From One Equation to a Family of Surfaces
A function \(F(x,y,z)\) assigns a number to each point in its domain in three-dimensional space. Fixing one output value \(c\) asks a geometric question: which points receive that value? The answer is a set of points, often a surface. As \(c\) changes, the set can change position, size, or even whether it exists at all.
This viewpoint extends the idea of contour curves. A contour curve in the plane collects points where a function has a fixed value. For a function of three variables, fixing the output usually produces a level surface in space.
The number \(c\) is the level, or function value, shared by every point in the set. A point \((a,b,d)\) belongs to the level set exactly when substituting its coordinates gives \(F(a,b,d)=c\). The equation does not necessarily describe a surface for every \(c\): the set can be empty, a point, a curve, or a surface, depending on \(F\) and the chosen level.
Worked Example: A Family of Planes
Worked Example: The family \(x+y+z=c\)
Let \(F(x,y,z)=x+y+z\). For a fixed value \(c\), its level set consists of all points satisfying \(x+y+z=c\). For \(c=2\), the equation is \(x+y+z=2\). The point \((2,0,0)\) belongs to this set because \(2+0+0=2\). The point \((1,1,0)\) also belongs because \(1+1+0=2\). In contrast, \((1,1,1)\) does not belong to this level set because \(1+1+1=3\), not \(2\).
For \(c=-1\), the level set is \(x+y+z=-1\). For example, \((-1,0,0)\) belongs because \(-1+0+0=-1\), and \((0,-1,0)\) belongs because \(0-1+0=-1\). These are two different level sets of the same function, one for each selected value of \(c\).
Each equation is linear in \(x,y,z\) and describes a plane. Both planes have the same normal vector \((1,1,1)\), so they are parallel. As a check on the first equation, substituting any listed point gives the required sum \(2\); for the second equation, the listed points each give the required sum \(-1\).
This example illustrates what “a family indexed by \(c\)” means. The function stays fixed, while the constant on the right-hand side selects one member of the family. The level value is not an additional coordinate of a point in space; it is the output value used to select the set.
Why the Planes Are Parallel
The parallel relationship in the previous example follows directly from how the equation changes with \(c\). A plane described by \(x+y+z=c\) has normal vector \((1,1,1)\): changing a point in a direction perpendicular to this vector leaves the sum \(x+y+z\) unchanged. Since that normal vector is the same for every value of \(c\), the planes have the same orientation.
We can also show explicitly how a point on one plane moves to another. Suppose \((x,y,z)\) satisfies \(x+y+z=c\), and choose another level \(d\). Translate the point by \(\frac{d-c}{3}(1,1,1)\). The new coordinates are \(x+\frac{d-c}{3}\), \(y+\frac{d-c}{3}\), and \(z+\frac{d-c}{3}\). Their sum is
So every point on the level-\(c\) plane is carried to a point on the level-\(d\) plane by the same translation. This explains, rather than merely suggests, why changing \(c\) shifts the plane without changing its orientation.
Worked Example: Spherical Level Sets
Worked Example: \(x^2+y^2+z^2=c\)
Let \(F(x,y,z)=x^2+y^2+z^2\). For \(c=9\), the level equation is \(x^2+y^2+z^2=9\). Since the left side is the square of the distance from \((x,y,z)\) to the origin, this set is a sphere centered at the origin with radius \(3\).
For instance, \((3,0,0)\) belongs because \(3^2+0^2+0^2=9+0+0=9\). The point \((1,2,2)\) belongs because \(1^2+2^2+2^2=1+4+4=9\). The point \((1,1,1)\) does not belong because \(1^2+1^2+1^2=3\), which is not \(9\).
The value of \(c\) matters. Each square in \(x^2+y^2+z^2\) is nonnegative, so their sum cannot be negative. If \(c<0\), the level set is empty. If \(c=0\), the sum can equal zero only when all three coordinates are zero, so the level set consists of the single point \((0,0,0)\). If \(c>0\), the equation describes a sphere of radius \(\sqrt{c}\). For example, at \(c=4\), the radius is \(\sqrt{4}=2\); the point \((0,2,0)\) checks because \(0^2+2^2+0^2=4\).
