2.6 Quadric Surfaces

Big idea

A first-degree equation in $x$, $y$ and $z$ produces a plane, and that exhausts the flat case. Allowing squares and cross terms produces a family called the quadric surfaces, and they are the three-dimensional analogues of the conic sections. Every ellipse, parabola and hyperbola you have met has a surface version here, and the surfaces are built so that their two-dimensional cross-sections are exactly those curves.

That last sentence is also the method. Free-hand sketching in three dimensions is unreliable, so instead you slice. Set $z$ equal to a constant and the equation collapses to a relation between $x$ and $y$, a curve you already know how to recognize. Do this for a few values of $z$, then repeat with $x$ constant and with $y$ constant. Those slices are called traces, and a handful of them determines the surface: the shape of each slice, and the way it changes as the slicing value moves, is the whole picture.

Identifying a quadric from its equation is therefore bookkeeping with signs. Count the squared terms and their signs, and note whether one variable appears to the first power instead. Three squares with the same sign give a closed surface; three with mixed signs give a hyperboloid or a cone; two squares and one linear term give a paraboloid, elliptic when the squares agree in sign and saddle-shaped when they do not.

Equations rarely arrive in standard position, so completing the square is the routine first step. It shifts the center away from the origin without changing the type of surface, exactly as it relocates a circle in the plane.

Decoder

The trace of a surface in a plane is the curve where the surface and that plane intersect, and a quadric surface is identified by the type of its traces in the three families of planes parallel to the coordinate planes.

The word “identified” is doing real work. Traces are not an approximation to the surface; for these equations they pin it down exactly. The slices of an elliptic paraboloid perpendicular to its axis are ellipses that grow, and slices parallel to the axis are parabolas; no other quadric has that combination. So the identification is a short decision procedure rather than an act of visualization.

One detail deserves care: an empty trace is information, not a failure. If a horizontal slice yields $x^2 + y^2 = -3$, the surface simply does not reach that height. That is how the hyperboloid of two sheets separates into two pieces and how the elliptic paraboloid ends at its vertex. A trace that degenerates to a single point or to a pair of lines is equally informative, locating a vertex or the tip of a cone.

The classic mistake is deciding the axis of symmetry from the wrong term. For the hyperboloids, the axis is the variable whose squared term has the sign that stands alone: with one minus sign among two pluses you get one sheet wrapped around that axis, and with one plus sign among two minuses you get two sheets opening along it. Getting that backwards turns one surface into the other.

Definitions and results

Traces. A trace is the intersection of a surface with a plane, usually one of $x = k$, $y = k$ or $z = k$. Substituting the constant leaves a two-variable equation whose graph is the trace.

Cylinders. If one variable is absent from the equation, the surface is a cylinder: take the curve drawn in the plane of the other two variables and slide it along the missing axis. So $x^2 + y^2 = 9$ is a circular cylinder about the $z$-axis and $z = x^2$ is a parabolic cylinder running along the $y$-axis.

Quadric surfaces. A quadric is the graph of a second-degree equation in three variables. After completing the square and, if necessary, rotating, every nondegenerate quadric is one of the six standard types below. The forms shown are centered at the origin with the $z$-axis as the distinguished direction; permuting the variables permutes the roles.

Ellipsoid. $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1$. Every trace is an ellipse or empty, and the surface is bounded, meeting the axes at $\pm a$, $\pm b$, $\pm c$. When $a = b = c$ it is a sphere.

Elliptic paraboloid. $z = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}$. Horizontal traces are ellipses for $z > 0$, a single point at $z = 0$ and empty for $z < 0$. Vertical traces are parabolas, all opening upward. The surface is a bowl with vertex at the origin.

Hyperbolic paraboloid. $z = \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}$. The trace in $y = 0$ is a parabola opening upward, the trace in $x = 0$ is a parabola opening downward, and horizontal traces are hyperbolas, opening along the $x$-direction above $z = 0$ and along the $y$-direction below it. At $z = 0$ the trace degenerates to two crossing lines. The result is a saddle.

Hyperboloid of one sheet. $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1$. Every horizontal trace is an ellipse, smallest at $z = 0$ and growing without bound, so the surface is a single connected tube around the $z$-axis. Vertical traces are hyperbolas.

Hyperboloid of two sheets. $-\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1$. Horizontal traces are empty for $|z| < c$, single points at $z = \pm c$ and ellipses beyond, so the surface is two separate bowls opening along the $z$-axis.

