From Circular Symmetry to Cylindrical Form
The expressions \(x^2+y^2\) and \(r^2\) measure the same squared distance from the \(z\)-axis. This makes cylindrical coordinates especially useful for surfaces whose height depends only on distance from that axis. A basic example is \(z=x^2+y^2\): its height changes with radius, but not with direction around the axis.
In cylindrical coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), and \(r\geq 0\). Consequently,
Thus the surface \(z=a(x^2+y^2)+c\) becomes \(z=ar^2+c\). The angle \(\theta\) does not appear: at a fixed radius, the height is the same in every direction around the \(z\)-axis.
The Standard Rotational Paraboloid
For constants \(a\) and \(c\), consider the graph
When \(a\ne 0\), this is a circular paraboloid with vertex on the \(z\)-axis at \((0,0,c)\). The term \(ar^2\) is zero at \(r=0\), and its magnitude increases as the distance \(r\) from the axis increases. The sign of \(a\) determines which way the surface opens.
- If \(a>0\), the surface opens upward and has its lowest point at height \(c\).
- If \(a<0\), the surface opens downward and has its highest point at height \(c\).
The special case \(a=0\) is not a paraboloid. It gives the horizontal plane \(z=c\), so it must be treated separately when finding slices.
Horizontal Cross-Sections
To find the intersection with a horizontal plane, set \(z=h\), where \(h\) is the chosen height. For \(a\ne 0\), the equation \(h=ar^2+c\) gives
A positive value of \(r^2\) gives a circle centered on the \(z\)-axis. If \(r^2=0\), the intersection is only the vertex. If \(r^2<0\), there are no real points in that slice. The sign of \(a\) therefore determines which heights meet the surface in circles.
| Coefficient and height | Horizontal intersection |
|---|---|
| \(a>0\), \(h>c\) | Circle of radius \(\sqrt{(h-c)/a}\) |
| \(a>0\), \(h=c\) | The vertex only |
| \(a>0\), \(h<c\) | No intersection |
| \(a<0\), \(h<c\) | Circle of radius \(\sqrt{(h-c)/a}\) |
| \(a<0\), \(h=c\) | The vertex only |
| \(a<0\), \(h>c\) | No intersection |
The radius formula in the table is used only in the rows where its radicand is positive. When \(a=0\), the equation is instead \(z=c\): the slice at \(h=c\) is the whole horizontal plane, and the slice at any \(h\ne c\) is empty.
Vertical Cross-Sections
A vertical plane through the \(z\)-axis contains a radial direction. Choose coordinates in that plane so that \(s\) measures signed horizontal distance from the axis. There, \(r^2=s^2\), and the intersection with the surface is the parabola
This cross-section opens upward when \(a>0\), downward when \(a<0\), and has its vertex at \(s=0,\ z=c\). Every vertical plane through the axis gives the same profile, which explains how a two-dimensional parabola generates a rotational paraboloid.
A vertical plane parallel to, but not through, the axis also cuts a parabola. For example, take the plane \(y=d\), where \(d\) is fixed, and let \(s=x\) within that plane. Since \(r^2=x^2+d^2=s^2+d^2\), the cross-section is
It has the same opening direction as the axial profile, but its vertex lies at height \(ad^2+c\), not at \(c\). This is a useful distinction: the vertex of the full surface is on the axis, while the vertex of an off-axis vertical slice is higher or lower depending on the sign of \(a\).
Why the Surface Has Rotational Symmetry
The symmetry follows directly from the equation. A rotation about the \(z\)-axis changes the horizontal coordinates but preserves their squared distance from the axis. To see this, for a rotation by angle \(\phi\), the new coordinates are
Expanding and combining terms gives
The cross terms cancel, and each squared trigonometric pair sums to \(1\). Therefore \(z=a(x^2+y^2)+c\) remains true after any rotation around the axis. This proves the claimed rotational symmetry.
Worked Example: The Surface \(z=x^2+y^2\)
Worked Example: The Surface \(z=x^2+y^2\)
Convert the equation to cylindrical form and identify two horizontal slices and an axial vertical slice.
First use \(x^2+y^2=r^2\):
Here \(a=1>0\) and \(c=0\), so the surface opens upward and has vertex \((0,0,0)\). At height \(h=4\), the slice satisfies \(4=r^2\), so \(r=2\); it is a circle of radius \(2\). At height \(h=-1\), the equation would require \(r^2=-1\), which has no real solution, so that slice is empty.
In the vertical plane \(y=0\), \(r^2=x^2\), and the cross-section is \(z=x^2\), an upward-opening parabola. As a check, a point on this trace with \(x=3\) has \(y=0\) and \(z=3^2=9\). Substitution into the original equation gives \(z=x^2+y^2=9+0=9\), as required.