The same function therefore produces different kinds of level sets for different values: no points for negative \(c\), one point for \(c=0\), and a sphere for positive \(c\). It is not enough to recognize the formula; one must also consider the allowed values of its left-hand side.
Worked Example: Cylindrical Level Sets
Worked Example: \(x^2+y^2=c\)
Let \(F(x,y,z)=x^2+y^2\). The variable \(z\) does not appear in this formula, so for a point to lie in a level set, its \(x\)- and \(y\)-coordinates must satisfy \(x^2+y^2=c\), while its \(z\)-coordinate can be any real number.
At \(c=4\), the equation \(x^2+y^2=4\) describes a circle of radius \(2\) in the \(xy\)-plane, extended in the \(z\)-direction. The resulting level set is a circular cylinder. For example, \((2,0,5)\) belongs because \(2^2+0^2=4+0=4\). The point \((0,-2,-3)\) also belongs because \(0^2+(-2)^2=0+4=4\). The point \((1,1,5)\) does not belong because \(1^2+1^2=2\), not \(4\).
For \(c<0\), the level set is empty because \(x^2+y^2\) cannot be negative. For \(c=0\), both \(x\) and \(y\) must be zero, but \(z\) can still be any real number; the level set is the \(z\)-axis, not a two-dimensional surface. For \(c>0\), the level set is a cylinder of radius \(\sqrt{c}\). These cases show why “level set” is the more general term: not every level set is a surface.
A Reliable Way to Read the Equation
To interpret a level-surface equation \(F(x,y,z)=c\), first identify the function \(F\) and the fixed value \(c\). Then describe the points that make the equation true. For a particular point, substitute all three coordinates and compare the computed function value with \(c\). For the whole set, look for the geometric structure represented by the equation and check whether the chosen \(c\) is possible.
Separate the formula \(F(x,y,z)\) from the chosen constant \(c\).
A point belongs exactly when substituting its coordinates gives \(F(x,y,z)=c\).
Use the equation to determine whether the set is a plane, sphere, cylinder, or another set, and note if it is empty or degenerate.
Common Mistake: Treating Every Level Set as a Surface
A common error is to assume that every equation \(F(x,y,z)=c\) gives a two-dimensional surface. For example, consider \(x^2+y^2+z^2=-1\). A mistaken argument might say, “This is a sphere of radius \(\sqrt{-1}\).” That is not a real radius, and it does not describe any points in \(\mathbb{R}^3\).
The correction is to check what values the function can take. Since \(x^2\geq 0\), \(y^2\geq 0\), and \(z^2\geq 0\), their sum satisfies \(x^2+y^2+z^2\geq 0\). It therefore cannot equal \(-1\), so the level set is empty. Similarly, the equation \(x^2+y^2+z^2=0\) does not give a sphere of positive radius: it has only the origin as a solution.
Why Level Surfaces Matter
Level surfaces organize the output values of a function across three-dimensional space. Rather than examining every point separately, we can ask which locations share one value and how the sets change as that value changes. The plane family \(x+y+z=c\), the growing spheres from \(x^2+y^2+z^2=c\), and the cylinders from \(x^2+y^2=c\) illustrate three distinct patterns.
This way of thinking is useful whenever a function represents a quantity assigned to points in space. Each fixed value picks out a level set, and a collection of such sets shows how the quantity is distributed. The central task here is to interpret the equation accurately: identify the function, fix the level, and describe the points that satisfy it.
Check Your Understanding
For each question, use the equation to identify the level set and justify your conclusion.
- For \(F(x,y,z)=x+y+z\), what equation describes the level set at \(c=3\)? Does \((1,1,1)\) belong to it?
- For \(F(x,y,z)=x^2+y^2+z^2\), describe the level set when \(c=16\), and give one point that belongs to it.
- Why is the level set of \(x^2+y^2+z^2\) empty when \(c<0\)?
- For \(F(x,y,z)=x^2+y^2\), describe the level set at \(c=0\). Is it a surface?
- What does the constant \(c\) select in an equation of the form \(F(x,y,z)=c\)?