Horizontal Slices and Level Curves

Horizontal Slices of a Surface

A graph $z=f(x,y)$ is a surface in three-dimensional space. One way to understand its shape is to intersect it with a horizontal plane. “Horizontal” means that the height coordinate $z$ is fixed while $x$ and $y$ are allowed to vary.

Choose a height $c$. The plane at that height has equation $z=c$. Where this plane meets the surface, the surface’s height must be $c$. Since the surface is defined by $z=f(x,y)$, its intersection points must satisfy $f(x,y)=c$.

This gives two related objects: a three-dimensional intersection lying in the plane $z=c$, and a planar set of $(x,y)$-coordinates satisfying $f(x,y)=c$. The planar set is called the level set of $f$ at height $c$. Thinking of it as a horizontal slice connects the equation to the geometry of the surface.

Definitions: Graph, Horizontal Plane, and Level Set

Let $f:D\to\mathbb{R}$ be a function, where $D$ is a subset of the $xy$-plane. Its graph consists of all points whose first two coordinates belong to $D$ and whose third coordinate is the function value.

\operatorname{Graph}(f) = \{(x,y,z)\in\mathbb{R}^3:(x,y)\in D,\ z=f(x,y)\}.

For a real number $c$, the horizontal plane at height $c$ is the set of points with third coordinate equal to $c$. Its intersection with the graph is called the horizontal trace of the graph at height $c$. The corresponding level set records the first two coordinates of that trace.

Definition: For $f:D\to\mathbb{R}$ and $c\in\mathbb{R}$, the level set of $f$ at height $c$ is $$ L_c=\{(x,y)\in D:f(x,y)=c\}. $$ The horizontal trace is $$ \operatorname{Graph}(f)\cap\{(x,y,z)\in\mathbb{R}^3:z=c\}. $$

The level set is drawn in the $xy$-plane, while the horizontal trace lies in three-dimensional space at height $c$. They contain the same horizontal positions, but they are not literally the same set of points: each point $(x,y)$ in the level set corresponds to the point $(x,y,c)$ in the trace.

Why the Intersection Gives the Level Set

The relationship follows directly from the definitions. A point on the graph has the form $(x,y,f(x,y))$. It lies in the plane $z=c$ precisely when its third coordinate equals $c$, or $f(x,y)=c$. Conversely, if $(x,y)$ belongs to $D$ and satisfies $f(x,y)=c$, then the graph contains $(x,y,c)$, which also lies in the plane $z=c$.

Theorem: For $f:D\to\mathbb{R}$ and $c\in\mathbb{R}$, $$ \operatorname{Graph}(f)\cap\{(x,y,z):z=c\} = \{(x,y,c):(x,y)\in D,\ f(x,y)=c\}. $$ Thus, deleting the fixed third coordinate $c$ from each intersection point gives exactly the level set $L_c$.

In particular, the horizontal trace is empty if and only if there are no points in the domain where $f$ takes the value $c$. This happens when $c$ is outside the range of $f$. When the level set is nonempty, its shape can vary: it may be a circle, a line, several branches, or another planar set. The name “level set” remains accurate even when the set is not a single curve.

Choose a height.
Fix a real number $c$, which specifies the horizontal plane $z=c$.

Use the graph equation.
On the graph, $z=f(x,y)$, so being at height $c$ requires $f(x,y)=c$.

Interpret the planar set.
The solutions $(x,y)$ in the domain form $L_c$; the corresponding three-dimensional points are $(x,y,c)$.

Worked Examples

Worked Example: A Circular Horizontal Trace

Let $f(x,y)=x^2+y^2$, with domain $D=\mathbb{R}^2$, and take the horizontal plane $z=4$. The graph is $z=x^2+y^2$. At an intersection with $z=4$, both expressions for $z$ must agree:

$$ x^2+y^2=4. $$

This equation describes a circle of radius $2$, centered at the origin in the $xy$-plane. Its points lift to the horizontal trace by adding height coordinate $4$. For example, $(2,0)$ is on the level set because $2^2+0^2=4$, and its point on the surface is $(2,0,4)$. The function value there is $f(2,0)=4+0=4$, so it lies in the plane $z=4$. Similarly, $(0,2)$, $(-2,0)$, and $(0,-2)$ each have function value $4$, and their corresponding surface points all have third coordinate $4$.

Conversely, any point on this surface and in the plane must satisfy $z=4$ and $z=x^2+y^2$, so $x^2+y^2=4$. Thus the circle gives all, not merely some, of the horizontal trace’s $xy$-positions.