Elliptic cone. $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \dfrac{z^2}{c^2}$. Horizontal traces are ellipses whose size grows in proportion to $|z|$, shrinking to the single point at the origin. Vertical traces through the axis are pairs of lines. The cone is the boundary case that both hyperboloids approach far from the origin.

Worked examples

Reading an ellipsoid off its traces

Identify $\dfrac{x^2}{4} + \dfrac{y^2}{9} + z^2 = 1$.

All three squares are present with the same sign and the right-hand side is positive, so this is an ellipsoid with semi-axes $2$, $3$ and $1$ along $x$, $y$ and $z$.

Confirm with traces. Setting $z = 0$ gives $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$, an ellipse, and $x = 0$ gives $\dfrac{y^2}{9} + z^2 = 1$, again an ellipse. Setting $z = 1$ forces $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 0$, true only at the single point $(0,0,1)$, while $z = 2$ would require that expression to equal $-3$, which is impossible. The surface is bounded between $z = -1$ and $z = 1$.

One sheet or two

Compare $x^2 + y^2 - \dfrac{z^2}{4} = 1$ with $-x^2 - y^2 + \dfrac{z^2}{4} = 1$.

For the first, a horizontal slice at height $z = k$ gives $x^2 + y^2 = 1 + \dfrac{k^2}{4}$, a circle of radius $\sqrt{1 + k^2/4}$. That is defined for every $k$, with the smallest circle of radius $1$ at $k = 0$, so the surface is one connected piece: a hyperboloid of one sheet around the $z$-axis. The trace in $x = 0$ is $y^2 - \dfrac{z^2}{4} = 1$, a hyperbola opening along the $y$-axis.

For the second, the same slice gives $x^2 + y^2 = \dfrac{k^2}{4} - 1$, which has no solutions when $|k| < 2$, reduces to a point when $|k| = 2$, and is a circle beyond. So this surface is two separate bowls, opening upward from $(0,0,2)$ and downward from $(0,0,-2)$. In both cases $z$ is the axis.

Completing the square

Identify $x^2 + 4y^2 - z^2 - 2x + 16y + 13 = 0$.

Group and complete the square in $x$ and in $y$:

$$ (x^2 - 2x) + 4(y^2 + 4y) - z^2 + 13 = 0 $$

$$ (x-1)^2 - 1 + 4(y+2)^2 - 16 - z^2 + 13 = 0 $$

$$ (x-1)^2 + 4(y+2)^2 - z^2 = 4 $$

Dividing by $4$ puts it in standard form:

$$ \frac{(x-1)^2}{4} + (y+2)^2 - \frac{z^2}{4} = 1 $$

Two positive squares and one negative, equal to $1$, so this is a hyperboloid of one sheet centered at $(1,-2,0)$ with axis parallel to the $z$-axis. Check the narrowest slice: at $z = 0$ the trace is an ellipse centered at $(1,-2)$ with semi-axes $2$ and $1$, and the point $(3,-2,0)$ should lie on the surface. Substituting into the original equation gives $9 + 16 - 0 - 6 - 32 + 13 = 0$, which confirms it.

Bowl against saddle

Compare $z = \dfrac{x^2}{4} + \dfrac{y^2}{9}$ with $z = \dfrac{x^2}{4} - \dfrac{y^2}{9}$.

In the first, both vertical traces open upward: $y = 0$ gives $z = x^2/4$ and $x = 0$ gives $z = y^2/9$. Horizontal traces are ellipses for $z > 0$ and empty below. This is an elliptic paraboloid, a bowl resting on the origin.

In the second, $y = 0$ still gives the upward parabola $z = x^2/4$, but $x = 0$ gives $z = -y^2/9$, which opens downward. The origin is a minimum along one direction and a maximum along the other, which is what a saddle point means. The trace at $z = 0$ is the pair of lines $y = \pm \dfrac{3x}{2}$, and slices above and below the origin are hyperbolas opening in perpendicular directions. Replacing $z$ by $z^2$ in the first equation would instead give the elliptic cone $z^2 = \dfrac{x^2}{4} + \dfrac{y^2}{9}$.

Practice

These problems ask you to compute traces, match an equation to a surface, complete the square to find a center and an axis, and describe how a cross-section changes as the plane moves.

Quiz

Five items on traces, cylinders, and identifying each standard quadric from its equation.