why the Cross-Sections Repeat
Consider an equation that does not involve \(z\), written as \(F(x,y)=1\). Fix any number \(c\) and intersect the surface with the horizontal plane \(z=c\). The plane condition sets \(z\) equal to \(c\), but the equation \(F(x,y)=1\) is unchanged. Therefore every such horizontal cross-section is the same curve in its own plane.
More directly, if \((x_0,y_0,z_0)\) lies on the surface, then \(F(x_0,y_0)=1\). For every real number \(t\), the point \((x_0,y_0,z_0+t)\) also satisfies \(F(x_0,y_0)=1\). Thus the surface contains the entire line through that point parallel to the \(z\)-axis. The cross-section curve is extended along these parallel lines, which explains the cylinder shape and its axis direction.
This also explains why a coordinate-plane trace alone does not show the whole surface. A trace is only the intersection with one plane. For a useful sketch, identify the cross-section perpendicular to the axis, then use the traces in coordinate planes to see how that cross-section sits relative to the axes.
Worked Example: A Circular Cylinder
Worked Example: A Circular Cylinder
Identify and sketch the surface \(x^2+y^2=9\). Find its axis, its cross-sections in planes \(z=c\), and its traces in the coordinate planes.
Step 1: Identify the type and axis. The equation does not involve \(z\), so the surface extends parallel to the \(z\)-axis. The equation can be written as \((x-0)^2+(y-0)^2=3^2\). Its cross-section is a circle centered at \((0,0)\) with radius \(3\), so the surface is a circular cylinder. Its axis is the \(z\)-axis, described by \(x=0,\ y=0\).
Step 2: Find a cross-section perpendicular to the axis. In any plane \(z=c\), the equation remains \(x^2+y^2=9\). Thus every horizontal cross-section is a circle of radius \(3\), centered at \((0,0,c)\) in that plane.
Step 3: Find the coordinate-plane traces. In the plane \(z=0\), the trace is the circle \(x^2+y^2=9\). In the plane \(x=0\), the equation becomes \(y^2=9\), so \(y=3\) or \(y=-3\), with \(z\) unrestricted. The trace consists of two lines parallel to the \(z\)-axis. In the plane \(y=0\), the equation becomes \(x^2=9\), so \(x=3\) or \(x=-3\), again with \(z\) unrestricted. This trace is also two lines parallel to the \(z\)-axis.
Sketch and check. Draw the circle of radius \(3\) in the \(xy\)-plane and extend it vertically, parallel to the \(z\)-axis. The axis passes through the circle’s center. A point on the circle, such as \((3,0)\), gives the cylinder line \((3,0,z)\); indeed, \(3^2+0^2=9\) for every \(z\). This confirms that the cylinder extends in the stated direction.
Worked Example: An Elliptic Cylinder
Worked Example: An Elliptic Cylinder
Analyze the surface \(\frac{x^2}{4}+\frac{y^2}{9}=1\). Determine its type, axis, semiaxis lengths, and coordinate-plane traces.
Step 1: Read the equation. The equation is independent of \(z\), so the surface extends parallel to the \(z\)-axis. Its cross-section in the \(xy\)-plane is an ellipse centered at the origin. The denominator \(4\) is \(2^2\), and \(9\) is \(3^2\). Therefore the semiaxis length in the \(x\)-direction is \(2\), and the semiaxis length in the \(y\)-direction is \(3\). Since these are unequal, the cylinder is elliptic rather than circular.
Step 2: Describe the axis and repeated cross-sections. The axis is the \(z\)-axis, given by \(x=0,\ y=0\). In every plane \(z=c\), the cross-section is the same ellipse, \(\frac{x^2}{4}+\frac{y^2}{9}=1\), centered at \((0,0,c)\).
Step 3: Compute the traces. In the plane \(z=0\), the trace is the ellipse \(\frac{x^2}{4}+\frac{y^2}{9}=1\). In the plane \(x=0\), substitute \(x=0\): \(\frac{y^2}{9}=1\), so \(y^2=9\) and \(y=\pm3\). Since \(z\) is unrestricted, this trace is two lines parallel to the \(z\)-axis. In the plane \(y=0\), substitute \(y=0\): \(\frac{x^2}{4}=1\), so \(x^2=4\) and \(x=\pm2\). These are two more lines parallel to the \(z\)-axis.
Sketch and check. Draw an ellipse centered at the origin in the \(xy\)-plane, longer in the \(y\)-direction, with endpoints \((\pm2,0)\) and \((0,\pm3)\). Extend the ellipse parallel to the \(z\)-axis. For example, \((0,3,5)\) lies on the surface because \(\frac{0^2}{4}+\frac{3^2}{9}=0+1=1\). Its \(z\)-coordinate can vary without changing the equation.
Worked Example: A Shifted Cylinder with Axis Parallel to the \(x\)-Axis
Worked Example: A Shifted Cylinder with Axis Parallel to the \(x\)-Axis
Analyze \(\frac{(y-1)^2}{4}+\frac{(z+2)^2}{9}=1\). Find its axis, cross-section, and traces in all three coordinate planes.