How the Coefficient Changes the Shape
For a fixed height difference from the vertex, the horizontal radius is
provided the height is on the side of the vertex where the surface exists. Thus, with the same height difference \(|h-c|\), increasing \(|a|\) makes the circular slice smaller. The surface rises or falls more rapidly as one moves away from its axis. Decreasing \(|a|\), while keeping it nonzero, produces a wider, flatter paraboloid.
The sign and magnitude play different roles. Changing the sign of \(a\) reverses the opening direction; changing its nonzero magnitude changes steepness. Adding \(c\) translates the surface vertically: it moves the vertex from height \(0\) to height \(c\), without changing the opening or the shape of the cross-sections.
Worked Example: A Wider Upward-Opening Paraboloid
Analyze \(z=\frac14(x^2+y^2)\), then find the horizontal slice at height \(z=4\).
Replacing \(x^2+y^2\) by \(r^2\) gives \(z=\frac14r^2\). The coefficient is positive, so the surface opens upward with vertex at the origin. Set \(z=4\):
The slice is a circle of radius \(4\). For comparison, \(z=r^2\) has radius \(2\) at the same height, since \(4=r^2\) gives \(r=2\). The smaller coefficient therefore gives the wider surface. As another check, at \(r=2\) the present surface has \(z=\frac14(2^2)=1\), and at \(r=4\) it has \(z=\frac14(4^2)=4\), agreeing with the slice calculation.
Worked Example: A Shifted, Downward-Opening Paraboloid
Analyze \(z=5-2(x^2+y^2)\) and find its slice at height \(z=-3\).
In cylindrical form the equation is \(z=5-2r^2\). Its vertex is \((0,0,5)\), and the negative coefficient means it opens downward. At \(z=-3\),
The horizontal intersection is a circle of radius \(2\). In the axial plane \(y=0\), the profile is \(z=5-2x^2\), a downward-opening parabola. Check the circle using \(r=2\): the surface gives \(z=5-2(2^2)=5-8=-3\), which is the specified height.
The Zero-Coefficient Case and a Common Pitfall
The formula \(r^2=(h-c)/a\) requires \(a\ne0\). Dividing by \(a\) when \(a=0\) is not a permissible step. For example, if \(z=0(x^2+y^2)+0\), the surface is simply the plane \(z=0\). Its horizontal intersection at height \(h=0\) is the entire plane, not just the origin; at any height \(h\ne0\), the intersection is empty.
Worked Example: Treating \(a=0\) Separately
Consider \(z=0(x^2+y^2)+2\). The equation simplifies directly to \(z=2\), a horizontal plane. At height \(h=2\), every point \((x,y,2)\) is on the surface, so the intersection is the whole plane \(z=2\). At height \(h=3\), no point lies on both \(z=2\) and \(z=3\), so the intersection is empty. These conclusions follow without dividing by the zero coefficient.
A common incorrect calculation is to write \(r^2=(h-c)/a\) for \(a=0\), then claim that the slice at \(h=c\) is just the vertex because the numerator is zero. The fraction is undefined: it is \(0/0\), not \(0\). Returning to the original equation \(z=c\) shows the correction—the slice at \(h=c\) is the whole plane.
A Reliable Way to Describe These Surfaces
When given an equation involving \(x^2+y^2\), first convert that expression to \(r^2\). Then identify the coefficient and any vertical shift before interpreting cross-sections. Keeping these steps separate prevents sign errors and makes the exceptional plane case visible.
Replace \(x^2+y^2\) by \(r^2\), giving an equation such as \(z=ar^2+c\).
If \(a=0\), the surface is the plane \(z=c\). If \(a\ne0\), the surface is a paraboloid.
The vertex is \((0,0,c)\); positive \(a\) opens upward and negative \(a\) opens downward.
Horizontal slices are circles, a point, or empty for \(a\ne0\); axial vertical slices are parabolas.
The cylindrical equation also gives a compact description of the entire surface: \(r\) measures distance from the axis, while the formula assigns a height to each such distance. In this way, circular horizontal slices and parabolic vertical profiles are two complementary views of the same geometry.
Check Your Understanding
Answer each question by translating the equation into cylindrical form and interpreting its slices.
- Write \(z=3(x^2+y^2)\) in cylindrical form. Which direction does it open?
- For \(z=2r^2+1\), what is the radius of the horizontal slice at height \(z=9\)?
- Describe the horizontal slice at height \(z=0\) for \(z=-r^2\).
- What surface is represented by \(z=0(x^2+y^2)-4\), and what is its intersection with the plane \(z=-4\)?
- For \(z=ar^2+c\) with \(a\ne0\), describe the axial vertical cross-section and its vertex.