Worked Example: A Straight-Line Level Set

Consider $f(x,y)=2x-y$ on $\mathbb{R}^2$, and slice its graph with the plane $z=3$. At an intersection, $z=2x-y$ and $z=3$, so the level-set equation is

$$ 2x-y=3. $$

For instance, when $x=0$, the equation gives $-y=3$, hence $y=-3$. Evaluating the function confirms $f(0,-3)=2(0)-(-3)=3$, so $(0,-3,3)$ is on both the graph and the plane. When $x=2$, the equation gives $4-y=3$, hence $y=1$. Indeed, $f(2,1)=2(2)-1=3$, and $(2,1,3)$ also lies on both.

The planar set is the line $y=2x-3$. Every point on this line satisfies $2x-y=3$, so its graph point has height $3$. Conversely, every point of the graph at height $3$ must satisfy this same line equation. The horizontal trace is therefore the lifted line $\{(x,y,3):y=2x-3\}$.

Worked Example: A Height with No Intersection

Again let $f(x,y)=x^2+y^2$, but now choose the plane $z=-1$. A point in the intersection would need to satisfy

$$ x^2+y^2=-1. $$

For real $x$ and $y$, each square is nonnegative, so $x^2+y^2\geq 0$. It cannot equal $-1$. Therefore the level set is empty, and the plane $z=-1$ does not meet the graph.

This conclusion also agrees with the surface’s heights: $f(x,y)=x^2+y^2\geq 0$ everywhere on its domain. A horizontal plane below height $0$ has no point in common with this graph.

Worked Example: A Level Set with Two Branches

Let $f(x,y)=x^2-y^2$ and choose $c=0$. The level-set equation is

x^2-y^2=0 \qquad\Longleftrightarrow\qquad (x-y)(x+y)=0.

A product is zero when at least one factor is zero, so the level set consists of the two lines $y=x$ and $y=-x$. To check points on the first line, take any real $t$: $f(t,t)=t^2-t^2=0$. On the second line, $f(t,-t)=t^2-(-t)^2=t^2-t^2=0$. Thus each corresponding point $(t,t,0)$ or $(t,-t,0)$ lies on both the graph and the plane $z=0$.

There are two crossing branches, rather than one isolated curve. This example illustrates why “level set” is the general term: it describes all points at a specified function value without assuming the set has one particular shape.

A Common Mistake: Confusing Horizontal and Vertical Slices

A horizontal slice fixes $z$, not $x$ or $y$. For example, with $z=x^2+y^2$, the incorrect calculation “set $x=4$, so the slice at height $4$ is $x=4$” confuses a vertical restriction with a horizontal one. Fixing $x=4$ would constrain the surface to a vertical plane and would give $z=16+y^2$, not a trace at a single height.

For a horizontal slice at height $4$, set $z=4$ and equate it to the graph formula: $4=x^2+y^2$. The resulting circle is the set of horizontal positions where the surface has height $4$. The correction is to fix the third coordinate, then use $z=f(x,y)$ to find which $(x,y)$ positions have that height.

How to Read the Geometry

A level set answers a specific geometric question: at which points in the domain does the surface have height $c$? It does not by itself display the height coordinate, because every point in the level set is understood to be lifted to the plane $z=c$ when describing the actual intersection.

The height $c$ matters. Changing $c$ changes the plane and may change the level set’s shape or whether it exists at all. Even for a fixed function, one height might give an empty set, another might give a single point or a curve, and another might produce several branches. The equation $f(x,y)=c$ captures the precise condition in each case.

When examining a proposed level set, two checks are useful. First, every listed point should belong to the domain and satisfy $f(x,y)=c$. Second, the description should account for every point in the domain satisfying that equation. The first check rules out extraneous points; the second ensures the trace has not been only partly described.

Key takeaway: Intersecting $z=f(x,y)$ with the horizontal plane $z=c$ means requiring $f(x,y)=c$. The solutions in the $xy$-plane form the level set, and each solution $(x,y)$ corresponds to the intersection point $(x,y,c)$.

Check Your Understanding

Use the relationship between a horizontal plane and a level set to answer these questions.

For a graph $z=f(x,y)$, what equation specifies the horizontal plane at height $c$?
Why must the $xy$-coordinates of an intersection point satisfy $f(x,y)=c$?
For $f(x,y)=x^2+y^2$, does the graph meet the plane $z=9$? State the planar level-set equation.
Can a level set be empty? Explain what that means for the horizontal plane and graph.
For $f(x,y)=x^2-y^2$, what points belong to the level set at height $0$?