Step 1: Read the axis and cross-section. The variable \(x\) is absent, so the cylinder extends parallel to the \(x\)-axis. In the \(yz\)-plane, the cross-section is an ellipse centered at \((y,z)=(1,-2)\). Its semiaxis lengths are \(2\) in the \(y\)-direction and \(3\) in the \(z\)-direction. Its axis is the line \(y=1,\ z=-2\), with \(x\) unrestricted.
Step 2: Find the trace in \(x=0\). Setting \(x=0\) does not change the equation. The trace is the full ellipse \(\frac{(y-1)^2}{4}+\frac{(z+2)^2}{9}=1\) in the \(yz\)-plane.
Step 3: Find the trace in \(y=0\). Substitute \(y=0\): \(\frac{(0-1)^2}{4}+\frac{(z+2)^2}{9}=1\). Thus \(\frac14+\frac{(z+2)^2}{9}=1\), so \(\frac{(z+2)^2}{9}=\frac34\), and \((z+2)^2=\frac{27}{4}\). Taking square roots gives \(z+2=\pm\frac{3\sqrt{3}}{2}\), or \(z=-2\pm\frac{3\sqrt{3}}{2}\). For each of these two \(z\)-values, \(x\) is unrestricted, so the trace consists of two lines parallel to the \(x\)-axis.
Step 4: Find the trace in \(z=0\). Substitute \(z=0\): \(\frac{(y-1)^2}{4}+\frac{(0+2)^2}{9}=1\). Hence \(\frac{(y-1)^2}{4}+\frac49=1\), so \(\frac{(y-1)^2}{4}=\frac59\), and \((y-1)^2=\frac{20}{9}\). Therefore \(y-1=\pm\frac{2\sqrt{5}}{3}\), giving \(y=1\pm\frac{2\sqrt{5}}{3}\). Each value allows every \(x\), so this trace also consists of two lines parallel to the \(x\)-axis.
Sketch and check. Draw the shifted ellipse in a \(yz\)-plane and extend it parallel to the \(x\)-axis. Its centerline is \(y=1,\ z=-2\). The point \((0,1,1)\) lies on the surface: \(\frac{(1-1)^2}{4}+\frac{(1+2)^2}{9}=0+1=1\). Varying its \(x\)-coordinate leaves this equation unchanged, as required for a cylinder with axis parallel to the \(x\)-axis.
A Reliable Sketching Procedure
When an equation appears to describe a circular or elliptic cylinder, use the following sequence. It keeps the cross-section, extension direction, and coordinate traces distinct.
The surface extends parallel to the coordinate axis associated with that variable.
Read the center and radius for a circle, or the center and two semiaxis lengths for an ellipse.
Use the two variables present in the equation and place the curve in the corresponding coordinate plane.
Every point on the cross-section generates a line in the missing-variable direction.
Use substitutions such as \(x=0\), \(y=0\), and \(z=0\) to confirm the surface’s placement and orientation.
Common Mistake: Confusing Semiaxes and Axis Direction
A frequent error is to assume that the largest denominator gives the cylinder’s axis direction. Consider \(\frac{x^2}{4}+\frac{y^2}{9}=1\). A mistaken conclusion might be: “The largest denominator is under \(y^2\), so the cylinder’s axis is parallel to the \(y\)-axis.” That is incorrect. The \(y\)-term is present in the equation, while \(z\) is absent; therefore the cylinder’s axis is parallel to the \(z\)-axis.
The denominator \(9\) tells us something different: the ellipse is longer in the \(y\)-direction, with semiaxis length \(3\), compared with length \(2\) in the \(x\)-direction. The missing variable determines the cylinder’s extension direction; the denominators determine the cross-section’s dimensions. Keeping these roles separate prevents both axis and scale errors.
Why Coordinate Traces Matter
Coordinate-plane traces provide exact checks on a sketch. For a cylinder whose equation is independent of \(z\), the trace in \(z=0\) is its defining cross-section, while traces in \(x=0\) and \(y=0\) show where the surface meets the vertical coordinate planes. For shifted cylinders, these traces need not pass through the origin, so substitution is especially useful.
A trace may be empty for some equations or may consist of lines rather than a full circle or ellipse. The type of trace depends on the coordinate plane being used and the position of the cross-section. Do not infer the whole surface from one trace: combine the traces with the missing-variable rule and the cross-section’s center and dimensions.
Check Your Understanding
Answer these questions by identifying both the cross-section and the direction in which it extends.
- For \(x^2+z^2=16\), which variable is missing, what is the cylinder’s axis direction, and what is the radius?
- For \(\frac{(x-2)^2}{25}+\frac{(z+1)^2}{4}=1\), identify the axis line and the two semiaxis lengths.
- What is the trace of \(x^2+y^2=1\) in the plane \(x=0\)?
- For \(\frac{y^2}{9}+\frac{z^2}{9}=1\), is the cylinder circular or elliptic, and what is its axis direction?
- In \(\frac{x^2}{4}+\frac{y^2}{9}=1\), what does the denominator \(9\) indicate, and what determines the cylinder’s axis direction?