Quadric Surfaces
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.6: “Quadric Surfaces” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-6-quadric-surfaces |
| Textbook used in class | Stewart, Calculus, Section 12.6: “Cylinders and Quadric Surfaces” (Examples 3, 4, 5, 6) |
Opening Scenario
A quadric surface is the 3D analogue of a conic section: a surface defined by a second-degree equation in $x$, $y$, $z$. Just as conics fall into three types (ellipse, parabola, hyperbola), quadrics fall into six main types. The key technique for identifying them is to look at traces: the curves you get when you slice the surface with coordinate planes.
Quick Reference
The six standard quadric surfaces (all centered at the origin):
| Name | Standard Equation | Key Feature |
|---|---|---|
| Ellipsoid | $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1$ | Elliptical traces in all planes; enclosed |
| Elliptic Paraboloid | $z = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}$ | Opens in $+z$; elliptical horizontal traces |
| Hyperbolic Paraboloid | $z = \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}$ | Saddle shape; hyperbolic horizontal traces |
| Elliptic Cone | $\dfrac{z^2}{c^2} = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}$ | Two cones; vertex at origin |
| Hyperboloid (one sheet) | $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$ | Connected; waist at $z=0$ |
| Hyperboloid (two sheets) | $-\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1$ | Two separate pieces |
The trace method. Set $x = 0$, $y = 0$, or $z = 0$ (or $x = k$, etc.) to get a 2D curve. Identify that curve (circle, ellipse, parabola, hyperbola) to build a mental picture of the surface.
Key Concepts
1. The Ellipsoid
The ellipsoid $x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$ generalizes a sphere (which has $a = b = c$). Every coordinate-plane trace is an ellipse (or circle). The surface is closed and bounded.
Example 1. Describe the traces of the ellipsoid $x^2/4 + y^2/9 + z^2 = 1$. (Stewart 12.6, Example 3.)
- $z = 0$: $x^2/4 + y^2/9 = 1$, an ellipse with semi-axes $2$ and $3$.
- $x = 0$: $y^2/9 + z^2 = 1$, an ellipse with semi-axes $3$ and $1$.
- $y = 0$: $x^2/4 + z^2 = 1$, an ellipse with semi-axes $2$ and $1$.
The surface is an ellipsoid fitting inside the box $[-2,2]\times[-3,3]\times[-1,1]$.
2. Paraboloids
The elliptic paraboloid $z = x^2/a^2 + y^2/b^2$ opens upward (in $+z$). Horizontal traces ($z = k > 0$) are ellipses; vertical traces ($x = 0$ or $y = 0$) are parabolas.
The hyperbolic paraboloid $z = y^2/b^2 - x^2/a^2$ is a saddle. A horizontal trace $z = k$ gives $y^2/b^2 - x^2/a^2 = k$: a hyperbola. The saddle has a minimum in the $y$-direction and a maximum in the $x$-direction at the origin.
Example 2. Identify the surface $z = x^2 - y^2$. (Stewart 12.6, Example 5.)
This is a hyperbolic paraboloid. At $z = 0$: $(x-y)(x+y) = 0$, two lines. For $z > 0$: hyperbola opening in $x$-direction. For $z < 0$: hyperbola opening in $y$-direction. The surface looks like a saddle or Pringles chip.
3. Hyperboloids
One-sheet hyperboloid $x^2/a^2 + y^2/b^2 - z^2/c^2 = 1$: connected surface. Horizontal traces are ellipses; vertical traces are hyperbolas. The “waist” at $z = 0$ is the smallest horizontal ellipse.
Two-sheet hyperboloid $-x^2/a^2 - y^2/b^2 + z^2/c^2 = 1$: two disconnected pieces (one for $z > c$, one for $z < -c$). Setting $z = 0$ gives no solution (no waist).
Identifying one-sheet vs. two-sheet: count the minus signs. One minus sign on the squared terms: one-sheet. Two minus signs: two-sheet. (This is how many “pairs of sheets” come apart.)
classifying based on the number of terms without checking signs. Both hyperboloids have three squared terms, but one has one negative coefficient and the other has two. Always look at the signs, not just the number of terms.
Common Errors Summary
| Error | Correction |
|---|---|
| Confusing the elliptic and hyperbolic paraboloid | Elliptic: $z = x^2/a^2 + y^2/b^2$ (same sign); hyperbolic: $z = x^2/a^2 - y^2/b^2$ (opposite signs) |
| Forgetting to check horizontal traces for orientation | Set $z = k$ and simplify; the shape of the horizontal trace distinguishes paraboloids from hyperboloids |
| Saying $z^2/c^2 = x^2/a^2 + y^2/b^2$ is an ellipse | A cone (two sides); an ellipse would require $c^2 \neq 0$ on the right side |
Common Misconceptions
the number of squared terms alone determines the type of quadric surface.
This is the concept-image-conflicts-definition error. Both the one-sheet and two-sheet hyperboloids have three squared terms, but they differ in the number of negative coefficients. The one-sheet hyperboloid $x^2/a^2 + y^2/b^2 - z^2/c^2 = 1$ has one negative coefficient; the two-sheet hyperboloid $-x^2/a^2 - y^2/b^2 + z^2/c^2 = 1$ has two. Counting terms rather than checking signs leads to systematic misclassification. The sign pattern on the squared terms, not the count of terms, distinguishes these surfaces.
Leveled Practice
Problem 1. Identify and describe the surface $4x^2 + y^2 + 4z^2 = 16$.
Show answer
Divide by $16$: $x^2/4 + y^2/16 + z^2/4 = 1$. All signs positive, all three variables present: ellipsoid with semi-axes $a = 2$, $b = 4$, $c = 2$.
Problem 2. Identify the surface $z^2 = x^2 + y^2/4$.
Show answer
Rewrite as $z^2 - x^2 - y^2/4 = 0$: elliptic cone with vertex at the origin. Horizontal trace $z = k$: $x^2 + y^2/4 = k^2$, an ellipse.